Probability Reading: EC 6.16.3 Peter J. Haas INFO 150 Fall - - PowerPoint PPT Presentation

Probability

Reading: EC 6.1–6.3 Peter J. Haas INFO 150 Fall Semester 2019

Lecture 14 1/ 19

Probability Introduction Definition of Probability Rules for Computing Probabilities Conditional Probability Bernoulli Trials

Lecture 14 2/ 19

Introduction

Probability: The rules of chance I Most things are uncertain! (Origins of theory circa 1650, in gambling) I Foundational to machine learning, game theory, statistical analysis, ...

Lecture 14 3/ 19

Introduction

Probability: The rules of chance I Most things are uncertain! (Origins of theory circa 1650, in gambling) I Foundational to machine learning, game theory, statistical analysis, ... Goals I Compute the probability of a complex event from probabilities of simpler events

Lecture 14 3/ 19

Introduction

Probability: The rules of chance I Most things are uncertain! (Origins of theory circa 1650, in gambling) I Foundational to machine learning, game theory, statistical analysis, ... Goals I Compute the probability of a complex event from probabilities of simpler events I Conquer your terrible intuition about uncertainty (see Kahneman and Tversky)

Lecture 14 3/ 19

Probability Can be Non-Intuitive

Which is more likely if you pick a word at random from a dictionary: a ”k” in the first position or a ”k” in the third position?

Lecture 14 4/ 19

Probability Can be Non-Intuitive

Which is more likely if you pick a word at random from a dictionary: a ”k” in the first position or a ”k” in the third position? How likely is it that two people in this room share the same birthday?

Lecture 14 4/ 19

Probability Can be Non-Intuitive

Which is more likely if you pick a word at random from a dictionary: a ”k” in the first position or a ”k” in the third position? How likely is it that two people in this room share the same birthday? ”Linda 31 years old, single, outspoken, and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations”. Rank the following statements in order of relative likeihood:

• 1. ”Linda is a bank teller”
• 2. ”Linda is on the executive board of UNESCO”
• 3. ”Linda is a bank teller and an active feminist”

Lecture 14 4/ 19

Probability Can be Non-Intuitive

Which is more likely if you pick a word at random from a dictionary: a ”k” in the first position or a ”k” in the third position? How likely is it that two people in this room share the same birthday? ”Linda 31 years old, single, outspoken, and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations”. Rank the following statements in order of relative likeihood:

• 1. ”Linda is a bank teller”
• 2. ”Linda is on the executive board of UNESCO”
• 3. ”Linda is a bank teller and an active feminist”

Which sequence of coin flips is more likely: HHTHTT or HHHHHH?

Lecture 14 4/ 19

Probability Can be Non-Intuitive

Which is more likely if you pick a word at random from a dictionary: a ”k” in the first position or a ”k” in the third position? How likely is it that two people in this room share the same birthday? ”Linda 31 years old, single, outspoken, and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations”. Rank the following statements in order of relative likeihood:

• 1. ”Linda is a bank teller”
• 2. ”Linda is on the executive board of UNESCO”
• 3. ”Linda is a bank teller and an active feminist”

Which sequence of coin flips is more likely: HHTHTT or HHHHHH? We will learn some mathematical tools for getting the answers right

Lecture 14 4/ 19

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in

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• f

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79.5%

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has to

be

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3

equally likely

Basic Terminology

Setting: The outcomes of an experiment I Sample space S: The set of possible outcomes I Initially we’ll focus on experiments where outcomes are equally likely

Lecture 14 5/ 19

Basic Terminology

Setting: The outcomes of an experiment I Sample space S: The set of possible outcomes I Initially we’ll focus on experiments where outcomes are equally likely Example: Roll two dice and add up the numbers I Attempt 1: sample space is S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} (Demo) I Attempt 2: Sample space is S = { double 1, double 2, double 3, double 4, double 5, double 6 {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {3, 4}, {3, 5}, {3, 6}, {4, 5}, {4, 6}, {5, 6}} (Demo) I Attempt 3: ordered pair representation (see table)

Lecture 14 5/ 19 Green 1 Green 2 Green 3 Green 4 Green 5 Green 6 Red 1 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) Red 2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) Red 3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) Red 4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) Red 5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) Red 6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Basic Terminology

