THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE

Transcription

1 THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT1301 PROBABILITY AND STATISTICS I EXAMPLE CLASS 2 Review Definition of conditional probability For any two events A and B, the conditional probability of A given the occurrence of B is written as P (A B) and is defined as P (A B) P (A B) = P (B) provided that P (B > 0). Multiplication theorem (a) For any two events A and B with P (B) > 0, P (A B) = P (B)P (A B). (b) For any three events A, B, C with P (B C) > 0, P (A B C) = P (C)P (B C)P (A B C). Independence (a) Two events A and B are called independent if and only if P (A B) = P (A)P (B). If P (A) > 0,then A and B are independent if P (B A) = P (B). (b) The events A 1, A 2,, A k are (mutually) independent if and only if the probability of the intersection of any combination of them is equal to the product of the probabilities of the corresponding 1

2 single events. For example, A 1, A 2, A 3 are independent if and only if Bayes Theorem (Bayes rule, Bayes law) P (A 1 A 2 ) = P (A 1 )P (A 2 ) P (A 1 A 3 ) = P (A 1 )P (A 3 ) P (A 2 A 3 ) = P (A 2 )P (A 3 ) P (A 1 A 2 A 3 ) = P (A 1 )P (A 2 )P (A 3 ) For any two event A and B with P (A) > 0 and P (B) > 0, P (B A) = P (A B) P (B) P (A). Bayes Theorem If B 1, B 2,..., B k are mutually exclusive and exhaustive events (i.e. a partition of the sample space), and A is any event with P (A) > 0, then for any B j, P (B j A) = P (A B j)p (B j ) P (A) = P (B j)p (A B j ), k P (B i )P (A B i ) i=1 where k can also be. Law of total probability (a) If 0 < P (B) < 1, then P (A) = P (A B)P (B) + P (A B c )P (B c ) for any A. (b) If B 1, B 2,..., B k are mutually exclusive and exhaustive events (i.e. a partition of the sample space), then for any event A, k P (A) = P (A B j )P (B j ), where k can also be. j=1 2

3 Problems Problem 1 A and B are two events. Suppose that P (A B) = 0.6, P (B A) = 0.3, and P (A B) = LetP (A) = a. (a) Express P (A B) and P (B) in terms of a. (b) Using the results of (a), or otherwise, find the value of a. (c) Are A and B independent events? Explain your answer briefly. (a) P (A B) = P (B A)P (A) = 0.3a P (A B)P (B) = P (A B) 0.6P (B) = 0.3a P (B) = 0.5a (b) P (A B) = P (A) + P (B) P (A B) 0.72 = a + 0.5a 0.3a a = 0.6 (c) P (A B) = 0.6 = P (A) and P (B A) = 0.3 = = P (B) therefore A and B are independent events. Problem 2 A and B are two events. Suppose that P (B c A) = 3 4, P (Ac B) = 3 5, and P (Ac ) = 2 5, where Ac and B c are complementary events of A and B respectively. Let P (B) = p, where 0 < p < 1. (a) Find P (A B c ). (b) Express P (A c B) in terms of p. 3

4 (c) Using the fact that A c B is the complementary event of A B c, or otherwise, find the value of p. (d) Are A and B mutually exclusive? Explain your answer. (a) (b) (c) P (A B c ) = P (B c A)P (A) = P (B c A)[1 P (A c )] = 3 4 [1 2 5 ] = 9 20 P (A c B) = P (A c B)P (B) = 3 5 p P (A c B) = 1 P (A B c ) P (A c ) + P (B) P (A c B) = 1 P (A B c ) p 3 5 p = p = 3 20 p = 3 8 (d) therefore A and B are not mutually exclusive. P (A B) + P (A c B) = P (B) P (A B) = 3 8 P (A B) = Problem 3 (a) Say that C 1, C 2,, C k are (mutually) independent events that have respective probabilities p 1, p 2,, p k. Argue that the probability of at least one of C 1, C 2,, C k happens is equal to 1 (1 p 1 )(1 p 2 ) (1 p k ) HINT:(C 1 C 2 C k ) c = C c 1 C c 2 C c k 4

