Probability & Statistics
Exam Pass Guide

Everything from Week 2 to Week 6 in simple words, with the professor's own examples, and all 19 sample questions from Week 7 solved step by step.

Weeks 2 to 719 solved questionsNo symbol left unexplained

How to use this page

Go from top to bottom. Each lesson has the same shape: idea in simple words, then the rule, then an example from the slides.
Solutions are hidden. Read the question out loud, let everyone try for one or two minutes, then tap Show solution.
At the end of each lesson there is a blue box "Now solve". It sends you to the matching Week 7 sample questions. Solve them right away, while the idea is fresh.
The Menu button at the top jumps to any lesson or any question. You can also make the text bigger there.

The one idea behind the whole course

A probability is just a number between 0 and 1 that says how likely something is. 0 means "never". 1 means "always". 0.5 means "half of the time". Everything else in this course is a tool to find that number.

Plan for the 4 hours (12:00 to 4:00)

Time	What	Sample questions
12:00	0 · Symbols (quick look) and 1 · Sets and basic probability	Q1, Q2, Q3, Q4
12:35	2 · Conditional probability, Bayes, trees	Q7, Q8, Q9
1:10	3 · Counting and independent trials	Q5, Q6
1:30	4 · Random variables, PMF, the 6 families	Q10, Q11, Q13, Q16, Q19
2:15	Break, 10 minutes
2:25	5 · CDF	Q12, Q17
2:50	6 · Expected value and variance	Q14, Q15 (a, b, c)
3:20	7 · Conditional PMF	Q15 (d), Q18
3:45	9 · Formula sheet and traps (read together)

If you run late: never skip lessons 4, 5, 6. Most sample questions come from them.

Symbols dictionary

Come back here any time a symbol looks strange

Sets and events

Symbol	Say it	Meaning
${}$	"the set of"	A list of things. Example: ${1, 2, 3}$
$S$	"sample space"	All possible results of the experiment
$\in$	"is in"	$2 \in {1, 2, 3}$ means 2 is inside the list
$\in /$	"is not in"	$5 \in / {1, 2, 3}$
$A \subset B$	"A is inside B"	Everything in A is also in B
$φ$	"empty set"	A set with nothing in it. It means "impossible"
$A \cup B$	"A or B"	In A, or in B, or in both
$A \cap B, A B$	"A and B"	In A and in B at the same time. $A B$ is a short way to write it
$A^{c}$	"not A"	Everything that is not in A
$A - B$	"A minus B"	In A but not in B

Probability

Symbol	Say it	Meaning
$P [A]$	"probability of A"	How likely A is. A number from 0 to 1
$P [A ∣ B]$	"probability of A given B"	How likely A is when we already know B happened. The bar \| means "given"
$n!$	"n factorial"	$n \cdot (n - 1) \dots 2 \cdot 1$ . Example: $4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24$ . Also $0! = 1$
$(k n)$	"n choose k"	Number of ways to pick k things out of n when order does not matter
$\sum$	"add up"	Add all the things that follow
$\leq, \geq$	"at most", "at least"	$\leq$ includes the number itself. $<$ does not

Random variables

Symbol	Say it	Meaning
$X$	"big X"	The name of a random number. Example: X = number of heads
$x$	"small x"	One value that X can take. Example: x = 2
$S_{X}$	"range of X"	The list of all values X can take
$P_{X} (x)$	"PMF of X"	$P [X = x]$ . The probability that X is exactly x
$F_{X} (x)$	"CDF of X"	$P [X \leq x]$ . The probability that X is x or less
$E [X], μ_{X}$	"expected value", "mu"	The average of X. Both symbols mean the same thing
$Var [X]$	"variance"	How spread out the values are
$σ_{X}$	"sigma", standard deviation	$Var [X]$ . The typical distance from the average
$P_{X ∣ B} (x)$	"PMF of X given B"	The PMF after we learn that B happened
$p$	"small p"	Probability of success in one try
$α, λ$	"alpha", "lambda"	Greek letters used in Poisson. $λ$ = average per second. $α = λ T$ = average in T seconds
$e$	"e"	A fixed number, about 2.718. Use the $e^{x}$ button on the calculator
$ϵ$	"epsilon"	A very, very small positive number. $a - ϵ$ means "just before a"

Sets and basic probability

Week 2

1.1Four words you must know

Word	Simple meaning	Die example
Experiment	Something we do and watch	Roll a die once
Outcome	One possible result	The die shows 4
Sample space $S$	The list of all possible outcomes	$S = {1, 2, 3, 4, 5, 6}$
Event	A group of outcomes we care about	"Roll is even" $= {2, 4, 6}$

In simple words

The sample space is the full menu. An outcome is one dish. An event is a group of dishes, like "all the spicy ones".

The slides say the sample space must be three things:

Finest-grain: every different result is written separately.
Mutually exclusive: only one outcome can happen each time.
Collectively exhaustive: no possible outcome is missing.

Example from the slides

Flip four coins (penny, nickel, dime, quarter). Each shows head (h) or tail (t). How many outcomes are in the sample space?

Show solution

Each coin has 2 results. There are 4 coins. Multiply:

2 \cdot 2 \cdot 2 \cdot 2 = 2^{4} = 16

The outcomes look like: hhhh, hhht, hhth, ..., tttt.

Answer16 outcomes

1.2The four set operations

Events are sets, so we combine them with set operations. The blue area is the result.

A ∪ BA OR B (or both)

A ∩ BA AND B together

AᶜNOT A

A − Bin A but not in B

Try them with a die. Let $A = {1, 2, 3}$ and $B = {2, 4, 6}$ .

Write	Read	Result
$A \cup B$	A or B	${1, 2, 3, 4, 6}$
$A \cap B$	A and B	${2}$
$A^{c}$	not A	${4, 5, 6}$
$A - B$	A but not B	${1, 3}$

Rules from the slides

A - B = A \cap B^{c} A^{c} = S - A

De Morgan's law (how "not" moves inside brackets):

(A \cup B)^{c} = A^{c} \cap B^{c}

(A \cap B)^{c} = A^{c} \cup B^{c}

De Morgan in simple words

"Not (tea or coffee)" is the same as "no tea and no coffee". When "not" goes inside, or flips to and, and and flips to or.

1.3Two exam words: mutually exclusive, collectively exhaustive

Word	Meaning	How to test
Mutually exclusive (for two sets: disjoint)	They share nothing. They cannot happen together.	Is $A \cap B = φ$ ? If you find even one common outcome, the answer is no.
Collectively exhaustive	Together they cover everything.	Is $A \cup B = S$ ? If even one outcome of S is missing, the answer is no.

A group of events that is both is called an event space. Every outcome falls in exactly one of them.

Picture

Cut a pizza into slices. The slices do not overlap (mutually exclusive) and together they make the whole pizza (collectively exhaustive). The slices are an event space.

Quick check with a die

$A = {2, 4, 6}$ and $B = {1, 2}$ . Mutually exclusive? Collectively exhaustive?

Show solution

$A \cap B = {2}$ . They share 2, so they are not mutually exclusive.

$A \cup B = {1, 2, 4, 6}$ . The outcomes 3 and 5 are missing, so they are not collectively exhaustive.

1.4The rules of probability

$P [A]$ means "the probability of event A". The whole theory stands on three basic rules, called axioms:

The 3 axioms

$P [A] \geq 0$ . A probability is never negative.
$P [S] = 1$ . Something from the sample space always happens.
If events are mutually exclusive, the probability of "one or the other" is the sum:
$P [A_{1} \cup A_{2} \cup \dots] = P [A_{1}] + P [A_{2}] + \dots$

From these we get the tools you will use in every question:

Your toolbox

Not rule:

P [A^{c}] = 1 - P [A]

Or rule (always works):

P [A \cup B] = P [A] + P [B] - P [A \cap B]

Or rule when disjoint (nothing shared, so $P [A \cap B] = 0$ ):

P [A \cup B] = P [A] + P [B]

Split rule (cut A into the part inside B and the part outside B):

P [A] = P [A B] + P [A B^{c}]

Equally likely outcomes:

P [A] = \frac{number of outcomes in A}{number of outcomes in S}

Also: $P [φ] = 0$ , and if $A \subset B$ then $P [A] \leq P [B]$ .

Why do we subtract in the "or" rule?

If you add $P [A]$ and $P [B]$ , the shared part $A \cap B$ is counted two times. So we take it away once.

Writing note: $P [A \cap B]$ , $P [A, B]$ and $P [A B]$ all mean the same thing: "A and B".

Example from the slides: fair die

Roll a six-sided die. All faces are equally likely. Find the probability of each outcome, and of: "roll 4 or higher", "roll an even number", "roll the square of an integer".

Show solution

6 outcomes, all equally likely. Each one has probability $\frac{1}{6}$ .
"4 or higher" $= {4, 5, 6}$ . That is 3 outcomes: $\frac{3}{6} = \frac{1}{2}$ .
"Even" $= {2, 4, 6}$ . That is 3 outcomes: $\frac{3}{6} = \frac{1}{2}$ .
"Square of an integer" means $1^{2} = 1$ or $2^{2} = 4$ . So ${1, 4}$ : $\frac{2}{6} = \frac{1}{3}$ .

Answer

\frac{1}{6}

each; then

\frac{1}{2}, \frac{1}{2}, \frac{1}{3}

1.5The table trick (for questions with two labels)

Many questions describe each outcome with two labels, for example voice/data and long/brief. Draw a small table. Rows are one label, columns are the other. Each row and each column must add up to its total, and everything adds up to 1.

Example from the slides: phone calls

A call is voice (V) or data (D). It is also long (L) or brief (B). Given: $P [V] = 0.7$ , $P [L] = 0.6$ , $P [V L] = 0.35$ . Find:

(1) $P [D L]$ (2) $P [D \cup L]$ (3) $P [V B]$ (4) $P [V \cup L]$ (5) $P [V \cup D]$ (6) $P [L B]$

Show solution

Step 1. Put what you know in the table (the bold numbers). Then fill the rest by subtracting.

	L	B	Total
V	0.35	0.35	0.7
D	0.25	0.05	0.3
Total	0.6	0.4	1

$P [D] = 1 - 0.7 = 0.3$ and $P [B] = 1 - 0.6 = 0.4$ .
$P [V B] = 0.7 - 0.35 = 0.35$ (row V must add to 0.7).
$P [D L] = 0.6 - 0.35 = 0.25$ (column L must add to 0.6).
$P [D B] = 0.3 - 0.25 = 0.05$ .

Step 2. Read the answers.

$P [D L] = 0.25$
$P [D \cup L] = P [D] + P [L] - P [D L] = 0.3 + 0.6 - 0.25 = 0.65$
$P [V B] = 0.35$
$P [V \cup L] = 0.7 + 0.6 - 0.35 = 0.95$
$P [V \cup D] = 1$ . Every call is voice or data.
$P [L B] = 0$ . A call cannot be long and brief together.

Trap

$P [V L]$ is "V and L" (one cell inside the table). $P [V \cup L]$ is "V or L" (a whole row plus a whole column, with the shared cell counted once).

Now solve these sample questions: Q1 · sets Q2 · heart card Q3 · table trick Q4 · disjoint events

Conditional probability, Bayes, trees

Week 3, first half

2.1Conditional probability: "given that"

In simple words

Someone gives you a hint: "B happened". Now you forget everything outside B. B becomes your new, smaller world. Then you ask: inside this world, how much is A?

Rule

P [A ∣ B] = \frac{P [ A B ]}{P [ B ]}

Read: "probability of A given B" equals "probability of both" divided by "probability of the hint". It only makes sense when $P [B] > 0$ .

Fast way when all outcomes are equally likely: count inside the new world.

P [A ∣ B] = \frac{outcomes in both A and B}{outcomes in B}

Example from the slides: the urn

An urn has two black balls numbered 1 and 2, and two white balls numbered 3 and 4. One ball is selected, all equally likely. A = "ball is black". B = "number is even". C = "number is more than 2". Find $P [A ∣ B]$ and $P [A ∣ C]$ .

Show solution

Write the sets. $S = {1, 2, 3, 4}$ , $A = {1, 2}$ , $B = {2, 4}$ , $C = {3, 4}$ .
$P [A ∣ B]$ : the new world is $B = {2, 4}$ . Inside it, the black balls are ${2}$ . So $P [A ∣ B] = \frac{1}{2}$ .
With the formula: $\frac{P [ A B ]}{P [ B ]} = \frac{1/4}{2/4} = \frac{1}{2}$
$P [A ∣ C]$ : the new world is $C = {3, 4}$ . No black ball is in it. So $P [A ∣ C] = 0$ .

Answer

P [A ∣ B] = \frac{1}{2}

P [A ∣ C] = 0

Trap

$P [A ∣ B]$ and $P [B ∣ A]$ are different. The thing after the bar is the hint. Always divide by the probability of the hint.