Setting: The outcomes of an experiment I Sample space S: The set of possible outcomes I Initially we’ll focus on experiments where outcomes are equally likely Example: Roll two dice and add up the numbers I Attempt 1: sample space is S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} (Demo) I Attempt 2: Sample space is S = { double 1, double 2, double 3, double 4, double 5, double 6 {1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {2, 3}, {2, 4}, {2, 5}, {2, 6}, {3, 4}, {3, 5}, {3, 6}, {4, 5}, {4, 6}, {5, 6}} (Demo) I Attempt 3: ordered pair representation (see table) We are interested in proportion of outcomes in which some event occurs I Ex: Sum on two dice equals 10 (proportion of outcomes is 3/36) I An outcome is successful if the event occurs I Technically, an event is a specified set of E outcomes with E ✓ S

Lecture 14 5/ 19 Green 1 Green 2 Green 3 Green 4 Green 5 Green 6 Red 1 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) Red 2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) Red 3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) Red 4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) Red 5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) Red 6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

Definition of Probability (Equally Likely Outcomes)

Definition Given an experiment with a sample space S of equally likely outcomes and an event E, the probability of the event, denoted Prob(E) is the ratio of the number of successful

• utcomes to the total number of outcomes:

Prob(E) = n(E) n(S)

Lecture 14 6/ 19

Definition of Probability (Equally Likely Outcomes)

Definition Given an experiment with a sample space S of equally likely outcomes and an event E, the probability of the event, denoted Prob(E) is the ratio of the number of successful

• utcomes to the total number of outcomes:

Prob(E) = n(E) n(S) Example 1: E = “sum of two dice equals 10” I E = {(4, 6), (5, 5), (6, 4)}, so Prob(E) =

3 36 = 1 12 ⇡ 0.083 Lecture 14 6/ 19

Definition of Probability (Equally Likely Outcomes)

Definition Given an experiment with a sample space S of equally likely outcomes and an event E, the probability of the event, denoted Prob(E) is the ratio of the number of successful

• utcomes to the total number of outcomes:

Prob(E) = n(E) n(S) Example 1: E = “sum of two dice equals 10” I E = {(4, 6), (5, 5), (6, 4)}, so Prob(E) =

3 36 = 1 12 ⇡ 0.083

Example 2: E = “two cards drawn randomly from a 52-card deck have same value” I S = {(AS, 2H), (3C, KD), . . .}, so n(S) = 52 · 51 I E = {(AS, AC), (3D, 3S), . . .}, so n(S) = 52 · 3 I Hence Prob(E) =

52·3 52·51 = 1 17 Lecture 14 6/ 19

Definition of Probability (Equally Likely Outcomes)

Definition Given an experiment with a sample space S of equally likely outcomes and an event E, the probability of the event, denoted Prob(E) is the ratio of the number of successful

• utcomes to the total number of outcomes:

Prob(E) = n(E) n(S) Example 1: E = “sum of two dice equals 10” I E = {(4, 6), (5, 5), (6, 4)}, so Prob(E) =

3 36 = 1 12 ⇡ 0.083

Example 2: E = “two cards drawn randomly from a 52-card deck have same value” I S = {(AS, 2H), (3C, KD), . . .}, so n(S) = 52 · 51 I E = {(AS, AC), (3D, 3S), . . .}, so n(S) = 52 · 3 I Hence Prob(E) =

52·3 52·51 = 1 17

Example 3: E = “last two tosses have same value when tossing 5 coins” I S = n(S) = I E = n(E) = I Prob(E) =

Lecture 14 6/ 19

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25--32

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HHORTT

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Probability of the Complement of an Event

Definition I The Complement ¯ E of an event E is the set of outcomes not in E: ¯ E = S E. Complement Rule I For any event E, Prob(E) + Prob( ¯ E) = 1.