5 (b) Let C 1, C 2, C 3 be independent events with probabilities 1, 1, 1, respectively. Compute (i) P (C 1 C 2 C 3 ) (ii) P [(C c 1 C c 2) C 3 ] (a) We first consider C 1 and C 2, if the probability of at least one of them happens is P (C 1 C 2 ). P (C 1 C 2 ) = 1 P ((C 1 C 2 ) c ) = 1 P (C c 1 C c 2) Since C 1 and C 2 are independent,p(c 1 C 2 ) = P (C 1 )P (C 2 ). And since P (C 1 C 2 ) = P (C 1 ) + P (C 2 ) P (C 1 C 2 ),then we get P (C c 1 C c 2) = P ((C 1 C 2 ) c ) = 1 P (C 1 C 2 ) = 1 P (C 1 ) P (C 2 ) + P (C 1 C 2 ) = 1 P (C 1 ) P (C 2 ) + P (C 1 )P (C 2 ) = (1 P (C 1 ))(1 P (C 2 )) = P (C c 1)P (C c 2) Therefore C c 1 and C c 2 are independent. Then we can calculate P (C 1 C 2 ) = 1 P (C c 1)P (C c 2) = 1 (1 p 1 )(1 p 2 ) Based on mathematical induction we can get similar result, which is that C1, c C2, c, Ck c are independent, P (C1 c C2 c Ck c) =. Therefore the probability of at least one of C 1, C 2,, C k happens is P (C 1 C 2 C k ) = P (C1)P c (C2) c P (Ck c ), which is equal to P (C 1 C 2 C k ) = 1 P [(C 1 C 2 C k ) c ] = 1 P (C c 1 C c 2 C c k) = 1 P (C c 1)P (C c 2) P (C c k) = 1 (1 p 1 )(1 p 2 ) (1 p k ) (b) (i) P (C 1 C 2 C 3 ) = 1 (1 P (C 1 ))(1 P (C 2 ))(1 P (C 3 )) = = 3 4 (ii) P [(C c 1 C c 2) C 3 ] = 1 [1 P (C c 1 C c 2)](1 P (C 3 )) = 1 [1 P (C c 1)P (C c 2)](1 P (C 3 )) = 1 ( )(1 1 4 ) = 1 2 5

6 Problem 4 A small plane have gone down, and the search is organized into three regions. Starting with the likeliest, they are: Region Initial Chance Plane is There Chance of Being Overlooked in the Search Mountains Praire Sea The last column gives the chance that if the plane is there, it will not be found, it will not be found. For example, if it went down at sea, there is 90% chance it will have disappeared, or otherwise not be found. Since the pilot is not equipped to long survive a crash in the mountains, it is particularly important to determine the chance that the plane went down in the mountains. (a) Before any search is started, what is this chance in the mountains? (b) The initial search was in the mountains, and the plane was not found. Now what is the chance the plane is nevertheless in the mountains? (c) The search was continued over the other two regions, and unfortunately the plane was not found anywhere. Finally now what is the chance that the plane is in the mountains? (d) Describing how and why the chances changed from (a) to (b) to (c). Let M, P, S be the events that the plan went down in the mountains, Prairie, sea respectively. Let OM, OP, OS be the events that the plane is not found in mountains, prairie, sea respectively. Then we have P (M) = 0.5 P (P ) = 0.3 P (S) = 0.2 (a) P (M) = 0.5 P (OM M) = 0.3 P (OP M) = P (OS M) = 1 P (OP P ) = 0.2 P (OM M) = P (OS P ) = 1 P (OS S) = 0.9 P (OM S) = P (OP S) = 1 (b) P (M OM) = = P (OM M)P (M) P (OM M)P (M) + P (OM P )P (P ) + P (OM S)P (S) (0.3)(0.5) (0.3)(0.5) =

7 (c) P (OM, OP, OS) = P (OM, OP, OS M)P (M) + P (OM, OP, OS P )P (S) + P (OM, OP, OS S)P (S) = P (OM M)P (M) + P (OP P )P (P ) + P (OS S)P (S) = (0.3)(0.5) + (0.2)(0.3) + (0.9)(0.2) = 0.39 P (M OM, OP, OS) = P (OM M)P (M) P (OM, OP, OS) = (0.3)(0.5) 0.39 = (d) Prior to any search, the probability is 0.5. After the initial search, the plane was not found in the mountains. Therefore given this information the probability decreased to However, after the continued search, the plane was also not found in other regions. Hence the probability increased to due to this additional information. Problem 5 Let S = {1, 2,, n} and suppose that A and B are, independently, equally likely to be any of the 2 n subsets ( including the null set and S itself) of S. (a) Show that P (A B) = ( 3 4 )n. HINT: Let N(B) denote the number of elements in B. Use n P (A B) = P (A B N(B) = i)p (N(B) = i) (b) Show that P (A B = φ) = ( 3 4 )n. (a) Denote N(B) as the number of elements in B. Then Using law of total probability, P (A B) = = i=0 P (A B N(B) = i) = 2i 2 = n 2i n ( ) n P (N(B) = i) = /2 n j n P (A B N(B) = i)p (N(B) = i) i=0 n 2 i n i=0 n = 2 2n (2 i ) i=0 ( ) n /2 n j = 2 2n (2 + 1) n = ( 3 4 )n (binomial theorem) 7

8 (b) P (A B = φ) = P (A B c ) = ( 3 4 )n (B c and B have equivalent probabilistic behavior) Problem 6 (The Monty Hall Problem) Suppose you are on a game show, and you are given the choice of three boxes. In one box is a key to a new BMW while empty in others. You pick a box A. Then the host, Monty Hall, who knows what are inside the boxes, opens another box, say box B, which is empty. He then says to you, Do you want to abandon your box and pick box C? Is it to your advantage to switch your choice? Let SB = {hostchooseb}, we have known P (SB A) = 1 2 P (SB A c, B) = 0 P (SB A c, B c ) = 1 P (A) = P (B) = P (C) = 1 3 P (SB, B c A c ) = P (SB B c, A c )P (B c A c ) = 1 2 P (SB, B c A c ) = P (SB B c, A)P (B c A) = 1 2 Therefore P (A SB, B c ) = P (SB, B c 1 A)P (A) P (SB, B c A c )P (A c ) + P (SB, B c A)P (A) = =