HW Solution 3 Due: July 15

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1 ECS 315: Probability and Random Processes 2010/1 HW Solution 3 Due: July 15 Lecturer: Prapun Suksompong, Ph.D. Instructions (a) A part of ONE question will be graded. Of course, you do not know which problem will be selected; so you should work on all of them. (b) Late submission will not be accepted. (c) Write down all the steps that you have done to obtain your answers. You may not get full credit even when your answer is correct without showing how you get your answer. Problem 1. Someone has rolled a fair die twice. You know that one of the rolls turned up a face value of six. What is the probability that the other roll turned up a six as well? Hint: Not 1 6. Take as sample space the set {(i, j) i, j = 1,..., 6}, where i and j denote the outcomes of the first and second rolls. A probability of 1/36 is assigned to each element of the sample space. The event of two sixes is given by A = {(6, 6)} and the event of at least one six is given by B = (1, 6),..., (5, 6), (6, 6), (6, 5),..., (6, 1). Applying the definition of conditional probability gives P (A B) = P (A B)/P (B) = 1/36 11/36. Hence the desired probability is 1/11. [Tijms, 2007, Example 8.1, p. 244] Problem 2. [Gubner, 2006, Q2.60] You have five computer chips, two of which are known to be defective. (a) You test one of the chips; what is the probability that it is defective? (b) Your friend tests two chips at random and reports that one is defective and one is not. Given this information, you test one of the three remaining chips at random; what is the conditional probability that the chip you test is defective? 3-1

2 (a) 2 5. (b) Among the three remaining chips, only one is defective. So, the conditional probability that the chosen chip is defective is 1 3. Problem 3. Prove the following facts about independence between two events: (a) An event with probability 0 or 1 is independent of any event (including itself). (In particular, and Ω are independent of any events.) (b) An event A is independent of itself if and only if P (A) is 0 or 1. (a) Suppose P (A) = 1. Then, 0 P (B \ A) P (A c ) = 0 and hence P (A B) = P (B) P (B \ A) = P (B) = P (B) P (A). Suppose P (A) = 0. Then, 0 P (A B) P (A) = 0 and hence P (A B) = 0 = P (A) P (B). (b) If A is independent of itself, then P (A) = P (A A) = P (A) P (A) which implies P (A) is 0 or 1. (The solutions of x = x 2 are x = 0 or 1.) Problem 4. Suppose A B and A B. If P (A) = 1/3, find P (B). Because A B, we have P (A B) = P (A). The LHS of this equality can be factored into P (A)P (B) by the independence between A and B. Hence, P (A)P (B) = P (A). Because P (A) > 0, we can divide both sides of the equation by P (A) and get P (B) = 1. Problem 5. Show that if A and B are independent events, then so are A and B c, A c and B, and A c and B c. To show that two events C 1 and C 2 are independent, we need to show that P (C 1 C 2 ) = P (C 1 )P (C 2 ). 3-2

3 (a) Note that Because A P (A B c ) = P (A \ B) = P (A) P (A B). B, the last term can be factored in to P (A)P (B) and hence P (A B c ) = P (A) P (A)P (B) = P (A)(1 P (B)) = P (A)P (B c ) (b) By interchanging the role of A and B in the previous part, we have P (A c B) = P (B A c ) = P (B) P (A c ). (c) From set theory, we know that A c B c = (A B) c. Therefore, P (A c B c ) = 1 P (A B) = 1 P (A) P (B) + P (A B). Because A B, we have P (A c B c ) = 1 P (A) P (B) + P (A)P (B) = (1 P (A)) (1 P (B)) = P (A c )P (B c ). Remark: By interchanging the roles of A and A c and/or B and B c, it follows that if any one of the four pairs is independent, then so are the other three. [Gubner, 2006, p.31] Problem 6. Anne and Betty go fishing. Find the conditional probability that Anne catches no fish given that at least one of them catches no fish. Assume they catch fish independently and that each has probability 0 < p < 1 of catching no fish. Let A be the event that Anne catches no fish and B be the event that Betty catches no fish. From the question, we know that A and B are independent. The event at least one of the two women catches nothing can be represented by A B. So we have P (A A B) = P (A (A B)) P (A B) = P (A) P (A) + P (B) P (A) P (B) = p 2p p = p. [Gubner, 2006, Q2.62] Problem 7. In this question, each experiment has equiprobable outcomes. (a) Let Ω = {1, 2, 3, 4}, A 1 = {1, 2}, A 2 = {1, 3}, A 3 = {2, 3}. (i) Determine whether P (A i A j ) = P (A i ) P (A j ) for all i j. (ii) Check whether P (A 1 A 2 A 3 ) = P (A 1 ) P (A 2 ) P (A 3 ). 3-3