2.2The multiplication rule

Move $P [B]$ to the other side of the last rule:

Rule

P [A B] = P [A ∣ B] P [B]

"Probability of both" = "probability of the first" × "probability of the second, given the first".

This is exactly what you do when you multiply along a branch of a tree diagram.

2.3Tree diagrams

Use a tree when the experiment happens in steps (first this, then that).

The 3 tree rules

Branches that leave the same point add up to 1.
Multiply along a path to get the probability of that path.
Add the paths that give what you want.

Example from the slides: two traffic lights

The first light is equally likely red or green. With probability 0.8 the second light has the same colour as the first. Find: $P [G_{2}]$ (second light is green), $P [W]$ (you wait for at least one light), and $P [G_{1} ∣ R_{2}]$ .

First light, then second light. Same colour again has probability 0.8.

Show solution

Second light green. Add the paths that end in $G_{2}$ :
$P [G_{2}] = P [G_{1} G_{2}] + P [R_{1} G_{2}] = 0.4 + 0.1 = 0.5$
Wait for at least one light. You wait if at least one light is red. The only path with no waiting is $G_{1} G_{2}$ . Use the "not" rule:
$P [W] = 1 - P [G_{1} G_{2}] = 1 - 0.4 = 0.6$
First green given second red. Use the conditional rule. The hint is $R_{2}$ .
$P [R_{2}] = 0.1 + 0.4 = 0.5$
$P [G_{1} ∣ R_{2}] = \frac{P [ G _{1} R _{2} ]}{P [ R _{2} ]} = \frac{0.1}{0.5} = 0.2$

Answer

P [G_{2}] = 0.5

P [W] = 0.6

P [G_{1} ∣ R_{2}] = 0.2

The "at least one" trick

"At least one" is hard to count directly. Its opposite, "none", is one single path. So: P[at least one] = 1 − P[none]. You will use this many times.

2.4Law of total probability

In simple words

You want $P [A]$ , but A can happen in several different situations $B_{1}, B_{2}, \dots$ (an event space, like the pizza slices). Find A inside each situation, weigh it by how likely that situation is, and add.

Rule

P [A] = i = 1 \sum m P [A ∣ B_{i}] P [B_{i}]

For three situations: $P [A] = P [A ∣ B_{1}] P [B_{1}] + P [A ∣ B_{2}] P [B_{2}] + P [A ∣ B_{3}] P [B_{3}]$ .

In a tree: this is "add all paths that end in A".

Example from the slides: resistors

Three machines make resistors. 80% of the resistors from $B_{1}$ are acceptable (A), 90% from $B_{2}$ , 60% from $B_{3}$ . Each hour $B_{1}$ makes 3000, $B_{2}$ makes 4000, $B_{3}$ makes 3000. They are all mixed. What is the probability that a resistor is acceptable? Also: what is the probability of picking a resistor from $B_{2}$ that is not acceptable (N)?

Multiply along each path. The numbers on the right add up to 1.

Show solution

Which machine? Total is $3000 + 4000 + 3000 = 10000$ . So $P [B_{1}] = 0.3$ , $P [B_{2}] = 0.4$ , $P [B_{3}] = 0.3$ .
Given the machine: $P [A ∣ B_{1}] = 0.8$ , $P [A ∣ B_{2}] = 0.9$ , $P [A ∣ B_{3}] = 0.6$ .
Add all paths that end in A:
$P [A] = 0.8 (0.3) + 0.9 (0.4) + 0.6 (0.3) = 0.24 + 0.36 + 0.18 = 0.78$
From $B_{2}$ and not acceptable: one path. $P [B_{2} N] = 0.4 \times 0.1 = 0.04$

Answer

P [A] = 0.78

and

P [B_{2} N] = 0.04

2.5Bayes' theorem: turning the question around

In simple words

You know "given the machine, how likely is a good resistor". The question asks the reverse: "given a good resistor, how likely is it from this machine?" Bayes flips the direction.

Rule

P [B ∣ A] = \frac{P [ A ∣ B ] P [ B ]}{P [ A ]}

In a tree: (the one path you want) ÷ (all paths that end in A).

Example: resistors again

A resistor is acceptable. What is the probability it came from machine $B_{3}$ ?

Show solution

P [B_{3} ∣ A] = \frac{P [ A ∣ B _{3} ] P [ B _{3} ]}{P [ A ]} = \frac{0.6 \times 0.3}{0.78} = \frac{0.18}{0.78} = \frac{3}{13} \approx 0.231

Top: the path $B_{3} A$ . Bottom: all three paths that end in A.

2.6Independent events

In simple words

A and B are independent when knowing one tells you nothing about the other. Example: a coin flip and a die roll.

The test

A and B are independent exactly when:

P [A B] = P [A] P [B]

Same thing, written another way: $P [A ∣ B] = P [A]$ and $P [B ∣ A] = P [B]$ . The hint changes nothing.

Biggest trap of this lesson

Independent is not the same as disjoint.

Disjoint: they cannot happen together. $P [A B] = 0$ .
Independent: they do not affect each other. $P [A B] = P [A] P [B]$ .

If A and B are disjoint (and both have probability more than 0), they are not independent. Why? If A happens, you know for sure B did not. That is a lot of information.

Example from the slides: three test lights

Three tests, each light is red (r) or green (g). All 8 sequences are equally likely. $R_{2}$ = second light red, $G_{2}$ = second light green, $R_{1}$ = first light red. (a) Are $R_{2}$ and $G_{2}$ independent? (b) Are $R_{1}$ and $R_{2}$ independent?

Show solution

$S = {r r r, r r g, r g r, r g g, g r r, g r g, g g r, g g g}$ . Each outcome has probability $\frac{1}{8}$ .

(a) $P [R_{2}] = \frac{4}{8} = \frac{1}{2}$ and $P [G_{2}] = \frac{1}{2}$ . The second light cannot be red and green together, so $P [R_{2} G_{2}] = 0$ .
Test: $P [R_{2}] P [G_{2}] = \frac{1}{4} \neq = 0$ . Not independent. (They are disjoint.)
(b) $R_{1} R_{2} = {r r r, r r g}$ , so $P [R_{1} R_{2}] = \frac{2}{8} = \frac{1}{4}$ .
Test: $P [R_{1}] P [R_{2}] = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ . Equal. Independent.

Example from the slides: circuit with two tests

A circuit has a mechanical test and an electrical test. Failures are independent. Mechanical failure has probability 0.05. Electrical failure has probability 0.2. Find the probability of each of the four results.

Show solution

Independent means: multiply. Let M = mechanical fails, E = electrical fails. Then $P [M^{c}] = 0.95$ , $P [E^{c}] = 0.8$ .

	E fails (0.2)	E passes (0.8)
M fails (0.05)	0.05 × 0.2 = 0.01	0.05 × 0.8 = 0.04
M passes (0.95)	0.95 × 0.2 = 0.19	0.95 × 0.8 = 0.76

Check: $0.01 + 0.04 + 0.19 + 0.76 = 1$ .

Now solve these sample questions: Q7 · tree diagram Q8 · conditional with a die Q9 · total probability and Bayes

Counting and independent trials

Week 3, second half

3.1The basic idea: multiply the choices

Fundamental principle of counting

If step 1 can go $n$ ways and step 2 can go $k$ ways, then both steps together can go $n \times k$ ways.

Example from the slides: flip a coin (2 results) and roll a die (6 results). Together: $2 \times 6 = 12$ outcomes.

Picture

3 shirts and 2 trousers give $3 \times 2 = 6$ outfits. For every shirt you can pick any of the trousers.

3.2The two questions that pick the formula

Before any counting question, ask:

Can things repeat? (Can the same item be used again?)
Does order matter? (Is AB different from BA?)

Situation	Formula	Example
Repeat allowed, order matters "sampling with replacement"	$m^{n}$	4-digit PIN: $1 0^{4}$
No repeat, order matters "k-permutation"	$(n)_{k} = \frac{n !}{( n - k )!}$	Deal 3 cards in order: $52 \cdot 51 \cdot 50$
No repeat, use all n, order matters	$n!$	Shuffle a deck: $52!$
No repeat, order does not matter "k-combination"	$(k n) = \frac{n !}{k ! ( n - k )!}$	Choose 7 cards as a hand: $(7 52)$
Arrange n items made of groups of identical items "multinomial"	$\frac{n !}{n _{0} ! n _{1} ! \dots n _{m - 1} !}$	3 A's, 5 B's, 2 C's for 10 students

What the letters mean

$m^{n}$ : $m$ = number of choices for each position, $n$ = number of positions.
$(n)_{k}$ and $(k n)$ : $n$ = how many things you choose from, $k$ = how many you pick.
$\frac{n !}{( n - k )!}$ is just the first $k$ numbers counting down from $n$ . Example: $\frac{52 !}{49 !} = 52 \cdot 51 \cdot 50$ .

How to compute "n choose k" by hand

(3 5) = \frac{5 !}{3 ! 2 !} = \frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{( 3 \cdot 2 \cdot 1 ) ( 2 \cdot 1 )} = \frac{120}{12} = 10

On the calculator it is the nCr button: 5 nCr 3 = 10. Useful fact: $(k n) = (n - k n)$ , so $(3 5) = (2 5)$ .

3.3Examples from the slides

Cards in order

Shuffle a deck and choose three cards in order. How many outcomes?

Show solution

No repeat (a card cannot come twice), order matters. So it is a k-permutation with $n = 52$ , $k = 3$ :

(52)_{3} = \frac{52 !}{( 52 - 3 )!} = 52 \cdot 51 \cdot 50 = 132600

Seven cards, no queens

We draw seven cards. What is the probability of getting a hand without any queens?

Show solution

A hand is a group. Order does not matter. So use "choose".
All possible hands: $(7 52)$ .
Hands with no queen: remove the 4 queens. Choose 7 from the other 48: $(7 48)$ .
All hands are equally likely, so divide:
$P = \frac{( 7 48 )}{( 7 52 )} \approx 0.550$

Binary code words with 4 bits

Each code word has 4 bits (0 or 1), like 0110. (1) How many code words are there? (2) How many have exactly two zeroes? (3) How many begin with a zero? (4) With N = 8 bits and exactly M = 3 ones, how many code words?

Show solution

Each of 4 positions has 2 choices, repeats allowed: $2^{4} = 16$ .
Choose which 2 of the 4 positions hold the zeroes: $(2 4) = 6$ .
The first bit is fixed (0). The other 3 bits are free: $2^{3} = 8$ .
Choose which 3 of the 8 positions hold the ones: $(3 8) = 56$ .

Answer16, 6, 8, 56

Grades for 10 students

10 students take an exam. 3 of them get A, 5 get B, 2 get C. How many ways are there to distribute the grades?

Show solution

We arrange 10 grades, but the 3 A's are identical, the 5 B's are identical, the 2 C's are identical. This is the multinomial formula:

\frac{10 !}{3 ! 5 ! 2 !} = \frac{3628800}{6 \cdot 120 \cdot 2} = 2520

Trap

"How many words with exactly two zeroes" is not $m^{n}$ . The word "exactly" means you are choosing positions, so use $(k n)$ .

3.4Independent trials

In simple words

Repeat the same small experiment $n$ times. Each time it is a success with probability $p$ or a failure with probability $1 - p$ . The tries do not affect each other. What is the chance of getting exactly $n_{1}$ successes?

Rule

P [n_{0} failures and n_{1} successes] = (n _{1} n) (1 - p)^{n_{0}} p^{n_{1}}

where $n = n_{0} + n_{1}$ is the total number of tries.

Read it in three pieces:

$p^{n_{1}}$ : the successes all happen.
$(1 - p)^{n_{0}}$ : the failures all happen.
$(n _{1} n)$ : the number of different orders in which they can happen.

Example from the slides: sending one bit five times

A phone sends the same bit 5 times. The receiver is right if 3 or more of the 5 are received correctly. Each bit is wrong with probability $q = 0.1$ . What is the probability $P [E]$ that the information is wrong?

Show solution

The information is wrong when 3, 4 or 5 of the bits are wrong. (Then fewer than 3 are correct.)
Here "success" = a bit is wrong, with $p = 0.1$ . There are $n = 5$ tries.
3 wrong: $(3 5) (0.1)^{3} (0.9)^{2} = 10 \times 0.001 \times 0.81 = 0.0081$
4 wrong: $(4 5) (0.1)^{4} (0.9)^{1} = 5 \times 0.0001 \times 0.9 = 0.00045$
5 wrong: $(5 5) (0.1)^{5} = 0.00001$
Add: $P [E] = 0.0081 + 0.00045 + 0.00001 = 0.00856$

Answer

P [E] = 0.00856

This rule comes back in lesson 4 with a new name: the binomial random variable.