Lecture 14 7/ 19

Probability of the Complement of an Event

Definition I The Complement ¯ E of an event E is the set of outcomes not in E: ¯ E = S E. Complement Rule I For any event E, Prob(E) + Prob( ¯ E) = 1. Example 1: What is probability that same result occurs more than once in 3 die rolls? I E = {(1, 6, 1), (2, 2, 4), (3, 3, 3), . . .} I ¯ E = set of rolls that are all different I n( ¯ E) = P(6, 3) = 120 and Prob( ¯ E) = 120

63 = 120 216

I Hence Prob(E) = 1 Prob( ¯ E) = 1 120

216 ⇡ 0.44 Lecture 14 7/ 19

Probability of the Complement of an Event

Definition I The Complement ¯ E of an event E is the set of outcomes not in E: ¯ E = S E. Complement Rule I For any event E, Prob(E) + Prob( ¯ E) = 1. Example 1: What is probability that same result occurs more than once in 3 die rolls? I E = {(1, 6, 1), (2, 2, 4), (3, 3, 3), . . .} I ¯ E = set of rolls that are all different I n( ¯ E) = P(6, 3) = 120 and Prob( ¯ E) = 120

63 = 120 216

I Hence Prob(E) = 1 Prob( ¯ E) = 1 120

216 ⇡ 0.44

Example 2: What is probability that, in a group of 6, at least 2 people were born in same month? I E = {(1, 2, 8, 8, 8, 3), (3, 1, 4, 4, 5, 5), (2, 9, 10, 3, 4, 10) . . .} I ¯ E = set of distinct birthday-month assignments I n( ¯ E) = P(12, 6) I Hence Prob(E) = 1 Prob( ¯ E) = 1 P(12,6)

126

= 1

665,280 2,985,984 ⇡ 0.78 Lecture 14 7/ 19

The Birthday Problem

What is the probability that in a room of n people, at least two share the same birthday? ◮ Assume 365 days in a year (ignore leap years) ◮ Assume that every birth day is equally likely (ignore birth seasonality) ◮ Compute the complement of the event Prob at least two identical birthdays = 1 − Prob all birthdays different = 1 − P(365, n) 365n

Lecture 14

Disjoint Events

Definition Two events E1 and E2 are disjoint (or mutually exclusive) if they cannot occur simultaneously in the experiment. Formally, E1 \ E2 = ;.

Lecture 14 8/ 19

Disjoint Events

Definition Two events E1 and E2 are disjoint (or mutually exclusive) if they cannot occur simultaneously in the experiment. Formally, E1 \ E2 = ;. Examples I Dice roll: E1 = “getting a 3” and E2 = “getting a 4” [disjoint] I E1 = “die is even” and E2 = “die is odd” [disjoint] I E1 = “3 on first roll” and E2 = “4 on second roll” [not disjoint, e.g., (3, 4)]

Lecture 14 8/ 19

Disjoint Events

Definition Two events E1 and E2 are disjoint (or mutually exclusive) if they cannot occur simultaneously in the experiment. Formally, E1 \ E2 = ;. Examples I Dice roll: E1 = “getting a 3” and E2 = “getting a 4” [disjoint] I E1 = “die is even” and E2 = “die is odd” [disjoint] I E1 = “3 on first roll” and E2 = “4 on second roll” [not disjoint, e.g., (3, 4)] Problem: Which of these of these pairs of events are disjoint? I In 4 coin tosses, E1 = “exactly 3 heads” and E2 = “exactly 2 tails” I When choosing 4 cards, E1 = “cards have same value” and E2 = “cards have same suit” I When choosing a committee of 3 from 8 men and 12 women, E1 = “committee has a woman” and E2 = “committee has a man”

Lecture 14 8/ 19

Disjoint

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The Sum Rule for Probability

Sum Rule If E1 and E2 are disjoint events, then Prob(E1 or E2) = Prob(E1) + Prob(E2)

Lecture 14 9/ 19

The Sum Rule for Probability

Sum Rule If E1 and E2 are disjoint events, then Prob(E1 or E2) = Prob(E1) + Prob(E2) Example: For a roll of two dice, E1 = “at least one 5” and E2 = “sum equals 5” I E1 = {(1, 5), (2, 5), (3, 5), (4, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (5, 5)}, so n(E1) = 11 I E2 = {(1, 4), (2, 3), (4, 1), (3, 2)}, so n(E2) = 4 I Or n(E1) = 62 52 = 11 and n(E2) = C(r 1, r n) = C(4, 3) = 4 I Hence Prob(E1 or E2) = Prob(E1) + Prob(E2) = 11

