2.2 Conditional Probability

Transcription

1 2.2 Conditional Probability Contents Conditional probability. Conditioning and its consequences. Multiple conditioning. Matching problems. Examples. Independent events. Types of independence. Notions of reliability. Examples. References: Ross (Chapter 3); Ben Arous notes (Sections II.5, II.4). Exercises: 39 42, 44 54, 37, 38 of Recueil d exercices. Probabilité et Statistique I Sections 2.2, 2.3 1

2 Conditional Probability Example : Two fair dice are rolled, one red and one green. Let A and B be the events that the total exceeds 8, and that the red die shows 6. If B is known to have occurred, how does P(A) change? Definition: Let A, B be events of a probability space (Ω, F, P), such that P(B) > 0. Then the conditional probability of A given B is P(A B) = P(A B). P(B) When P(B) = 0, we adopt the convention P(A B) = P(A B)P(B), both sides having the value zero. Thus P(A) = P(A B) + P(A B c ) = P(A B)P(B) + P(A B c )P(B c ) even if P(B) = 0 or P(B c ) = 0. Probabilité et Statistique I Sections 2.2, 2.3 2

3 Conditional Probability II Theorem 2.2 (Law of total probability): Let {B i } i=1 be pairwise disjoint events of a probability space (Ω, F, P), and let the event A satisfy A i=1 B i. Then P(A) = P(A B i ) = i=1 P(A B i )P(B i ). Theorem 2.3 (Bayes): Suppose that the conditions above hold, and that P(A) > 0. Then i=1 P(B j A) = P(A B j )P(B j ) i=1 P(A B, for any j N. i)p(b i ) In particular, these results are true if {B i } i=1 is a partition of Ω. Probabilité et Statistique I Sections 2.2, 2.3 3

4 Example 2.11: Cars are manufactured in towns called Farad, Gilbert and Henry. Of 1000 made in Farad 20% are defective, of 2000 made in Gilbert 10% are defective, and of 3000 made in Henry 5% are defective. You buy a car from a distant dealer. If D is the event that the car is defective, find (a) P(F H c ), (b) P(D H c ), (c) P(D), and P(F D). Assume you are equally likely to have bought any of the 6000 cars produced. Example 2.12: You undergo a test for a rare disease which occurs by chance in 1 in every 100,000 people. The test is fairly reliable: if you have the disease it will correctly say so with probability 0.95; if you do not have the disease, the test will wrongly say that you do with probability If the test says you do have the disease, what is the probability that this is a correct diagnosis? Probabilité et Statistique I Sections 2.2, 2.3 4

5 Conditioning is a key idea of probability, and gives elegant solutions to many problems. Example 2.13 (The fly): A room has four walls, a floor, and a ceiling. A fly moves between these surfaces. If it leaves the floor or ceiling then it is equally likely to alight on any one of the four walls or the surface it has just left. If it leaves a wall then it is equally likely to alight on any one of the other three walls, or the floor, or the ceiling. Initially it is on the ceiling. Find F k, the probability that the fly is on the floor after k moves. Example 2.14 (Gambler s ruin): You enter a casino with $k, and on each spin of the roulette wheel you bet $1 at evens on the event R that the result is red. The wheel is not fair, so P(R) = p < 1 2. If you lose all $k you must leave, and if you ever possess $K $k, you leave immediately. What is the probability that you leave with nothing? Probabilité et Statistique I Sections 2.2, 2.3 5

6 Conditional Probability III Theorem 2.4: Let (Ω, F, P) be a probability space, and let B F have P(B) > 0 and Q(A) = P(A B). Then (Ω, F, Q) is a probability space. In particular, 1. if A F, then 0 Q(A) 1; 2. Q(Ω) = 1; 3. if {A i } i=1 are pairwise disjoint (that is, A i A j =, i j), then Q ( i=1 A i ) = Q(A i ). j=1 Thus conditioning enables us to construct many different probability distributions, starting from a given one. Probabilité et Statistique I Sections 2.2, 2.3 6

7 Multiple Conditioning Theorem 2.5 (Prediction decomposition): Let A 1,...,A n be events of a probability space. Then P(A 1 A 2 ) = P(A 2 A 1 )P(A 1 ) P(A 1 A 2 A 3 ) = P(A 3 A 1 A 2 )P(A 2 A 1 )P(A 1 ). n P(A 1 A n ) = P(A i A 1 A i 1 )P(A 1 ) i=2 Example 2.15: Two fair dice are thrown. Define the events A, B, C to be the total is at most 6, the total is even, and the first die shows 4. (a) How does knowledge that B or C has occurred affect the probability of A? (b) Compute P(A B C). Probabilité et Statistique I Sections 2.2, 2.3 7