Now solve these sample questions: Q5 · words from letters Q6 · binary code

Random variables and PMF

Week 4 · the heart of the exam

4.1What is a random variable?

In simple words

A random variable is a rule that turns each outcome into a number. Outcomes are often words or letters (like "hht"). We want numbers, so we can do maths with them.

Capital letter $X$ : the name of the random number. Example: X = "number of heads".
Small letter $x$ : one value it can take. Example: $x = 2$ .
${X = x}$ is an event: "X came out equal to x".
$S_{X}$ : the range. The list of all values X can take.
Discrete random variable: you can list its values one by one (0, 1, 2, ...). The whole course is about these.

Example from the slides: three phone calls

We observe three calls. Each is voice (v) or data (d), equally likely. X = number of voice calls. Y = number of data calls. R = XY (X times Y). Write the sample space and the values of X, Y, R.

Show solution

$2 \cdot 2 \cdot 2 = 8$ outcomes, each with probability $\frac{1}{8}$ .

Outcome	Prob.	X (voice)	Y (data)	R = XY
ddd	1/8	0	3	0
ddv	1/8	1	2	2
dvd	1/8	1	2	2
dvv	1/8	2	1	2
vdd	1/8	1	2	2
vdv	1/8	2	1	2
vvd	1/8	2	1	2
vvv	1/8	3	0	0

So R can only be 0 or 2: $S_{R} = {0, 2}$ .

4.2The PMF

Definition

P_{X} (x) = P [X = x]

PMF = probability mass function. It is a price list: for each value x, it tells you the probability that X is exactly x.

Reading the symbol: the small $X$ below the P says which random variable. The $(x)$ is the value you ask about.

How to build a PMF (always the same 3 steps):

List the values X can take.
For each value, add the probabilities of all outcomes that give it.
Write it in the "brace form", and finish with 0 otherwise.

Example: PMF of R (three calls)

From the table above: R = 0 for 2 outcomes (ddd, vvv). R = 2 for the other 6 outcomes.

P_{R} (r) = ⎩ ⎨ ⎧ 1/4 3/4 0 r = 0 r = 2 otherwise

$P_{R} (0) = \frac{2}{8} = \frac{1}{4}$ and $P_{R} (2) = \frac{6}{8} = \frac{3}{4}$ . "Otherwise" means every other number has probability 0.

Example from the slides: two free throws

Wilt Chamberlain shoots two free throws. Each shot is good (g) or bad (b), equally likely. A good shot is 1 point. Find the PMF of X, the number of points.

Show solution

Outcomes: bb, bg, gb, gg. Each has probability $\frac{1}{4}$ .
Points: bb gives 0. bg and gb give 1. gg gives 2.
Add up: $P_{X} (0) = \frac{1}{4}$ , $P_{X} (1) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ , $P_{X} (2) = \frac{1}{4}$ .

P_{X} (x) = ⎩ ⎨ ⎧ 1/4 1/2 1/4 0 x = 0 x = 1 x = 2 otherwise

Example from the slides: GPA

A student takes two courses. In each one the grade is B (3 points) with probability 0.6 or C (2 points) with probability 0.4, independently. G = average of the two. Find the values of G and their probabilities.

Show solution

Outcome	Probability	G
BB	0.6 × 0.6 = 0.36	3
BC	0.6 × 0.4 = 0.24	2.5
CB	0.4 × 0.6 = 0.24	2.5
CC	0.4 × 0.4 = 0.16	2

So $P_{G} (2) = 0.16$ , $P_{G} (2.5) = 0.48$ , $P_{G} (3) = 0.36$ .

4.3The three PMF rules

Theorem from the slides

$P_{X} (x) \geq 0$ . No negative probabilities.
$x \in S_{X} \sum P_{X} (x) = 1$ . All the probabilities add up to 1.
$P [B] = x \in B \sum P_{X} (x)$ . For any event B, add the PMF of the values inside B.

How these rules are used in the exam

Rule 2 finds an unknown constant c: write the sum, set it equal to 1, solve for c.
Rule 3 answers "find $P [X \geq 2]$ " or "find P[X is even]": pick the values that fit, add their probabilities.

Example from the slides: find the constant

P_{N} (n) = {c / n 0 n = 1, 2, 3 otherwise

Find (1) the constant c, (2) $P [N = 1]$ , (3) $P [N \geq 2]$ , (4) $P [N > 3]$ .

Show solution

Put in each value: $P_{N} (1) = \frac{c}{1}$ , $P_{N} (2) = \frac{c}{2}$ , $P_{N} (3) = \frac{c}{3}$ .
They must add to 1:
$c + \frac{c}{2} + \frac{c}{3} \frac{6 c + 3 c + 2 c}{6} \frac{11 c}{6} = 1 = 1 = 1 \Rightarrow c = \frac{6}{11}$
$P [N = 1] = c = \frac{6}{11}$
$P [N \geq 2] = P_{N} (2) + P_{N} (3) = \frac{3}{11} + \frac{2}{11} = \frac{5}{11}$
$P [N > 3] = 0$ . N can never be bigger than 3.

Answer

c = \frac{6}{11}

; then

\frac{6}{11}, \frac{5}{11}, 0

4.4The six famous families

Some stories come back again and again, so their PMFs have names. In the exam your first job is to recognise the story. Then you copy the formula and put in the numbers.

In all of them: $p$ = probability of success in one try, and $1 - p$ = probability of failure.

1. Bernoulli (p)

Story: one try. X = 1 if success, X = 0 if failure.

P_{X} (x) = ⎩ ⎨ ⎧ 1 - p p 0 x = 0 x = 1 otherwise

E [X] = p

Var [X] = p (1 - p)

2. Binomial (n, p)

Story: do a fixed number n of tries. X = how many successes.

P_{X} (x) = (x n) p^{x} (1 - p)^{n - x}

for $x = 0, 1, \dots, n$ , and 0 otherwise.

Read: (which x tries succeed) × (x successes) × (n − x failures). It is the independent trials rule from lesson 3.

E [X] = n p

Var [X] = n p (1 - p)

3. Geometric (p)

Story: keep trying until the first success. X = number of tries (the success is counted).

P_{X} (x) = {p (1 - p)^{x - 1} 0 x = 1, 2, 3, \dots otherwise

Read: first $x - 1$ failures, then one success.

E [X] = 1/ p

Var [X] = (1 - p) / p^{2}

4. Pascal (k, p)

Story: keep trying until the k-th success. X = number of tries. ("k-th" means number k in order: 2nd, 3rd, ...)

P_{X} (x) = (k - 1 x - 1) p^{k} (1 - p)^{x - k}

for $x = k, k + 1, \dots$ , and 0 otherwise.

Read: the last try is the k-th success. Before it, the other $k - 1$ successes sit somewhere in the first $x - 1$ tries. With $k = 1$ it becomes the geometric.

E [X] = k / p

Var [X] = k (1 - p) / p^{2}

5. Discrete uniform (k, l)

Story: every whole number from k to l is equally likely. (A fair die is uniform (1, 6).)

P_{X} (x) = ⎩ ⎨ ⎧ \frac{1}{l - k + 1} 0 x = k, k + 1, \dots, l otherwise

$l - k + 1$ is just how many values there are. From 5 to 15 there are $15 - 5 + 1 = 11$ values.

E [X] = (k + l) /2

Var [X] = (l - k) (l - k + 2) /12

6. Poisson (α)

Story: count how many arrivals (calls, hits, queries) happen in a time interval.

P_{X} (x) = ⎩ ⎨ ⎧ \frac{α ^{x} e ^{- α}}{x !} 0 x = 0, 1, 2, \dots otherwise

α = λ T

$λ$ (lambda) = average arrivals per second. $T$ = length of the interval. $α$ (alpha) = average number of arrivals in that interval.

E [X] = α

Var [X] = α

4.5Which family is it? Look for these words

Words in the question	Family
One try, yes or no	Bernoulli (p)
"Sends n times", "100 bits", "number of successes / errors in n tries"	Binomial (n, p)
"Until the first error / success", "number of tries needed"	Geometric (p)
"Until the third error", "until the k-th success"	Pascal (k, p)
"Uniformly distributed between ... and ...", "equally likely"	Discrete uniform (k, l)
"Average rate", "per second", "in an interval of T seconds", "arrivals"	Poisson (α = λT)

The quick question to ask yourself

What is fixed, and what is counted?

Number of tries is fixed, I count successes: binomial.
Number of successes is fixed (1 or k), I count tries: geometric or Pascal.
A time interval is fixed, I count arrivals: Poisson.

Traps

"Success" is whatever the question counts. In the modem question the "success" is an error.
Geometric and Pascal start at $x = 1$ and $x = k$ . Binomial and Poisson start at $x = 0$ .
Poisson: if the time interval changes, $α$ changes. Always compute $α = λ T$ again.
"At least one" = $1 - P_{X} (0)$ . "At least two" = $1 - P_{X} (0) - P_{X} (1)$ .

4.6Poisson examples from the slides

Web site hits (Week 4)

Hits at a web site are Poisson with $λ = 2$ hits per second. (a) What is the probability of no hits in 0.25 seconds? (b) What is the probability of no more than two hits in one second?

Show solution

(a) $T = 0.25$ , so $α = λ T = 2 \times 0.25 = 0.5$ .
"No hits" means $x = 0$ :
$P_{X} (0) = \frac{0. 5 ^{0} e ^{- 0.5}}{0 !} = e^{- 0.5} \approx 0.607$
(Remember: any number to the power 0 is 1, and $0! = 1$ .)
(b) New interval: $T = 1$ , so $α = 2 \times 1 = 2$ .
"No more than two" means 0, 1 or 2:
$P_{X} (0) + P_{X} (1) + P_{X} (2) = e^{- 2} + \frac{2 ^{1} e ^{- 2}}{1 !} + \frac{2 ^{2} e ^{- 2}}{2 !} = e^{- 2} (1 + 2 + 2) = 5 e^{- 2} \approx 0.677$

Answer(a) 0.607 (b) 0.677

Database queries (Week 5)

The number of queries in any 10-second interval is Poisson with $α = 5$ . (a) Probability of no queries in a 10-second interval? (b) Probability of at least two queries in a 2-second interval?

Show solution

(a) $α = 5$ : $P_{K} (0) = e^{- 5} \approx 0.0067$
(b) The interval changed. 5 queries in 10 seconds means $λ = \frac{5}{10} = 0.5$ per second. For 2 seconds: $α = 0.5 \times 2 = 1$ .
"At least two" = 1 − (zero or one):
$P [K \geq 2] = 1 - P_{K} (0) - P_{K} (1) = 1 - e^{- 1} - \frac{1 ^{1} e ^{- 1}}{1 !} = 1 - 2 e^{- 1} \approx 0.264$

Answer(a) 0.0067 (b) 0.264

Calls at a switching office (Week 5)

Calls arrive with an average of $λ = 0.25$ calls per second. What is the PMF of the number of calls in $T = 2$ seconds?

Show solution

$α = λ T = 0.25 \times 2 = 0.5$ . Put it in the Poisson formula:

P_{J} (j) = ⎩ ⎨ ⎧ \frac{( 0.5 ) ^{j} e ^{- 0.5}}{j !} 0 j = 0, 1, 2, \dots otherwise

For the sketch: $P_{J} (0) \approx 0.607$ , $P_{J} (1) \approx 0.303$ , $P_{J} (2) \approx 0.076$ , $P_{J} (3) \approx 0.013$ . The bars get smaller and smaller.

Now solve these sample questions: Q10 · find the constant c Q11 · paging (binomial)Q13 · ticket agency Q16 · M&Ms (uniform)Q19 · modem (3 families)

CDF: the running total

Week 5, first half

5.1What is the CDF?

Definition

F_{X} (x) = P [X \leq x]

CDF = cumulative distribution function. "Cumulative" means "collected so far". $F_{X} (x)$ is the probability that X is x or less.

In simple words

The PMF is the money you get on each day. The CDF is the total in your pocket so far. To get the CDF, walk from left to right and keep adding the PMF.

	PMF $P_{X} (x)$	CDF $F_{X} (x)$
Asks	X is exactly x?	X is x or less?
Symbol	$P [X = x]$	$P [X \leq x]$
Graph	Separate bars	A staircase going up

5.2From PMF to CDF

Example from the slides: CDF of R

R has $P_{R} (0) = \frac{1}{4}$ and $P_{R} (2) = \frac{3}{4}$ . Find and sketch the CDF.

PMF of R: one bar for each possible value.

Walk from left to right and ask "how much probability have I passed so far?"

Before 0: nothing yet. Total = 0.
From 0 up to just before 2: we passed the bar at 0. Total = $\frac{1}{4}$ .
From 2 onward: we also passed the bar at 2. Total = $\frac{1}{4} + \frac{3}{4} = 1$ .