36 + 4 36 = 5 12 Lecture 14 9/ 19

The Generalized Sum Rule

Generalized Sum Rule If E1 and E2 are any events, then Prob(E1 or E2) = Prob(E1) + Prob(E2) Prob(E1 and E2) Example: When choosing 3 cards, what is probability of either 3 face cards (E1) or 3 cards of same suit (E2) (or both)? I C(52, 3) ways of selecting 3 cards, 13 cards per suit, 4 suits, 3 face cards per suit (jack, queen, king) I Prob(E1 or E2) = C(12,3)

C(52,3) + 4·C(13,3) C(52,3) 4·1 C(52,3) = 1,360 22,100 ⇡ 0.0615 Lecture 14 10/ 19

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Independent Events

Definition Two events E1 and E2 are independent if the occurrence of one is not influenced by

• ccurrence (or nonoccurrence) of the other.

Lecture 14 11/ 19

Independent Events

Definition Two events E1 and E2 are independent if the occurrence of one is not influenced by

• ccurrence (or nonoccurrence) of the other.

Examples I Toss coin twice: E1 = “heads on first toss” and E2 = “tails on second toss” [independent] I Choosing 2 cards from a deck: E1 = “first card is an ace” and E2 = “second card is an ace” [not independent]

Lecture 14 11/ 19

Independent Events, Continued

Definition Two events E1 and E2 are independent if the occurrence of one is not influenced by

• ccurrence (or nonoccurrence) of the other.

Problem: Which of these pairs of events are independent? I In 4 die rolls, E1 = “first 2 rolls sum to 7” and E2 = “last 2 rolls sum to 10” I When choosing a committee of 3 from 8 men and 12 women, E1 = “committee has a woman” and E2 = “committee has a man” I In a household with 4 children, E1 = “first child is male” and E2 = “at least half the children are female”

Lecture 14 12/ 19

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The Product Rule for Probability

Product Rule If E1 and E2 are independent events, then Prob(E1 and E2) = Prob(E1) · Prob(E2) Example: For two rolls of a “loaded” die with Prob(6) = 1/2 and Prob(i) = 1/10 for i 6= 6, E1 = “5 on first roll” and E2 = “6 on second roll” I Prob(E1 and E2) = Prob(E1) · Prob(E2) =

1 10 · 1 2 = 1 20 Lecture 14 13/ 19

The Product Rule, Continued

Product Rule If E1 and E2 are independent events, then Prob(E1 and E2) = Prob(E1) · Prob(E2) Problem: For each example, compute Prob(E1 and E2) and Prob(E1) · Prob(E2) I In 4 die rolls, E1 = “first 2 rolls sum to 7” and E2 = “last 2 rolls sum to 10” I When choosing a committee of 3 from 8 men and 12 women, E1 = “committee has a woman” and E2 = “committee has a man” I In a household with 4 children, E1 = “first child is male” and E2 = “at least half the children are female”

Lecture 14 14/ 19

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Conditional Probability

Definition Given events E1 and E2, the conditional probability of E1 given E2, denoted by Prob(E1|E2), is the probability that E1 happens given that E2 occurs. If E1 and E2 are independent, then Prob(E1|E2) = Prob(E1).

Lecture 14 15/ 19

Conditional Probability

Definition Given events E1 and E2, the conditional probability of E1 given E2, denoted by Prob(E1|E2), is the probability that E1 happens given that E2 occurs. If E1 and E2 are independent, then Prob(E1|E2) = Prob(E1). Example: Choose a committee of 3 from 8 men and 12 women, let W = “committee has a woman” and M = “committee has a man”, and compute Prob(W |M) I Reduced sample space S0 is set of (equally likely) committees that have a man: n(M) = n(S) n( ¯ M) = C(20, 3) C(12, 3) = 920 I Those that also contain a woman are of the form {M, W , W } or {M, M, W } I Hence Prob(W |M) = C(8,1)·C(12,2)+C(8,2)·C(12,1)