4 (iii) Are A 1, A 2, and A 3 independent? (b) Let Ω = {1, 2, 3, 4, 5, 6}, A 1 = {1, 2, 3, 4}, A 2 = A 3 = {4, 5, 6}. (i) Check whether P (A 1 A 2 A 3 ) = P (A 1 ) P (A 2 ) P (A 3 ). (ii) Check whether P (A i A j ) = P (A i ) P (A j ) for all i j. (iii) Are A 1, A 2, and A 3 independent? (a) We have P (A i ) = 1 2 and P (A i A j ) = 1 4. (i) P (A i A j ) = P (A i )P (A j ) for any i j. (ii) A 1 A 2 A 3 =. Hence, P (A 1 A 2 A 3 ) = 0, which is not the same as P (A 1 ) P (A 2 ) P (A 3 ). (iii) No. (b) We have P (A 1 ) = 4 6 = 2 3 and P (A 2) = P (A 3 ) = 3 6 = 1 2. (i) A 1 A 2 A 3 = {4}. Hence, P (A 1 A 2 A 3 ) = 1. 6 P (A 1 ) P (A 2 ) P (A 3 ) = = Hence, P (A 1 A 2 A 3 ) = P (A 1 ) P (A 2 ) P (A 3 ). (ii) P (A 2 A 3 ) = P (A 2 ) = 1 2 P (A 2)P (A 3 ) P (A 1 A 2 ) = p(4) = 1 6 P (A 1)P (A 2 ) P (A 1 A 3 ) = p(4) = 1 6 P (A 1)P (A 3 ) Hence, P (A i A j ) P (A i ) P (A j ) for all i j. (iii) No. Problem 8. Suppose A, B, and C are independent. Show that A Start with expressing B as a disjoint union B = (B C c ) (B C). Then, by intersecting both sides by A, we still have disjoint union: A B = (A B C c ) (A B C). (B C c ). Hence, P (A B) = P (A B C c ) + P (A B C). 3-4

5 Therefore, P (A B C c ) = P (A B) P (A B C). By the independence among A, B, and C, we have P (A B C c ) = P (A)P (B) P (A)P (B)P (C) = P (A) (P (B) P (B C)) = P (A)P (B C c ), where we use the independence property again to get the second-to-last equality. Remarks (a) It is now easy to show that A role of B and C. (B c C). Simply follow the same proof but switch the (b) One of the properties that follows from the fact that A, B, and C are independent is that A (B C). This follow easily from the definition: P (A (B C)) = P (A B C) = P (A) P (B) P (C) = P (A) P (B C). (c) A useful fact that we shall come back to later is that if A C 1, C 2,..., C n, then A n C i. To see this, note that A C i = (A C i ) where the union on the RHS is a disjoint union. Hence, ( ) n n P A C i = P (A C i ) = P (A) P (C i ) = P (A) C i for all disjoint events n P (C i ), where the second equality is obtained by applying the independence between A and C i. Now, applying the fact that the C i are disjoint, we finally get ( ) ( n ) P A C i = P (A) P C i ; that is A n C i. (d) Because we now know that A (B C c ), A conclude, using the previous remark, that A union of B C c, B c C, and B C. (e) Because A (B c C), and A (B C), we can (B C) because B C is a disjoint (B C), we know that A is independent of (B C) c which is B c C c. 3-5