F_{R} (r) = ⎩ ⎨ ⎧ 0 1/4 1 r < 0 0 \leq r < 2 r \geq 2

CDF of R. Filled dot = the value AT that point. Empty dot = not included.

What every CDF looks like (theorem from the slides)

It starts at 0 on the far left and ends at 1 on the far right.
It never goes down.
It jumps at each value X can take. The height of the jump at $x_{i}$ is $P_{X} (x_{i})$ .
Between jumps it is flat (a horizontal line).

How to read the brace form

Look at the line " $\frac{1}{4}$ for $0 \leq r < 2$ ". The left side has $\leq$ , so the jump happens at 0 and the value at 0 is already $\frac{1}{4}$ (filled dot). The right side has $<$ , so at 2 this line is finished (empty dot) and the next line takes over.

5.3From CDF back to PMF

Rule: the PMF is the jump

X can only take the values where the CDF jumps. At each one:

P_{X} (x_{i}) = - (height after the jump) (height before the jump)

Example: a CDF goes 0 → 0.2 at $x = - 1$ , then 0.2 → 0.7 at $x = 0$ , then 0.7 → 1 at $x = 1$ . So $P_{X} (- 1) = 0.2$ , $P_{X} (0) = 0.7 - 0.2 = 0.5$ , $P_{X} (1) = 1 - 0.7 = 0.3$ . Check: $0.2 + 0.5 + 0.3 = 1$ .

5.4Reading probabilities from a CDF

This is asked a lot. Learn this table. "Just before a" is written $a - ϵ$ in the slides ( $ϵ$ is a tiny positive number).

You want	Do this	In words
$P [X \leq a]$	$F_{X} (a)$	Height at a (the filled dot)
$P [X < a]$	$F_{X} (a - ϵ)$	Height just before a
$P [X > a]$	$1 - F_{X} (a)$	1 minus the height at a
$P [X \geq a]$	$1 - F_{X} (a - ϵ)$	1 minus the height just before a
$P [X = a]$	jump at a	How much it jumps at a (0 if no jump)
$P [a < X \leq b]$	$F_{X} (b) - F_{X} (a)$	Height at b minus height at a

The safest method

If the table feels hard: first write the PMF from the jumps. Then every question becomes "add the PMF of the values that fit". It always works and it is hard to get wrong.

Trap: < and ≤ are different

For discrete random variables, $P [X < 1]$ and $P [X \leq 1]$ are different when X can be exactly 1. The difference is $P [X = 1]$ , the jump at 1.

The opposite of " $\leq 1$ " is " $> 1$ ". The opposite of " $< 1$ " is " $\geq 1$ ". So $P [X \geq 2] = 1 - P [X \leq 1]$ , not $1 - P [X \leq 2]$ .

Now solve these sample questions: Q12 · CDF in brace form Q17 · CDF from a graph

Part (d) of Q12 asks for the expected value. That is lesson 6: value × probability, then add.

Expected value and variance

Week 5, second half

6.1Three kinds of average for a list of numbers

Name	How
Mean	Add all the numbers, divide by how many there are
Median	Sort them, take the middle one (for an even count: halfway between the two middle ones)
Mode	The number that appears most often

Example from the slides: quiz grades

10 students have grades: 9, 5, 10, 8, 4, 7, 5, 5, 8, 7. Find the mean, the median and the mode.

Show solution

Sort first: 4, 5, 5, 5, 7, 7, 8, 8, 9, 10
Mean: the sum is 68. $68 \div 10 = 6.8$
Median: 10 numbers, so the middle two are the 5th and 6th: 7 and 7. Median = 7
Mode: 5 appears three times, more than any other. Mode = 5

AnswerMean 6.8, median 7, mode 5

For a random variable the same three ideas use probabilities:

Mode $x_{m o d}$ : the value with the biggest PMF. $P_{X} (x_{m o d}) \geq P_{X} (x)$ for all x.
Median $x_{m e d}$ : the value that splits the probability in two equal sides. $P [X < x_{m e d}] = P [X > x_{m e d}]$ .
Mean: called the expected value. See next.

6.2Expected value E[X]

Definition

E [X] = μ_{X} = x \in S_{X} \sum x P_{X} (x)

Take each value, multiply by its probability, and add. $E [X]$ and $μ_{X}$ ("mu") are two names for the same number.

In simple words

It is the long-run average. Repeat the experiment thousands of times and average the results: you get close to $E [X]$ . It does not have to be a value X can really take. The average of a die is 3.5, but you never roll 3.5.

Example from the slides: E[R]

$P_{R} (0) = \frac{1}{4}$ , $P_{R} (2) = \frac{3}{4}$ . What is $E [R]$ ?

Show solution

E [R] = 0 \cdot \frac{1}{4} + 2 \cdot \frac{3}{4} = \frac{3}{2} = 1.5

Example from the slides: cost of a call

A call is voice with $P [V] = 0.7$ or data with $P [D] = 0.3$ . Voice costs 25 cents, data costs 40 cents. C = cost of one call. Find (1) the PMF $P_{C} (c)$ , (2) $E [C]$ .

Show solution

C is 25 when the call is voice, and 40 when it is data:
$P_{C} (c) = ⎩ ⎨ ⎧ 0.7 0.3 0 c = 25 c = 40 otherwise$
$E [C] = 25 (0.7) + 40 (0.3) = 17.5 + 12 = 29.5$ cents

Answer

E [C] = 29.5

cents

6.3Variance and standard deviation

Why we need them

Class A: two students both got 5. Class B: one got 0, one got 10. Both classes have average 5, but B is far more spread out. Variance is the number that measures this spread.

Definitions

Var [X] = E [(X - μ_{X})^{2}] = x \in S_{X} \sum (x - μ_{X})^{2} P_{X} (x)

Variance = the average of the squared distances from the centre $μ_{X}$ .

σ_{X} = Var [X]

Standard deviation $σ_{X}$ ("sigma") = square root of the variance. It is the typical distance from the centre, in the same units as X. Also $Var [X] = σ_{X}^{2}$ .

The shortcut you will use in the exam

Var [X] = E [X^{2}] - (E [X])^{2}

where $E [X^{2}] = x \in S_{X} \sum x^{2} P_{X} (x)$ : square each value first, then multiply by its probability, then add.

The 4-step recipe (same every time)

$E [X]$ : value × probability, add.
$E [X^{2}]$ : value² × probability, add.
$Var [X] = E [X^{2}] - (E [X])^{2}$ .
$σ_{X} = Var [X]$ .

Traps

$E [X^{2}]$ and $(E [X])^{2}$ are different. In the first you square inside (before averaging). In the second you square the final average.
A negative value squared is positive: $(- 3)^{2} = 9$ .
Variance can never be negative. If you get a negative number, check your work.
In the long formula you multiply by the probability $P_{X} (x)$ , not by $μ_{X}$ .

Example from the slides: Var[R]

$P_{R} (0) = \frac{1}{4}$ , $P_{R} (2) = \frac{3}{4}$ , and $E [R] = 1.5$ . What is the variance of R?

Show solution

$E [R^{2}] = 0^{2} \cdot \frac{1}{4} + 2^{2} \cdot \frac{3}{4} = 0 + 3 = 3$
$Var [R] = E [R^{2}] - (E [R])^{2} = 3 - (1.5)^{2} = 3 - 2.25 = 0.75$
Check with the long formula: $(0 - 1.5)^{2} \cdot \frac{1}{4} + (2 - 1.5)^{2} \cdot \frac{3}{4} = 0.5625 + 0.1875 = 0.75$ . Same.

Answer

Var [R] = 0.75

, so

σ_{R} = 0.75 \approx 0.866

Example from the slides: voltage V

P_{V} (v) = {1/7 0 v = - 3, - 2, \dots, 3 otherwise

What is the variance of V?

Show solution

The values are −3, −2, −1, 0, 1, 2, 3. Seven values, each with probability $\frac{1}{7}$ .
$E [V] = \frac{1}{7} (- 3 - 2 - 1 + 0 + 1 + 2 + 3) = 0$ . The values balance around 0.
$E [V^{2}] = \frac{1}{7} (9 + 4 + 1 + 0 + 1 + 4 + 9) = \frac{28}{7} = 4$
$Var [V] = 4 - 0^{2} = 4$
Check with the uniform formula, $k = - 3$ , $l = 3$ : $\frac{( l - k ) ( l - k + 2 )}{12} = \frac{6 \cdot 8}{12} = 4$ . Same.

Answer

Var [V] = 4

Example from the slides: two calls

P_{N} (n) = ⎩ ⎨ ⎧ 0.1 0.4 0.5 0 n = 0 n = 1 n = 2 otherwise

Find (1) $E [N]$ , (2) the second moment $E [N^{2}]$ , (3) $Var [N]$ , (4) $σ_{N}$ .

Show solution

$E [N] = 0 (0.1) + 1 (0.4) + 2 (0.5) = 1.4$
$E [N^{2}] = 0^{2} (0.1) + 1^{2} (0.4) + 2^{2} (0.5) = 0.4 + 2 = 2.4$
$Var [N] = 2.4 - (1.4)^{2} = 2.4 - 1.96 = 0.44$
$σ_{N} = 0.44 \approx 0.663$

Answer1.4, 2.4, 0.44, 0.663

6.4Moments (just names)

Name	Formula	What it is
1st moment	$E [X]$	The average
2nd moment	$E [X^{2}]$	Square, then average
n-th moment	$E [X^{n}]$	Same idea with power n
2nd central moment	$E [(X - μ_{X})^{2}]$	The variance

If the exam says "find the second moment", it just means $E [X^{2}]$ .

6.5Variance of aX + b

Theorem

Var [a X + b] = a^{2} Var [X]

Adding a number $b$ changes nothing. Multiplying by $a$ multiplies the variance by $a^{2}$ .

Why

If every student steps 3 metres forward, the gaps between them stay the same: adding does not change the spread. If you zoom a photo 2 times, every gap becomes 2 times bigger, and variance (a squared distance) becomes $2^{2} = 4$ times bigger.

Special cases from the slides: $Var [b] = 0$ (a constant has no spread) and $Var [X + b] = Var [X]$ .

Example: if $Var [X] = 4$ , then $Var [X + 10] = 4$ , $Var [3 X] = 9 \cdot 4 = 36$ , $Var [2 X - 5] = 4 \cdot 4 = 16$ .

6.6Ready-made E[X] and Var[X] for the six families

If the question names the family, do not build a table. Use these directly.

Family	$E [X]$	$Var [X]$
Bernoulli (p)	$p$	$p (1 - p)$
Binomial (n, p)	$n p$	$n p (1 - p)$
Geometric (p)	$\frac{1}{p}$	$\frac{1 - p}{p ^{2}}$
Pascal (k, p)	$\frac{k}{p}$	$\frac{k ( 1 - p )}{p ^{2}}$
Poisson (α)	$α$	$α$
Uniform (k, l)	$\frac{k + l}{2}$	$\frac{( l - k ) ( l - k + 2 )}{12}$

How to remember with less effort

Binomial = n Bernoullis. So multiply Bernoulli by n: $n p$ and $n p (1 - p)$ .
Pascal = k geometrics in a row. So multiply geometric by k: $k / p$ and $k (1 - p) / p^{2}$ .
Poisson: mean and variance are the same number $α$ .
Uniform mean is the midpoint: (smallest + largest) ÷ 2.

Using the table backwards

Sometimes the exam gives you $E [X]$ and asks for the PMF. Find the parameter from the table, then write the family formula. Example: geometric with $E [X] = 30$ means $\frac{1}{p} = 30$ , so $p = \frac{1}{30}$ .

Now solve these sample questions: Q14 · PMF from E[X]Q15 · CDF, PMF, standard deviation

Conditional PMF

Week 6

7.1The idea: a PMF after a hint

In simple words

This is lesson 2 (conditional probability) meeting lesson 4 (PMF). Someone gives a hint about X, for example "X is 5 or more". The values that do not fit the hint become impossible. The values that fit share all the probability between them.

Definition

P_{X ∣ B} (x) = P [X = x ∣ B]

Read: "the PMF of X given B". B is the hint (the "conditioning event").

The rule you actually use

P_{X ∣ B} (x) = ⎩ ⎨ ⎧ \frac{P _{X} ( x )}{P [ B ]} 0 x \in B otherwise

Keep the values inside B and divide each by $P [B]$ . Everything outside B becomes 0.

The 3-step recipe

List the values of X that fit the hint. That is B.
Find $P [B]$ : add the PMF of those values.
Divide the PMF of each kept value by $P [B]$ . Check: the new numbers add up to 1.