920

= 864

920 = 108 115 Lecture 14 15/ 19

Conditional Probability

Definition Given events E1 and E2, the conditional probability of E1 given E2, denoted by Prob(E1|E2), is the probability that E1 happens given that E2 occurs. If E1 and E2 are independent, then Prob(E1|E2) = Prob(E1). Example: Choose a committee of 3 from 8 men and 12 women, let W = “committee has a woman” and M = “committee has a man”, and compute Prob(W |M) I Reduced sample space S0 is set of (equally likely) committees that have a man: n(M) = n(S) n( ¯ M) = C(20, 3) C(12, 3) = 920 I Those that also contain a woman are of the form {M, W , W } or {M, M, W } I Hence Prob(W |M) = C(8,1)·C(12,2)+C(8,2)·C(12,1)

920

= 864

920 = 108 115

Observation I We calculated Prob(W |M) as n(W \M)

n(M)

I Dividing top and bottom by n(S)—total number possible committees—we have Prob(W |M) = n(W \M)/n(S)

n(M)/n(S)

= Prob(W and M)

Prob(M) Lecture 14 15/ 19

General Product Rule

General Product Rule If E1 and E2 are any events, then Prob(E1 and E2) = Prob(E1) · Prob(E2|E1) = Prob(E2) · Prob(E1|E2) Example: Draw 2 marbles from bag with 3 red, 5 white, 8 green

• 1. What is probability that both are red?

I Prob(R1 and R2) = Prob(R1) · Prob(R2|R1) = 3 16 · 2 15 = 1 40

• 2. What is probability that one is green and one is white?

I Apply sum rule and then product rule I Prob(W1 and G2) + Prob(G1 and W2) =

Prob(W1) · Prob(G2|W1) + Prob(G1) · Prob(W2|G1) =

5 16 · 8 15 + 8 16 · 5 15 = 1 3 Lecture 14 16/ 19

Drug Testing for Athletes

Properties of a drug test I For steroid users, test is correct with probability 0.995 and wrong with probability 0.005 I For non-steroid users, test is correct with probability 0.98 and wrong with probability 0.02 2% of athletes use steroids Analysis of the test I P = “test is positive for steroid use” and S = “athlete has used steroids” I What is Prob(P and S)? I What is Prob(P and ¯ S)? I What is Prob(P)? I What is Prob(S|P)?

Lecture 14 17/ 19 Prob(E1 and E2) = Prob(E1) · Prob(E2|E1) = Prob(E2) · Prob(E1|E2)

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Bayes Rule

Bayes’ Rule Prob(E2|E1) = Prob(E1|E2)·Prob(E2)

Prob(E1)

, which is often written in the form Prob(E2|E1) / Prob(E1|E2) · Prob(E2) I Prob(E2) is called the prior probability, Prob(E1|E2) is called the likelihood, Prob(E2|E1) is called the posterior probability I This is the fundamental equation of statistical machine learning

Lecture 14 18/ 19 Prob(E1 and E2) = Prob(E1) · Prob(E2|E1) = Prob(E2) · Prob(E1|E2)

!

proportional

to

Bernoulli Trials

We will now look at outcomes that are not equally likely I Look at experiment based on repetitions of simple experiment (“trials”)

Lecture 14 19/ 19

Bernoulli Trials

We will now look at outcomes that are not equally likely I Look at experiment based on repetitions of simple experiment (“trials”) Example: Baseball player gets a hit with probability 1/3 every time he steps to the

• plate. What is the probability that he gets exactly one hit in 4 tries (event E)?

Lecture 14 19/ 19

Bernoulli Trials

We will now look at outcomes that are not equally likely I Look at experiment based on repetitions of simple experiment (“trials”) Example: Baseball player gets a hit with probability 1/3 every time he steps to the

• plate. What is the probability that he gets exactly one hit in 4 tries (event E)?

I Assume that tries are independent I Define H = “gets a hit” and N = “does not get a hit” I Outcome NNNN more likely than HHHH I Can use product rule for any given sequence: Prob(HNNN) = Prob(H) · Prob(N) · Prob(N) · Prob(N) = ⇣ 1 3 ⌘ · ⇣ 2 3 ⌘ · ⇣ 2 3 ⌘ · ⇣ 2 3 ⌘ = ⇣ 1 3 ⌘⇣ 2 3 ⌘3 I Final solution: Prob(E) = Prob(HNNN or NHNN or NNHN or NNNH} = ⇣ 1 3 ⌘⇣ 2 3 ⌘3 + ⇣ 1 3 ⌘⇣ 2 3 ⌘3 + ⇣ 1 3 ⌘⇣ 2 3 ⌘3 + ⇣ 1 3 ⌘⇣ 2 3 ⌘3 = 4 ⇣ 1 3 ⌘⇣ 2 3 ⌘3 I Q: What is probability of at least one hit in 4 times at bat?