8 Matchings Example 2.16: n men attend a dinner. Each leaves his hat in the cloakroom. When they leave, bien arrosé, they choose their hats at random. (a) What is the probability that no-one has the correct hat? (b) What is the probability that exactly r choose the correct hats? (c) What happens when n is very large? Note: Variations on this example have many applications. Probabilité et Statistique I Sections 2.2, 2.3 8

9 2.3 Independence Probabilité et Statistique I Sections 2.2, 2.3 9

10 Independent Events Intuitively, saying A and B are independent means that occurrence of one does not affect the occurrence of the other. That is, P(A B) = P(A), so knowledge that B has occurred leaves P(A) unchanged. Example 2.17 (Dice again): Two fair dice are thrown. Let A and B be the events the total is even and the first die shows an even number. Compute P(A) and P(A B). Example 2.18 (Boys): A family has two children. (a) The first is known to be a boy. What is the probability that the second is boy? (b) One is known to be a boy. What is the probability that the other is a boy? Probabilité et Statistique I Sections 2.2,

11 Independence II Definition: Let (Ω, F, P) be a probability space. Two events A, B F are independent iff P(A B) = P(A)P(B). In line with our intuition, this implies that P(A B) = P(A B) P(B) and by symmetry P(B A) = P(B). = P(A)P(B) P(B) = P(A), Example 2.19: A deck of cards is well shuffled and a card drawn at random. Are the events A the card is an ace, and H the card is a heart independent? What about the events A and K the card is a king? Probabilité et Statistique I Sections 2.2,

12 Total, Pairwise and Conditional Independence Definition: (a) Events A 1,...,A n are independent (or mutually independent) if for any set of indices F {1,...,n}, it is true that ( ) P A i = P(A i ). i F i F (b) Events A 1,...,A n are pairwise independent if P(A i A j ) = P(A i ) P(A j ), 1 i < j n. (c) Events A 1,...,A n are conditionally independent given B if ( ) P A i B = P(A i B) i F i F for any set of indices F {1,...,n}. Probabilité et Statistique I Sections 2.2,

13 Note: Independence is a key idea which greatly simplifies many probability calculations. In practice it is wise to check the basis of claims that events are independent, as unsuspected dependence can make a big difference to computed probabilities. Note: Mutual independence implies pairwise independence, but the converse is true only when n = 2. Note: Mutual independence implies conditional independence but the converse is true only if B = Ω. Example 2.20: A family has two children. Show that the events the first child is a boy, the second child is a boy, and there is exactly one boy are pairwise but not mutually independent. Probabilité et Statistique I Sections 2.2,

14 Example 2.21 (Birthdays): n people are in a room. What is the probability that they all have different birthdays? Probability n Probabilité et Statistique I Sections 2.2,

15 Example 2.22: In any given year the probability that a male driver makes a claim on his insurance is µ, independently of other years. The probability for a female driver is λ < µ. An insurance company has equal numbers of male and female drivers, and selects one at random. (a) What is the probability that (s)he makes a claim this year? (b) What is the probability that (s)he makes a claim in two consecutive years? (c) If the insurance company selects a claimant at random, what is the probability that this claimant makes a claim in the following year? (d) Show that knowledge that a claim has been made in one year increases the probability of a claim in a second year. Probabilité et Statistique I Sections 2.2,

16 Series and Parallel Systems An electrical system is composed of components labelled 1,...,n, which fail independently. Let A i be the event that the ith component fails, and suppose that P(A i ) = p i. The event B, system failure occurs if current cannot pass from one side of the system to the other. If the components are arranged in parallel, then n P P (B) = P(A 1 A n ) = p i. If the components are arranged in series, then P S (B) = P(A 1 A n ) = 1 i=1 n (1 p i ). If p i > p > 0, for all i, and n, then P P (B) 0, P S (B) 1. i=1 Probabilité et Statistique I Sections 2.2,

17 Reliability Example 2.23 (Chernobyl): A nuclear power station depends on a safety system whose components are arranged as shown in the figure (blackboard). Components fail independently with probability p, and the system fails if electrical current cannot pass from A to B. (a) What is the probability that the system fails? (b) The components are manufactured in batches, which can be good or bad. For a good batch, p = 10 6, while for a bad batch p = The probability that a batch is good is What are the probabilities that the system fails (i) if the components come from different batches? (ii) if all the components come from the same batch? Probabilité et Statistique I Sections 2.2,