Warm-up: a die

Roll a fair die. Hint: "the number is even". Find the conditional PMF.

Show solution

$B = {2, 4, 6}$ .
$P [B] = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1}{2}$ .
Each kept value: $\frac{1/6}{1/2} = \frac{1}{3}$ . The values 1, 3, 5 get 0.

P_{X ∣ B} (x) = {1/3 0 x = 2, 4, 6 otherwise

Example from the slides: fax machines

The length X of a fax (in pages) has PMF

P_{X} (x) = ⎩ ⎨ ⎧ 0.15 0.1 0 x = 1, 2, 3, 4 x = 5, 6, 7, 8 otherwise

The company has two machines: one for faxes shorter than five pages, one for faxes with five or more pages. What is the PMF of the fax length in the second machine?

Show solution

The second machine gets the long faxes. The hint is $L = {X \geq 5} = {5, 6, 7, 8}$ .
$P [L] = 0.1 + 0.1 + 0.1 + 0.1 = 0.4$ .
Each kept value: $\frac{0.1}{0.4} = 0.25$ .

P_{X ∣ L} (x) = {0.25 0 x = 5, 6, 7, 8 otherwise

Check: $4 \times 0.25 = 1$ .

Example from the slides: waiting for a bus

X is the time in whole minutes you wait for a bus:

P_{X} (x) = {1/20 0 x = 1, 2, \dots, 20 otherwise

The bus has not arrived by the eighth minute. What is the conditional PMF of your waiting time X?

Show solution

"Not arrived by minute 8" means the wait is more than 8: $B = {X > 8} = {9, 10, \dots, 20}$ . That is 12 values.
$P [B] = 12 \times \frac{1}{20} = \frac{12}{20}$ .
Each kept value: $\frac{1/20}{12/20} = \frac{1}{12}$ .

P_{X ∣ X > 8} (x) = {1/12 0 x = 9, 10, \dots, 20 otherwise

A conditional PMF is still a PMF (theorem from the slides)

$P_{X ∣ B} (x) \geq 0$ for any x in B.
$x \in B \sum P_{X ∣ B} (x) = 1$ .
For any event C inside B: $P [C ∣ B] = x \in C \sum P_{X ∣ B} (x)$ .

7.2The other direction: mixing several PMFs

In simple words

Sometimes the question gives you one PMF for each situation (rainy day or sunny day, image page or text page). You want the one overall PMF. This is the law of total probability from lesson 2, written with PMFs.

Rule

P_{X} (x) = i = 1 \sum m P_{X ∣ B_{i}} (x) P [B_{i}]

For each situation $B_{i}$ : (PMF in that situation) × (probability of that situation). Then add.

Example from the slides: the 70-year-old

X = number of additional years a 70-year-old person lives. With high blood pressure (event H), X is geometric (p = 0.1). With regular blood pressure (event R), X is geometric (p = 0.05). Find $P_{X ∣ H} (x)$ and $P_{X ∣ R} (x)$ . If 40% of all 70-year-olds have high blood pressure, what is the PMF of X?

Show solution

Geometric formula: $p (1 - p)^{x - 1}$ for $x = 1, 2, \dots$ Put in each p:
$P_{X ∣ H} (x) = {0.1 (0.9)^{x - 1} 0 x = 1, 2, \dots otherwise$

$P_{X ∣ R} (x) = {0.05 (0.95)^{x - 1} 0 x = 1, 2, \dots otherwise$
$P [H] = 0.4$ , so $P [R] = 1 - 0.4 = 0.6$ .
Mix them: $P_{X} (x) = P_{X ∣ H} (x) P [H] + P_{X ∣ R} (x) P [R]$
$P_{X} (x) = 0.04 (0.9)^{x - 1} + 0.03 (0.95)^{x - 1}$
for $x = 1, 2, \dots$ , and 0 otherwise.
(because $0.4 \times 0.1 = 0.04$ and $0.6 \times 0.05 = 0.03$ )

7.3Conditional expected value and variance

Nothing new here. Do the same recipe as lesson 6, but with the conditional PMF in place of the PMF.

Rules

E [X ∣ B] = μ_{X ∣ B} = x \in B \sum x P_{X ∣ B} (x)

E [X^{2} ∣ B] = x \in B \sum x^{2} P_{X ∣ B} (x)

Var [X ∣ B] = E [X^{2} ∣ B] - (E [X ∣ B])^{2}

And to get the overall average from the averages in each situation:

E [X] = i = 1 \sum m E [X ∣ B_{i}] P [B_{i}]

Example from the slides: long faxes

For the fax PMF above, find the conditional expected value, the conditional variance and the conditional standard deviation for the long faxes.

Show solution

From before: $P_{X ∣ L} (x) = 0.25$ for $x = 5, 6, 7, 8$ .

$E [X ∣ L] = 0.25 (5 + 6 + 7 + 8) = 0.25 \times 26 = 6.5$ pages
$E [X^{2} ∣ L] = 0.25 (25 + 36 + 49 + 64) = 0.25 \times 174 = 43.5$
$Var [X ∣ L] = 43.5 - (6.5)^{2} = 43.5 - 42.25 = 1.25$
$σ_{X ∣ L} = 1.25 \approx 1.12$ pages

Answer6.5, 1.25, 1.12

Big example from the slides: web pages

N = number of packets needed to send a web page. If the page has images (event I), N is uniform between 1 and 50. If the page is just text (event T), N is uniform between 1 and 5. A page has images with probability 1/4. Find:

(1) $P_{N ∣ I} (n)$ (2) $P_{N ∣ T} (n)$ (3) $P_{N} (n)$ (4) $P_{N ∣ N \leq 10} (n)$ (5) $E [N ∣ N \leq 10]$ (6) $Var [N ∣ N \leq 10]$

Show solution

Uniform (1, 50) has 50 values:
$P_{N ∣ I} (n) = {1/50 0 n = 1, 2, \dots, 50 otherwise$
Uniform (1, 5) has 5 values:
$P_{N ∣ T} (n) = {1/5 0 n = 1, 2, \dots, 5 otherwise$
Mix them with $P [I] = \frac{1}{4}$ and $P [T] = \frac{3}{4}$ .
For $n = 1$ to 5, both situations are possible: $\frac{1}{50} \cdot \frac{1}{4} + \frac{1}{5} \cdot \frac{3}{4} = 0.005 + 0.15 = 0.155$
For $n = 6$ to 50, only images are possible: $\frac{1}{50} \cdot \frac{1}{4} = 0.005$
$P_{N} (n) = ⎩ ⎨ ⎧ 0.155 0.005 0 n = 1, 2, \dots, 5 n = 6, 7, \dots, 50 otherwise$
Check: $5 (0.155) + 45 (0.005) = 0.775 + 0.225 = 1$ .
The hint is $B = {N \leq 10}$ . $P [B] = 5 (0.155) + 5 (0.005) = 0.775 + 0.025 = 0.8$ .
Divide: $\frac{0.155}{0.8} = 0.19375$ and $\frac{0.005}{0.8} = 0.00625$ .
$P_{N ∣ N \leq 10} (n) = ⎩ ⎨ ⎧ 0.19375 0.00625 0 n = 1, \dots, 5 n = 6, \dots, 10 otherwise$
$E [N ∣ N \leq 10] = 0.19375 (1 + 2 + 3 + 4 + 5) + 0.00625 (6 + 7 + 8 + 9 + 10)$
$= 0.19375 (15) + 0.00625 (40) = 2.906 + 0.25 \approx 3.156$
$E [N^{2} ∣ N \leq 10] = 0.19375 (1 + 4 + 9 + 16 + 25) + 0.00625 (36 + 49 + 64 + 81 + 100)$
$= 0.19375 (55) + 0.00625 (330) = 10.656 + 2.0625 \approx 12.719$
$Var [N ∣ N \leq 10] = 12.719 - (3.156)^{2} \approx 2.76$

Answer(5) about 3.16 (6) about 2.76

Traps

Do not forget to divide by $P [B]$ . Without it the new PMF does not add to 1.
Read the hint carefully: " $X > 0$ ", " $X \geq 5$ ", " $X \geq μ_{X}$ " decide which values are kept.
For $Var [X ∣ B]$ , use the conditional PMF in both $E [X ∣ B]$ and $E [X^{2} ∣ B]$ .

Now solve these sample questions: Q15 · part (d)Q18 · binomial with a condition

All 19 sample questions, solved

Week 7 · same order as the professor's slides

How to practise

Cover the solution. Try alone first. Then open it and compare step by step.
Tap Mark done when everyone understands the question. 0 of 19 done
Each question has a tag that tells you which lesson it comes from.

Q1Sets · Lesson 1

An integrated circuit factory has three machines X, Y and Z. Test one circuit produced by each machine. Either a circuit is acceptable (a) or it fails (f). An observation is a sequence of three test results for the circuits from X, Y and Z, in that order. For example, aaf means the circuits from X and Y pass and the circuit from Z fails.

(a) What are the elements of the sample space?
(b) What are the elements of the sets $Z_{F}$ = {circuit from Z fails} and $X_{A}$ = {circuit from X is acceptable}?
(c) Are $Z_{F}$ and $X_{A}$ mutually exclusive?
(d) Are $Z_{F}$ and $X_{A}$ collectively exhaustive?
(e) What are the elements of the sets C = {more than one circuit acceptable} and D = {at least two circuits fail}?
(f) Are C and D mutually exclusive?
(g) Are C and D collectively exhaustive?

Key idea: the 1st letter is machine X, the 2nd is Y, the 3rd is Z. Mutually exclusive = share nothing. Collectively exhaustive = together they cover all of S.

Show solution

(a) Three positions, 2 choices each: $2^{3} = 8$ outcomes.
$S = {aaa, aa f, a f a, a f f, f aa, f a f, f f a, f f f}$
(b) $Z_{F}$ : the third letter is f. $X_{A}$ : the first letter is a.
$Z_{F} X_{A} = {aa f, a f f, f a f, f f f} = {aaa, aa f, a f a, a f f}$
(c) Do they share anything? $Z_{F} \cap X_{A} = {aa f, a f f}$ . This is not empty. No, not mutually exclusive.
(d) Do they cover all of S? $Z_{F} \cup X_{A} = {aaa, aa f, a f a, a f f, f a f, f f f}$ . The outcomes faa and ffa are missing. No, not collectively exhaustive.
(e) "More than one acceptable" = 2 or 3 a's. "At least two fail" = 2 or 3 f's.
$C D = {aaa, aa f, a f a, f aa} = {a f f, f a f, f f a, f f f}$
(f) $C \cap D = φ$ . They share nothing. Yes, mutually exclusive.
(g) $C \cup D$ has all 8 outcomes, so it equals S. Yes, collectively exhaustive.

Why (f) and (g) work: with 3 circuits, either at least 2 are acceptable or at least 2 fail. Never both, and always one of them.

Q2Equally likely · Lesson 1

Shuffle a deck of cards and turn over the first card. What is the probability that the first card is a heart?

Key idea: all 52 cards are equally likely. Count and divide.

Show solution

A deck has 52 cards in 4 suits (hearts, diamonds, clubs, spades). Each suit has 13 cards.

P [heart] = \frac{hearts}{all cards} = \frac{13}{52} = \frac{1}{4}

Answer

\frac{1}{4} = 0.25

Q3Table trick · Lesson 1

There are two types of cellular phones: handheld phones (H) and mobile phones (M). Phone calls can be classified by the speed of the user as fast (F) or slow (W). Monitor a call and observe the type of phone and the speed of the user. The probability model has: $P [F] = 0.5$ , $P [H F] = 0.2$ , $P [M W] = 0.1$ .

What is the sample space of the experiment? Calculate: (a) $P [W]$ (b) $P [M F]$ (c) $P [H]$

Key idea: two labels (phone type and speed), so draw a 2 × 2 table.

Show solution

Sample space. Each call has a type and a speed: $S = {H F, H W, M F, M W}$ .

Table. The bold numbers are given. Fill the others by subtracting.

	F (fast)	W (slow)	Total
H	0.2	0.4	0.6
M	0.3	0.1	0.4
Total	0.5	0.5	1

(a) A call is fast or slow, never both: $P [W] = 1 - P [F] = 1 - 0.5 = 0.5$
(b) Column F must add to 0.5: $P [M F] = P [F] - P [H F] = 0.5 - 0.2 = 0.3$
(c) Column W must add to 0.5: $P [H W] = P [W] - P [M W] = 0.5 - 0.1 = 0.4$ .
Row H: $P [H] = P [H F] + P [H W] = 0.2 + 0.4 = 0.6$

Answer(a) 0.5 (b) 0.3 (c) 0.6

Q4Disjoint and independent · Lessons 1, 2

In an experiment, A and B are events with probabilities $P [A \cup B] = \frac{3}{4}$ and $P [A] = \frac{1}{4}$ . Furthermore, A and B are disjoint.