Lecture 14 19/ 19

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Bernoulli Trials: General Case

Theorem (Bernoulli Trials) For a sequence of n Bernoulli trials with success probability p, the probability of exactly k successes is C(n, k) · pk · (1 p)nk.

Lecture 14 20/ 19

Bernoulli Trials: General Case

Theorem (Bernoulli Trials) For a sequence of n Bernoulli trials with success probability p, the probability of exactly k successes is C(n, k) · pk · (1 p)nk. Proof

• 1. An outcome is an ordered list of length n with entries from {S, F}
• 2. The number of such lists is C(n, k) as per Lecture 13, Slide 24
• 3. By the product rule, each such list has probability pk(1 p)nk
• 4. Result follows from the sum rule

Lecture 14 20/ 19

Bernoulli Trials: General Case

Theorem (Bernoulli Trials) For a sequence of n Bernoulli trials with success probability p, the probability of exactly k successes is C(n, k) · pk · (1 p)nk. Proof

• 1. An outcome is an ordered list of length n with entries from {S, F}
• 2. The number of such lists is C(n, k) as per Lecture 13, Slide 24
• 3. By the product rule, each such list has probability pk(1 p)nk
• 4. Result follows from the sum rule

Example 1: Probability of exactly five 6’s in 10 rolls of a die I C(10, 5) = 252 ordered lists of length 10 having five S’s and five F’s I Probability of each such list, e.g., SSSSSFFFFF, is 1

6

5 5

6

5 I So answer is C(10, 5) 1

6

5 5

6

5

Lecture 14 20/ 19

Bernoulli Trials: General Case

Theorem (Bernoulli Trials) For a sequence of n Bernoulli trials with success probability p, the probability of exactly k successes is C(n, k) · pk · (1 p)nk. Proof

• 1. An outcome is an ordered list of length n with entries from {S, F}
• 2. The number of such lists is C(n, k) as per Lecture 13, Slide 24
• 3. By the product rule, each such list has probability pk(1 p)nk
• 4. Result follows from the sum rule

Example 1: Probability of exactly five 6’s in 10 rolls of a die I C(10, 5) = 252 ordered lists of length 10 having five S’s and five F’s I Probability of each such list, e.g., SSSSSFFFFF, is 1

6

5 5

6

5 I So answer is C(10, 5) 1

6

5 5

6

5 Example 2: Given a biased coin with Prob(heads) = 4/7, what is the probability of at least 8 heads in 10 tosses? I Disjoint cases: 8 heads, 9 heads, 10 heads I using Theorem and sum rule, we get C(10, 8) 4

7

8 3

7

2 + C(10, 9) 4

7

9 3

7

1 + C(10, 10) 4

7

10 3

7

0 ⇡ 0.1255

Lecture 14 20/ 19

Bernoulli Trials: General Case

Theorem (Bernoulli Trials) For a sequence of n Bernoulli trials with success probability p, the probability of exactly k successes is C(n, k) · pk · (1 p)nk. Proof

• 1. An outcome is an ordered list of length n with entries from {S, F}
• 2. The number of such lists is C(n, k) as per Lecture 13, Slide 24
• 3. By the product rule, each such list has probability pk(1 p)nk
• 4. Result follows from the sum rule

Example 1: Probability of exactly five 6’s in 10 rolls of a die I C(10, 5) = 252 ordered lists of length 10 having five S’s and five F’s I Probability of each such list, e.g., SSSSSFFFFF, is 1

6

5 5

6

5 I So answer is C(10, 5) 1

6

5 5

6

5 Example 2: Given a biased coin with Prob(heads) = 4/7, what is the probability of at least 8 heads in 10 tosses? I Disjoint cases: 8 heads, 9 heads, 10 heads I using Theorem and sum rule, we get C(10, 8) 4

7

8 3

7

2 + C(10, 9) 4

7

9 3

7

1 + C(10, 10) 4

7

10 3

7

0 ⇡ 0.1255

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