6 (f) Observe that any event involving only B and C will be a union of one or more of B C c, B c C, B C, and B c C c. Because these four event are disjoint and we have shown that A is independent of all these four events, it follow from fact (c) above that A is independent of any event that involves only B and C. Formally, we say that A is independent of any event generated by B and C. Problem 9. Suppose {A} {B 1, B 2, B 1 B 2 }. Show that (a) A (b) A B c 1 B c 2 (c) A c B 1 (d) A (e) A B 1 B c 2 B c 1 B 2 (f) A (B 1 B 2 ) Remark: The question does not say that B 1 B 2. (a) The independence between the two collections gives A result from the previous question to show that A B1. c (b) The independence between the two collections gives A result from the previous question to show that A B2. c (c) The independence between the two collections gives A result from the previous question to show that A c B 1. B 1. B 2. B 1. We can then use the We can then use the We can then use the (d) Note that P (A B 1 B c 2) = P ((A B 1 ) \B 2 ) = P (A B 1 ) P (A B 1 B 2 ) = P (A) P (B 1 ) P (A) P (B 1 B 2 ), where we apply the given conditions that A B 1 and A (B 1 B 2 ). Therefore, P (A B 1 B c 2) = P (A) (P (B 1 ) P (B 1 B 2 )) = P (A) P (B 1 \ B 2 ) = P (A) P (B 1 B c 2). (e) This follows from the previous part by interchanging the role of B 1 and B

7 (f) We first express B 1 B 2 as B 1 B 2 = (B 1 B c 2) (B 1 B 2 ) (B c 1 B 2 ). Note that we have a disjoint union. Observe that we still have disjoint union when we expand A (B 1 B 2 ) as Hence, Remarks: A (B 1 B 2 ) = (A B 1 B c 2) (A B 1 B 2 ) (A B c 1 B 2 ). P (A (B 1 B 2 )) = P (A B 1 B c 2) + P (A B 1 B 2 ) + P (A B c 1 B 2 ). Using the results from the previous parts, we have P (A (B 1 B 2 )) = P (A) P (B 1 B c 2) + P (A) P (B 1 B 2 ) + P (A) P (B c 1 B 2 ) = P (A) (P (B 1 B c 2) + P (B 1 B 2 ) + P (B c 1 B 2 )) = P (A) P (B 1 B 2 ) Observe that the last part can be replaced by applying the fact, which we proved earlier, that if A C i for all disjoint events C 1, C 2,..., C n, then A n C i. (a) Because A (B 1 B 2 ), we know that A is independent of (B 1 B 2 ) c which is B c 1 B c 2. (b) Observe that any event involving only B 1 and B 2 will be a union of one or more of B 1 B c 2, B c 1 B 2, B 1 B 2, and B c 1 B c 2. Because these four event are disjoint and we have shown that A is independent of all these four events, it follow that A is independent of any event that involves only B 1 and B 2. Formally, we say that A is independent of any event generated by B 1 and B 2. Problem 10. A certain binary communication system has a bit-error rate of 0.1; i.e., in transmitting a single bit, the probability of receiving the bit in error is 0.1. To transmit messages, a three-bit repetition code is used. In other words, to send the message 1, 111 is transmitted, and to send the message 0, 000 is transmitted. At the receiver, if two or more 1s are received, the decoder decides that message 1 was sent; otherwise, i.e., if two or more zeros are received, it decides that message 0 was sent. Assuming bit errors occur independently, find the probability that the decoder puts out the wrong message. Let p = 0.1 be the bit error rate. Error event E occurs if there are at least two bit errors. Therefore ( ) ( ) 3 3 P (E) = p 2 (1 p) + p 3 = p 2 (3 2p). 2 3 When p = 0.1, P (E) [Gubner, 2006, Q2.62] 3-7