Find $P [A \cap B]$ , $P [B]$ , $P [A \cap B^{c}]$ and $P [A \cup B^{c}]$ . Are A and B independent?

Key idea: "disjoint" means they share nothing, so $P [A \cap B] = 0$ . Draw two circles that do not touch.

Show solution

$P [A \cap B]$ . Disjoint means no overlap: $P [A \cap B] = 0$
$P [B]$ . For disjoint events $P [A \cup B] = P [A] + P [B]$ . So $\frac{3}{4} = \frac{1}{4} + P [B]$ , which gives $P [B] = \frac{1}{2}$
$P [A \cap B^{c}]$ means "in A and not in B". A is completely outside B, so all of A is "not in B". $P [A \cap B^{c}] = P [A] = \frac{1}{4}$
$P [A \cup B^{c}]$ means "in A, or not in B". A is already inside "not B", so adding A changes nothing: $A \cup B^{c} = B^{c}$ .
$P [A \cup B^{c}] = P [B^{c}] = 1 - \frac{1}{2} = \frac{1}{2}$
Independent? Test: is $P [A B] = P [A] P [B]$ ?
Left side: 0. Right side: $\frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}$ . Not equal. Not independent.

Answer

0, \frac{1}{2}, \frac{1}{4}, \frac{1}{2}

. A and B are not independent.

Remember: disjoint events (with probability above 0) are never independent.

Q5Counting · Lesson 3

Consider a language containing four letters: A, B, C, D. How many three-letter words can you form in this language? How many four-letter words can you form if each letter appears only once in each word?

Key idea: ask "can letters repeat?" In the first part yes. In the second part no.

Show solution

Three-letter words. Nothing says letters cannot repeat (AAB is a word). 3 positions, 4 choices each. This is $m^{n}$ with $m = 4$ , $n = 3$ :
$4 \cdot 4 \cdot 4 = 4^{3} = 64$
Four-letter words, each letter once. No repeats, order matters, and we use all 4 letters. First position: 4 choices, then 3, then 2, then 1:
$4 \cdot 3 \cdot 2 \cdot 1 = 4! = 24$

Answer64 three-letter words and 24 four-letter words

Q6Counting · Lesson 3

Consider a binary code with 5 bits (0 or 1) in each code word. An example of a code word is 01010. How many different code words are there? How many code words have exactly three 0's?

Key idea: "how many in total" is $m^{n}$ . "Exactly three 0's" means choosing positions, so use "n choose k".

Show solution

All code words. 5 positions, 2 choices each: $2^{5} = 32$
Exactly three 0's. Choose which 3 of the 5 positions hold the zeros. The other two are 1's automatically.
$(3 5) = \frac{5 !}{3 ! 2 !} = \frac{120}{6 \cdot 2} = 10$

Answer32 code words. 10 of them have exactly three 0's.

Q7Tree diagram · Lesson 2

A machine produces photo detectors in pairs. Tests show that the first photo detector is acceptable with probability 3/5. When the first photo detector is acceptable, the second photo detector is acceptable with probability 4/5. If the first photo detector is defective, the second photo detector is acceptable with probability 2/5. Use a tree diagram to find the following.

(a) What is the probability that exactly one photo detector of a pair is acceptable?
(b) What is the probability that both photo detectors in a pair are defective?

Key idea: branches from one point add to 1. Multiply along a path. Add the paths you want.

Show solution

Build the tree. Fill in the missing branches with "1 minus":

First detector: acceptable $\frac{3}{5}$ , so defective $1 - \frac{3}{5} = \frac{2}{5}$ .
After an acceptable first: second acceptable $\frac{4}{5}$ , so defective $\frac{1}{5}$ .
After a defective first: second acceptable $\frac{2}{5}$ , so defective $\frac{3}{5}$ .

A = acceptable, D = defective. 1 means first detector, 2 means second.

(a) Exactly one acceptable. Two paths (green in the tree): "first good, second bad" and "first bad, second good".
$P [A_{1} D_{2}] + P [D_{1} A_{2}] = \frac{3}{5} \cdot \frac{1}{5} + \frac{2}{5} \cdot \frac{2}{5} = \frac{3}{25} + \frac{4}{25} = \frac{7}{25}$
(b) Both defective. One path:
$P [D_{1} D_{2}] = \frac{2}{5} \cdot \frac{3}{5} = \frac{6}{25}$

Answer(a)

\frac{7}{25} = 0.28

(b)

\frac{6}{25} = 0.24

Check: the four paths add to $\frac{12 + 3 + 4 + 6}{25} = 1$ .

Q8Conditional probability · Lesson 2

You have a six-sided die that you roll once. Let $R_{i}$ denote the event that the roll is i. Let $G_{j}$ denote the event that the roll is greater than j. Let E denote the event that the roll of the die is even-numbered.

(a) What is $P [R_{3} ∣ G_{1}]$ , the conditional probability that 3 is rolled given that the roll is greater than 1?
(b) What is the conditional probability that 6 is rolled given that the roll is greater than 3?
(c) What is $P [G_{3} ∣ E]$ , the conditional probability that the roll is greater than 3 given that the roll is even?
(d) Given that the roll is greater than 3, what is the conditional probability that the roll is even?

Key idea: the hint (after the bar) becomes the new world. Count inside it.

Show solution

First translate the names into sets:

$R_{3} = {3}$ , $R_{6} = {6}$
$G_{1} = {2, 3, 4, 5, 6}$ (greater than 1), $G_{3} = {4, 5, 6}$ (greater than 3)
$E = {2, 4, 6}$

(a) New world $G_{1}$ has 5 outcomes. The roll 3 is one of them.
$P [R_{3} ∣ G_{1}] = \frac{P [ R _{3} G _{1} ]}{P [ G _{1} ]} = \frac{1/6}{5/6} = \frac{1}{5}$
(b) New world $G_{3} = {4, 5, 6}$ has 3 outcomes. The roll 6 is one of them.
$P [R_{6} ∣ G_{3}] = \frac{1/6}{3/6} = \frac{1}{3}$
(c) New world $E = {2, 4, 6}$ . Greater than 3 inside it: ${4, 6}$ , two outcomes.
$P [G_{3} ∣ E] = \frac{P [ G _{3} E ]}{P [ E ]} = \frac{2/6}{3/6} = \frac{2}{3}$
(d) Now the hint is $G_{3} = {4, 5, 6}$ . Even inside it: ${4, 6}$ , two outcomes.
$P [E ∣ G_{3}] = \frac{P [ E G _{3} ]}{P [ G _{3} ]} = \frac{2/6}{3/6} = \frac{2}{3}$

Answer(a)

\frac{1}{5}

(b)

\frac{1}{3}

(c)

\frac{2}{3}

(d)

\frac{2}{3}

(c) and (d) are equal here only because E and $G_{3}$ both have 3 outcomes. In general $P [A ∣ B]$ and $P [B ∣ A]$ are different.

Q9Total probability and Bayes · Lesson 2

A company has three machines $B_{1}$ , $B_{2}$ and $B_{3}$ for making 1 kΩ resistors. It has been observed that 80% of resistors produced by $B_{1}$ are within 50 Ω of the nominal value. Machine $B_{2}$ produces 90% of resistors within 50 Ω of the nominal value. The percentage for machine $B_{3}$ is 60%. Each hour, machine $B_{1}$ produces 3000 resistors, $B_{2}$ produces 4000 resistors, and $B_{3}$ produces 3000 resistors. All of the resistors are mixed together at random in one bin and packed for shipment.

What is the probability that the company ships a resistor that is within 50 Ω of the nominal value?
What is the probability that an acceptable resistor comes from machine $B_{3}$ ?

Key idea: first question = add all paths that end in "acceptable" (total probability). Second question = one path divided by all of them (Bayes).

Show solution

Let A = "the resistor is within 50 Ω" (acceptable).

Probability of each machine. Total per hour: $3000 + 4000 + 3000 = 10000$ .
$P [B_{1}] = 0.3$ , $P [B_{2}] = 0.4$ , $P [B_{3}] = 0.3$
Given the machine: $P [A ∣ B_{1}] = 0.8$ , $P [A ∣ B_{2}] = 0.9$ , $P [A ∣ B_{3}] = 0.6$ .

Multiply along each path. The numbers on the right add up to 1.

Law of total probability (add the green paths):
$P [A] = (0.8) (0.3) + (0.9) (0.4) + (0.6) (0.3) = 0.24 + 0.36 + 0.18 = 0.78$
Bayes (the $B_{3}$ path over all paths that end in A):
$P [B_{3} ∣ A] = \frac{P [ A ∣ B _{3} ] P [ B _{3} ]}{P [ A ]} = \frac{0.18}{0.78} = \frac{3}{13} \approx 0.231$

Answer

P [A] = 0.78

and

P [B_{3} ∣ A] = \frac{3}{13} \approx 0.231

Q10PMF rules · Lesson 4

The random variable V has PMF

P_{V} (v) = {c v^{2} 0 v = 1, 2, 3, 4 otherwise

(a) Find the value of the constant c.
(b) Find $P [V \in {u^{2} ∣ u = 1, 2, 3, \dots}]$ .
(c) Find the probability that V is an even number.
(d) Find $P [V > 2]$ .

Key idea: all probabilities add to 1 (this gives c). Then for each part, pick the values that fit and add their probabilities.

Show solution

(a) Put each value into $c v^{2}$ :

v 1 2 3 4

$P_{V} (v) = c v^{2}$ c 4c 9c 16c

They must add to 1: $c + 4 c + 9 c + 16 c = 30 c = 1$ , so $c = \frac{1}{30}$ .

v 1 2 3 4

$P_{V} (v)$ 1/30 4/30 9/30 16/30
(b) The strange set ${u^{2} ∣ u = 1, 2, 3, \dots}$ means "all perfect squares": $1, 4, 9, 16, \dots$ Which values of V are perfect squares? Only 1 and 4.
$P_{V} (1) + P_{V} (4) = \frac{1}{30} + \frac{16}{30} = \frac{17}{30}$
(c) Even values of V: 2 and 4.
$P_{V} (2) + P_{V} (4) = \frac{4}{30} + \frac{16}{30} = \frac{20}{30} = \frac{2}{3}$
(d) Values bigger than 2: 3 and 4.
$P_{V} (3) + P_{V} (4) = \frac{9}{30} + \frac{16}{30} = \frac{25}{30} = \frac{5}{6}$

v	1	2	3	4
$P_{V} (v) = c v^{2}$	c	4c	9c	16c

v	1	2	3	4
$P_{V} (v)$	1/30	4/30	9/30	16/30

Answer(a)

c = \frac{1}{30}

(b)

\frac{17}{30}

(c)

\frac{2}{3}

(d)

\frac{5}{6}

Q11Binomial · Lesson 4

When a conventional paging system transmits a message, the probability that the message will be received by the pager it is sent to is p. To be confident that a message is received at least once, a system transmits the message n times.

(a) Assuming all transmissions are independent, what is the PMF of K, the number of times the pager receives the same message?
(b) Assume p = 0.8. What is the minimum value of n that produces a probability of 0.95 of receiving the message at least once?

Key idea: a fixed number n of tries, and we count successes: binomial. "At least once" = 1 − P[zero times].

Show solution

(a) n independent tries. Each is a success (received) with probability p. K counts the successes. So K is binomial (n, p):
$P_{K} (k) = (k n) p^{k} (1 - p)^{n - k}$
for $k = 0, 1, \dots, n$ , and 0 otherwise.
(b) "At least once" means $K \geq 1$ . Its opposite is $K = 0$ (never received):
$P [K \geq 1] = 1 - P_{K} (0) = 1 - (1 - p)^{n} = 1 - (0.2)^{n}$
We need this to be at least 0.95. Try values of n:

n $(0.2)^{n}$ $1 - (0.2)^{n}$ ≥ 0.95?

1 0.2 0.80 No

2 0.04 0.96 Yes

n	$(0.2)^{n}$	$1 - (0.2)^{n}$	≥ 0.95?
1	0.2	0.80	No
2	0.04	0.96	Yes

Answer(a) binomial (n, p), formula above (b) minimum

n = 2

Where does $P_{K} (0) = (1 - p)^{n}$ come from? Put $k = 0$ in the formula: $(0 n) = 1$ and $p^{0} = 1$ , so only $(1 - p)^{n}$ is left. It means "fail n times in a row".

Q12CDF · Lessons 5, 6

The random variable X has CDF

F_{X} (x) = ⎩ ⎨ ⎧ 0 0.2 0.7 1 x < - 1 - 1 \leq x < 0 0 \leq x < 1 x \geq 1

(a) Draw a graph of the CDF.
(b) Write $P_{X} (x)$ , the PMF of X. Be sure to write the value of $P_{X} (x)$ for all x from −∞ to ∞.
(c) Find $P [X < 1]$ and $P [X = 1]$ .
(d) Find the expected value of the random variable X.

Key idea: the CDF jumps at −1, 0 and 1. Those are the values of X. The size of each jump is the PMF.

Show solution

(a) A staircase. Height 0 until −1, then 0.2 until 0, then 0.7 until 1, then 1 forever. Filled dot on the left of each step, empty dot on the right.
Answer for (a): a staircase with jumps at −1, 0 and 1.
(b) Jump sizes:
- at $x = - 1$ : $0.2 - 0 = 0.2$
- at $x = 0$ : $0.7 - 0.2 = 0.5$
- at $x = 1$ : $1 - 0.7 = 0.3$
$P_{X} (x) = ⎩ ⎨ ⎧ 0.2 0.5 0.3 0 x = - 1 x = 0 x = 1 otherwise$
"For all x from −∞ to ∞" just means: do not forget the line 0 otherwise. Check: $0.2 + 0.5 + 0.3 = 1$ .
(c) $P [X < 1]$ : values smaller than 1 are −1 and 0. $0.2 + 0.5 = 0.7$ . (Same as the height just before 1.)
$P [X = 1] = P_{X} (1) = 0.3$ . (The jump at 1.)
(d) Value × probability, then add:
$E [X] = (- 1) (0.2) + (0) (0.5) + (1) (0.3) = - 0.2 + 0 + 0.3 = 0.1$

Answer(c)

P [X < 1] = 0.7

P [X = 1] = 0.3

(d)

E [X] = 0.1

Q13"At least once" · Lesson 4

You manage a ticket agency that sells concert tickets. You assume that people will call three times in an attempt to buy tickets and then give up. Each time a caller tries, the probability that they get through to your agency is p (independent each time). You want to be able to serve at least 95% of the people who want tickets. What is the minimum value of p necessary to meet your goal?

Key idea: a person is served if they get through at least once in 3 tries. Use 1 − P[fail all three].

Show solution

One try fails with probability $1 - p$ . The tries are independent, so failing all 3 has probability $(1 - p)^{3}$ .
Served = not failing all three:
$P [served] = 1 - (1 - p)^{3}$
We need at least 0.95:
$1 - (1 - p)^{3} (1 - p)^{3} \geq 0.95 \leq 0.05$
Take the cube root of both sides (calculator: $0.0 5^{1/3}$ ):
$1 - p p \leq 3 0.05 \approx 0.368 \geq 1 - 0.368 = 0.632$

AnswerMinimum

p \approx 0.632

Check: with $p = 0.632$ , $(0.368)^{3} \approx 0.05$ , so 95% are served.

Q14Families backwards · Lessons 4, 6

Suppose that $E [X] = 30$ is the expected value of geometric random variable X. What is the PMF of X? (Write the equation)

Suppose that $E [X] = 5$ is the expected value of Poisson random variable X. What is the PMF of X? (Write the equation)

Key idea: use the $E [X]$ formula of the family to find its parameter. Then write the family's PMF with that number.

Show solution

Geometric. $E [X] = \frac{1}{p} = 30$ , so $p = \frac{1}{30}$ and $1 - p = \frac{29}{30}$ . Put it in $p (1 - p)^{x - 1}$ :
$P_{X} (x) = ⎩ ⎨ ⎧ \frac{1}{30} (\frac{29}{30})^{x - 1} 0 x = 1, 2, 3, \dots otherwise$
Poisson. $E [X] = α = 5$ . Put it in $\frac{α ^{x} e ^{- α}}{x !}$ :
$P_{X} (x) = ⎩ ⎨ ⎧ \frac{5 ^{x} e ^{- 5}}{x !} 0 x = 0, 1, 2, \dots otherwise$

To get full marks, always write the range of x (geometric starts at 1, Poisson starts at 0) and the "0 otherwise" line.

Q15CDF, variance, conditional PMF · Lessons 5, 6, 7

The random variable X has CDF

F_{X} (x) = ⎩ ⎨ ⎧ 0 0.4 0.8 1 x < - 3 - 3 \leq x < 5 5 \leq x < 7 x \geq 7

(a) Draw a graph of the CDF.
(b) Write $P_{X} (x)$ , the PMF of X.
(c) Find the standard deviation of the random variable X.
(d) Find $P_{X ∣ B} (x)$ , where the condition $B = {X > 0}$ . What are $E [X ∣ B]$ and $Var [X ∣ B]$ ?

Key idea: this one question uses three lessons. Jumps give the PMF. Then the 4-step recipe. Then the 3-step conditional recipe.

Show solution

(a) Staircase with jumps at −3, 5 and 7.
Answer for (a): jumps at −3, 5 and 7.
(b) Jump sizes: at −3: $0.4 - 0 = 0.4$ . At 5: $0.8 - 0.4 = 0.4$ . At 7: $1 - 0.8 = 0.2$ .
$P_{X} (x) = ⎩ ⎨ ⎧ 0.4 0.4 0.2 0 x = - 3 x = 5 x = 7 otherwise$
(c) The 4-step recipe.

x −3 5 7 Sum

$P_{X} (x)$ 0.4 0.4 0.2 1

$x \cdot P_{X} (x)$ −1.2 2 1.4 2.2

$x^{2} \cdot P_{X} (x)$ 9(0.4) = 3.6 25(0.4) = 10 49(0.2) = 9.8 23.4

$E [X] = 2.2$ and $E [X^{2}] = 23.4$ .
$Var [X] = E [X^{2}] - (E [X])^{2} = 23.4 - (2.2)^{2} = 23.4 - 4.84 = 18.56$
$σ_{X} = 18.56 \approx 4.31$
(d) The 3-step conditional recipe.
Values that fit $X > 0$ : 5 and 7. (−3 is thrown away.)
$P [B] = 0.4 + 0.2 = 0.6$
Divide: $\frac{0.4}{0.6} = \frac{2}{3}$ and $\frac{0.2}{0.6} = \frac{1}{3}$
$P_{X ∣ B} (x) = ⎩ ⎨ ⎧ 2/3 1/3 0 x = 5 x = 7 otherwise$
Now the recipe again with the new PMF:
$E [X ∣ B] = 5 \cdot \frac{2}{3} + 7 \cdot \frac{1}{3} = \frac{10}{3} + \frac{7}{3} = \frac{17}{3} \approx 5.67$

$E [X^{2} ∣ B] = 25 \cdot \frac{2}{3} + 49 \cdot \frac{1}{3} = \frac{50}{3} + \frac{49}{3} = \frac{99}{3} = 33$

$Var [X ∣ B] = 33 - (\frac{17}{3})^{2} = \frac{297}{9} - \frac{289}{9} = \frac{8}{9} \approx 0.889$

x	−3	5	7	Sum
$P_{X} (x)$	0.4	0.4	0.2	1
$x \cdot P_{X} (x)$	−1.2	2	1.4	2.2
$x^{2} \cdot P_{X} (x)$	9(0.4) = 3.6	25(0.4) = 10	49(0.2) = 9.8	23.4

Answer(c)

σ_{X} \approx 4.31

(d)

E [X ∣ B] = \frac{17}{3} \approx 5.67

Var [X ∣ B] = \frac{8}{9} \approx 0.889

Q16Discrete uniform · Lesson 4

In a package of M&Ms, Y, the number of yellow M&Ms, is uniformly distributed between 5 and 15.

(a) What is the PMF of Y?
(b) What is $P [Y < 10]$ ?
(c) What is $P [Y > 12]$ ?
(d) What is $P [8 \leq Y \leq 12]$ ?

Key idea: uniform = every value has the same probability. So each part is "count the values that fit, multiply by that probability".

Show solution

(a) Discrete uniform with $k = 5$ , $l = 15$ . Number of values: $l - k + 1 = 15 - 5 + 1 = 11$ . Each has probability $\frac{1}{11}$ .
$P_{Y} (y) = {1/11 0 y = 5, 6, \dots, 15 otherwise$
(b) $Y < 10$ : the values 5, 6, 7, 8, 9. That is 5 values. $P [Y < 10] = \frac{5}{11}$
(c) $Y > 12$ : the values 13, 14, 15. That is 3 values. $P [Y > 12] = \frac{3}{11}$
(d) $8 \leq Y \leq 12$ : the values 8, 9, 10, 11, 12. That is 5 values. $P [8 \leq Y \leq 12] = \frac{5}{11}$

Answer(b)

\frac{5}{11}

(c)

\frac{3}{11}

(d)

\frac{5}{11}

Trap: from 5 to 15 there are 11 values, not 10. Count on your fingers if you need to. Also watch < against ≤: "less than 10" does not include 10.

Q17CDF from a graph · Lesson 5

Discrete random variable Y has the CDF $F_{Y} (y)$ as shown:

The graph from the slide: jumps at y = 1, 2 and 3.

Use the CDF to find the following probabilities:
(a) $P [Y < 1]$
(b) $P [Y \leq 1]$
(c) $P [Y > 2]$
(d) $P [Y \geq 2]$
(e) $P [Y = 1]$
(f) $P [Y = 3]$
(g) $P_{Y} (y)$

Key idea: do part (g) first. Read the jumps to get the PMF. Then every other part is easy.

Show solution

Read the jumps (this is part g).
- at $y = 1$ : from 0 up to 0.25. Jump = 0.25
- at $y = 2$ : from 0.25 up to 0.5. Jump = 0.25
- at $y = 3$ : from 0.5 up to 1. Jump = 0.5
$P_{Y} (y) = ⎩ ⎨ ⎧ 0.25 0.25 0.5 0 y = 1 y = 2 y = 3 otherwise$
(a) $P [Y < 1]$ : no value of Y is below 1. Answer: 0. (Height just before 1.)
(b) $P [Y \leq 1]$ : only $y = 1$ fits. Answer: 0.25. (Height at 1.)
(c) $P [Y > 2]$ : only $y = 3$ fits. Answer: 0.5. ( $1 - F_{Y} (2) = 1 - 0.5$ .)
(d) $P [Y \geq 2]$ : $y = 2$ and $y = 3$ fit. $0.25 + 0.5 =$ 0.75. (1 minus the height just before 2.)
(e) $P [Y = 1]$ = jump at 1 = 0.25
(f) $P [Y = 3]$ = jump at 3 = 0.5

Answer(a) 0 (b) 0.25 (c) 0.5 (d) 0.75 (e) 0.25 (f) 0.5

Q18Binomial with a condition · Lessons 4, 6, 7

The binomial random variable X has PMF

P_{X} (x) = (x 5) (\frac{1}{2})^{5}

1. Find the standard deviation of X.
2. Find $P [μ_{X} - σ_{X} \leq X \leq μ_{X} + σ_{X}]$ , the probability that X is within one standard deviation of the expected value.
3. Find $P_{X ∣ B} (x)$ , where the condition $B = {X \geq μ_{X}}$ .
4. What are $E [X ∣ B]$ and $Var [X ∣ B]$ ?

Key idea: recognise binomial with $n = 5$ and $p = \frac{1}{2}$ . Make a small table of the PMF first. It answers everything.

Show solution

Why $n = 5$ , $p = \frac{1}{2}$ ? The binomial formula is $(x n) p^{x} (1 - p)^{n - x}$ . With $p = \frac{1}{2}$ : $(\frac{1}{2})^{x} (\frac{1}{2})^{5 - x} = (\frac{1}{2})^{5} = \frac{1}{32}$ . So $P_{X} (x) = (x 5) \cdot \frac{1}{32}$ .

x	0	1	2	3	4	5
$(x 5)$	1	5	10	10	5	1
$P_{X} (x)$	1/32	5/32	10/32	10/32	5/32	1/32

Part 1. Use the ready-made binomial formulas:
$μ_{X} = n p = 5 \cdot \frac{1}{2} = 2.5$

$Var [X] = n p (1 - p) = 5 \cdot \frac{1}{2} \cdot \frac{1}{2} = 1.25$

$σ_{X} = 1.25 \approx 1.118$
Part 2. The interval is from $2.5 - 1.118 = 1.382$ to $2.5 + 1.118 = 3.618$ . Which whole numbers are inside? Only 2 and 3.
$P [1.382 \leq X \leq 3.618] = P_{X} (2) + P_{X} (3) = \frac{10}{32} + \frac{10}{32} = \frac{20}{32} = \frac{5}{8}$
Part 3. The hint is $X \geq 2.5$ . The values that fit: 3, 4, 5.
$P [B] = \frac{10}{32} + \frac{5}{32} + \frac{1}{32} = \frac{16}{32} = \frac{1}{2}$
Divide each by $\frac{1}{2}$ (same as multiplying by 2):
$P_{X ∣ B} (x) = ⎩ ⎨ ⎧ 10/16 5/16 1/16 0 x = 3 x = 4 x = 5 otherwise$
Part 4. The recipe with the new PMF:
$E [X ∣ B] = 3 \cdot \frac{10}{16} + 4 \cdot \frac{5}{16} + 5 \cdot \frac{1}{16} = \frac{30 + 20 + 5}{16} = \frac{55}{16} \approx 3.44$

$E [X^{2} ∣ B] = 9 \cdot \frac{10}{16} + 16 \cdot \frac{5}{16} + 25 \cdot \frac{1}{16} = \frac{90 + 80 + 25}{16} = \frac{195}{16} \approx 12.19$

$Var [X ∣ B] = \frac{195}{16} - (\frac{55}{16})^{2} = \frac{3120 - 3025}{256} = \frac{95}{256} \approx 0.371$

Answer1)

σ_{X} \approx 1.118

\frac{5}{8}

3) see PMF above 4)

E [X ∣ B] = \frac{55}{16} \approx 3.44

Var [X ∣ B] = \frac{95}{256} \approx 0.371

Q19Geometric, binomial, Pascal · Lesson 4

Each time a modem transmits one bit, the receiving modem analyzes the signal that arrives and decides whether the transmitted bit is 0 or 1. It makes an error with probability p, independent of whether any other bit is received correctly.

(1) If the transmission continues until the receiving modem makes its first error, what is the PMF of X, the number of bits transmitted?
(2) If p = 0.1, what is the probability that X = 10?
(3) If the modem transmits 100 bits, what is the PMF of Y, the number of errors?
(4) If p = 0.01 and the modem transmits 100 bits, what is the probability of Y = 2 errors at the receiver? What is the probability that Y ≤ 2?
(5) If the transmission continues until the receiving modem makes three errors, what is the PMF of Z, the number of bits transmitted?
(6) If p = 0.25, what is the probability of Z = 12 bits transmitted?

Key idea: here the "success" is an error. Read the words: "until first error" is geometric, "100 bits, number of errors" is binomial, "until three errors" is Pascal.

Show solution

(1) "Until the first error", counting bits: geometric (p).
$P_{X} (x) = {p (1 - p)^{x - 1} 0 x = 1, 2, 3, \dots otherwise$
(2) $p = 0.1$ , $x = 10$ : nine correct bits, then one error.
$P_{X} (10) = 0.1 (0.9)^{9} \approx 0.0387$
(3) A fixed number of bits (100), counting errors: binomial (100, p).
$P_{Y} (y) = (y 100) p^{y} (1 - p)^{100 - y}$
for $y = 0, 1, \dots, 100$ , and 0 otherwise.
(4) $p = 0.01$ .
$P_{Y} (2) = (2 100) (0.01)^{2} (0.99)^{98} = 4950 \times 0.0001 \times 0.3735 \approx 0.185$
" $Y \leq 2$ " means 0, 1 or 2 errors. Find each one and add:
$P_{Y} (0) = (0.99)^{100} \approx 0.366$

$P_{Y} (1) = (1 100) (0.01)^{1} (0.99)^{99} = 100 \times 0.01 \times 0.3697 \approx 0.370$

$P [Y \leq 2] = 0.366 + 0.370 + 0.185 \approx 0.921$
(5) "Until three errors", counting bits: Pascal (k = 3, p).
$P_{Z} (z) = (2 z - 1) p^{3} (1 - p)^{z - 3}$
for $z = 3, 4, 5, \dots$ , and 0 otherwise.
(6) $p = 0.25$ , $z = 12$ :
$P_{Z} (12) = (2 11) (0.25)^{3} (0.75)^{9} = 55 \times 0.015625 \times 0.0751 \approx 0.0645$

Answer(2) 0.0387 (4) 0.185 and 0.921 (6) 0.0645

Calculator help: $(2 100) = \frac{100 \cdot 99}{2} = 4950$ and $(2 11) = \frac{11 \cdot 10}{2} = 55$ . Use the $x^{y}$ button for powers like $0.9 9^{98}$ .

One-page formula sheet and traps

Read this together at the end, and again tonight

Events

P [A^{c}] = 1 - P [A]

P [A \cup B] = P [A] + P [B] - P [A B]

P [A ∣ B] = \frac{P [ A B ]}{P [ B ]}

P [A B] = P [A ∣ B] P [B]

P [A] = i \sum P [A ∣ B_{i}] P [B_{i}]

(total probability)

P [B ∣ A] = \frac{P [ A ∣ B ] P [ B ]}{P [ A ]}

(Bayes)

Independent: P [A B] = P [A] P [B]

Disjoint: P [A B] = 0

Counting

Repeat?	Order matters?	Formula
Yes	Yes	$m^{n}$
No	Yes	$\frac{n !}{( n - k )!}$
No	No	$(k n) = \frac{n !}{k ! ( n - k )!}$
Groups of identical items		$\frac{n !}{n _{0} ! n _{1} ! \dots}$

Random variables

P_{X} (x) = P [X = x] x \sum P_{X} (x) = 1

P [B] = x \in B \sum P_{X} (x)

F_{X} (x) = P [X \leq x]

P_{X} (x_{i}) = jump of F_{X} at x_{i}

E [X] = x \sum x P_{X} (x)

E [X^{2}] = x \sum x^{2} P_{X} (x)

Var [X] = E [X^{2}] - (E [X])^{2}

σ_{X} = Var [X]

Var [a X + b] = a^{2} Var [X]

P_{X ∣ B} (x) = \frac{P _{X} ( x )}{P [ B ]} for x \in B, 0 otherwise

P_{X} (x) = i \sum P_{X ∣ B_{i}} (x) P [B_{i}]

E [X] = i \sum E [X ∣ B_{i}] P [B_{i}]

Var [X ∣ B] = E [X^{2} ∣ B] - (E [X ∣ B])^{2}

The six families in one table

Family	Story	PMF	x values	$E [X]$	$Var [X]$
Bernoulli (p)	One try	$1 - p, p$	0, 1	$p$	$p (1 - p)$
Binomial (n, p)	Successes in n tries	$(x n) p^{x} (1 - p)^{n - x}$	0, ..., n	$n p$	$n p (1 - p)$
Geometric (p)	Tries until 1st success	$p (1 - p)^{x - 1}$	1, 2, ...	$\frac{1}{p}$	$\frac{1 - p}{p ^{2}}$
Pascal (k, p)	Tries until k-th success	$(k - 1 x - 1) p^{k} (1 - p)^{x - k}$	k, k+1, ...	$\frac{k}{p}$	$\frac{k ( 1 - p )}{p ^{2}}$
Uniform (k, l)	All equally likely	$\frac{1}{l - k + 1}$	k, ..., l	$\frac{k + l}{2}$	$\frac{( l - k ) ( l - k + 2 )}{12}$
Poisson (α)	Arrivals in time T, $α = λ T$	$\frac{α ^{x} e ^{- α}}{x !}$	0, 1, 2, ...	$α$	$α$

Swipe the table sideways to see all columns.

What does the question want? Pick the tool

The question says	Do this
"Elements of the sample space / of the set"	List the outcomes one by one
"Mutually exclusive?"	Check if they share any outcome
"Collectively exhaustive?"	Check if together they give all of S
Two labels for each outcome (H/M and F/W)	Draw the 2 × 2 table
"Given that", "if we know that"	Conditional: divide by the probability of the hint
Things happen in steps, "use a tree diagram"	Tree: multiply along, add paths
"Comes from machine ...", reversed condition	Bayes: one path ÷ all paths
"Independent?"	Test $P [A B] = P [A] P [B]$
"How many words / code words"	Counting table: repeat? order?
"Find the constant c"	Sum of the PMF = 1
"At least one"	1 − P[none]
"Minimum n" or "minimum p"	Write 1 − P[none] ≥ target, then solve or try values
A CDF is given	Jumps give the PMF. Then answer from the PMF
"Standard deviation"	$E [X]$ , $E [X^{2}]$ , subtract, square root
"Where the condition B = {...}"	Keep values in B, divide by $P [B]$
"E[X] = ... , what is the PMF?"	Find the parameter from the E formula, write the family PMF

The 14 traps that cost marks

Disjoint is not independent. Disjoint: $P [A B] = 0$ . Independent: $P [A B] = P [A] P [B]$ .
$P [A ∣ B]$ is not $P [B ∣ A]$ . Divide by the probability of the hint.
$P [A B]$ is "and" (overlap). $P [A \cup B]$ is "or" (subtract the overlap once).
In a tree, the branches from one point must add to 1. Fill the missing branch with "1 minus".
"Exactly k zeros" is $(k n)$ , not $m^{n}$ .
Always finish a PMF with "0 otherwise" and write which x values it is for.
Geometric and Pascal start at $x = 1$ and $x = k$ . Binomial and Poisson start at 0.
Poisson: new time interval means new $α = λ T$ .
From k to l there are $l - k + 1$ values. From 5 to 15 that is 11.
$<$ and $\leq$ give different answers for discrete variables. $P [X \geq 2] = 1 - P [X \leq 1]$ .
$E [X^{2}]$ is not $(E [X])^{2}$ . Square the values first, then average.
Negative numbers squared are positive. Variance is never negative.
Standard deviation is the square root of the variance. Do not stop at the variance.
Conditional PMF: divide by $P [B]$ , then check the new numbers add to 1.

In the exam room

Read the question twice. Underline the words that pick the tool: "given", "until", "at least", "exactly", "uniformly", "per second".
Write the formula first, then put in the numbers. You get marks for the method even if the arithmetic slips.
Check with a rule: probabilities are between 0 and 1, a PMF adds to 1, a variance is not negative.
Stuck on a CDF or a condition? Write the PMF as a small table. Almost every question becomes "add the right cells".
Do the questions you know first. Come back to the hard ones.

Last word

If you can solve the 19 sample questions alone, with the solutions closed, you are ready. Good luck to all of you.

Outcome	Prob.	X (voice)	Y (data)	R = XY
ddd	1/8	0	3	0
ddv	1/8	1	2	2
dvd	1/8	1	2	2
dvv	1/8	2	1	2
vdd	1/8	1	2	2
vdv	1/8	2	1	2
vvd	1/8	2	1	2
vvv	1/8	3	0	0

Outcome	Prob.	X (voice)	Y (data)	R = XY
ddd	1/8	0	3	0
ddv	1/8	1	2	2
dvd	1/8	1	2	2
dvv	1/8	2	1	2
vdd	1/8	1	2	2
vdv	1/8	2	1	2
vvd	1/8	2	1	2
vvv	1/8	3	0	0

Probability & StatisticsExam Pass Guide

How to use this page

Plan for the 4 hours (12:00 to 4:00)

Symbols dictionary

Sets and events

Probability

Random variables

Sets and basic probability

1.1Four words you must know

1.2The four set operations

1.3Two exam words: mutually exclusive, collectively exhaustive

1.4The rules of probability

1.5The table trick (for questions with two labels)

Conditional probability, Bayes, trees

2.1Conditional probability: "given that"

2.2The multiplication rule

2.3Tree diagrams

2.4Law of total probability

2.5Bayes' theorem: turning the question around

2.6Independent events

Counting and independent trials

3.1The basic idea: multiply the choices

3.2The two questions that pick the formula

3.3Examples from the slides

3.4Independent trials

Random variables and PMF

4.1What is a random variable?

4.2The PMF

4.3The three PMF rules

4.4The six famous families

1. Bernoulli (p)

2. Binomial (n, p)

3. Geometric (p)

4. Pascal (k, p)

5. Discrete uniform (k, l)

6. Poisson (α)

4.5Which family is it? Look for these words

4.6Poisson examples from the slides

CDF: the running total

5.1What is the CDF?

5.2From PMF to CDF

5.3From CDF back to PMF

5.4Reading probabilities from a CDF

Expected value and variance

6.1Three kinds of average for a list of numbers

6.2Expected value E[X]

6.3Variance and standard deviation

6.4Moments (just names)

6.5Variance of aX + b

6.6Ready-made E[X] and Var[X] for the six families

Conditional PMF

7.1The idea: a PMF after a hint

7.2The other direction: mixing several PMFs

7.3Conditional expected value and variance

All 19 sample questions, solved

How to practise

One-page formula sheet and traps

Events

Counting

Random variables

The six families in one table

What does the question want? Pick the tool

The 14 traps that cost marks

In the exam room

Probability & Statistics
Exam Pass Guide

Outcome	Prob.	X (voice)	Y (data)	R = XY
ddd	1/8	0	3	0
ddv	1/8	1	2	2
dvd	1/8	1	2	2
dvv	1/8	2	1	2
vdd	1/8	1	2	2
vdv	1/8	2	1	2
vvd	1/8	2	1	2
vvv	1/8	3	0	0