Denker FALL Probability- Assignment 6

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1 Denker FALL Probability- Assignment 6 Due Date: Thursday, Oct. 7, 2010 Write the final answer to the problems on this assignment attach the worked out solutions! Problem 1: A box contains n + 1 coins, the i-th coin when flipped will turn head with probability (i 1)/n for i = 1, 2,..., n + 1. A coin is romly selected then repeatedly flipped. What is the conditional probability that the i-th coin was selected given that the first n trials all result in heads? Solution: Let H be the event that the coin is flipped n times all flips showed head. Let E i denote the event that the i-th coin is chosen. We have Using Bayes formula (general form) P(E i ) = 1 n + 1 ( ) n i 1 P(H E i ) =. n P(E i H) = = = P(H E i )P(E i ) n+1 k=1 P(E k)p(h E k ) ( i 1 ) n n( k 1 n n+1 k=1 (i 1)n n k=0 kn. ) n Problem 2: An urn contains 8 balls of which 3 are white. Four players successively draw a ball from the box until a white ball is drawn. Find the probability for each player to draw the white ball. Solution:Let E i denote the event that the i-th player draws the first white ball. The players draw according to their ordering. The experiments are independent, so Player 1: P(E 1 ) = = 3 7.

2 Player 2: P(E 2 ) = = 2 7 Player 3: P(E 3 ) = = 5 28 Player 4: P(E 4 ) = = 3 28 Problem 3: The value of a stock is $50 at present. It is decided to sell the stock either at a value of $80 or at a value of $30. The change of the value of the stock is either up by $1 with probability 0.6 or down by $1 with probability 0.4, independently for each change of the value. What is the probability to sell the stock with a win of $30? Solution: The probability is the same as for a gambler who starts with the fortune of $20 wins all when reaching $50. The probability of going up is 0.6 hence by the formula for a gambler to win P(stock is sold with $80) = 1 ( ) = 1 approximately. 1 (2/3) 50 Problem 4: Each newborn to a couple is a boy with probability 0.49, independently of the sex of other children. If the couple has 6 children, what is the probability that there are exactly 3 boys three girls? Solution: There are ( 6 3) possible distributions of three boys three girls among the children ordered by age. Each such sample has probability (0.49) 3 (0.51 ) 3 by independence. Hence ( ) 6 P(there are exacly three boys among 6 children) = (0.49) 3 (0.51) 3 =

3 Problem 5: A true-false question is to be posed to a husb wife team on a quiz show. Both will independently answer a question correctly with probability q. What strategy of the team is better: Choose one of them romly let that person answer the question; or let both answer the question: if they agree take this answer; or if not, flip a coin to decide which answer is taken. Solution: First strategy: Let E be the event that the answer is correct, H the event that husb is chosen to answer. P(E) = P(E H)P(H) + P(E H c )P(H c ) = 1 (q + q) = q. 2 Second strategy: Let E be as before. Let A h be the event that the husb answers correctly likewise W w the event that the wife answers correctly. Let T be the event that the coin is tossed the husb is chosen to answer. Then E = A h A w T A h A c w T c A w A c h. All sets are A h A w, T A h A c w T c A w A c h pairwise disjoint, so by independence P(E) = q q(1 q) q(1 q) = q2 + q q 2 = q. Answer: No one is better. Problem 6: Let a sample space be given by all natural numbers from 1 to 120, the probability be given by the equal chance principle (so P({i}) = 1 for each i S). Let 120 E be the event that the number is divisible by 4 F the event that the number is divisible by 3. Show that E F are independent. If G is the event that the number is divisible by 5, are the three events E, F G independent? Solution: There are 30 numbers among 1, which divisible by four. The event E that a number is divisible by 4 satisfies P(E) = 1. Let F be the event that a number is 4 divisible by 3. Then P(F) = 1. E F is the event that a number is divisible by 3 by 4. 3 There are 10 such numbers, so P(E F) = 1 = 1 1. Hence E F are independent. In the same way one shows that E G F G are independent. For E F G note that a number is divisible by 2,3 5 if it is divisible by 30. There are 4 such numbers, so P(EFG) = 1 30 = , E, F G are independent. Problem 7: A coin is flipped n times independently, Find the probability that it ls head k times, where k = 0, 1, 2,..., n.

4 Solution: Let E k be the event that the coin ls head k times. We assume that the probability that the coin ls head is p. Then ( ) n P(E k ) = p k (1 p) n k. k Problem 8: Two people go target shooting, the probability of hitting the target being p q. They shoot simultaneously at the same target. What is the probability that both hit the target; one of them hits the target but the other not? NOTE: What independence assumptions are made here? Solution: We assume that the two people shoot independently, meaning that the probabilities of outcomes are independent of the other person s performance. Let A be the event that the first person hits the target, B the event that the other person hits the target. Then P(A) = p P(B) = q. It follows that P(AB) = P(A)P(B) = pq P((A B c ) (B A c )) = P(A B c ) + P(B A c ) = p(1 q) + q(1 p) = p + q 2pq. Problem 9: Genes related to a certain disease are denoted by A a. The disease is present if both genes received from parents are of type a. A child receives one gene from each parent at rom independently. If each of the parents is of type (A, a) (i.e. has one of each genotypes) or of type (A, A) what is the probability that a newborn child develops the disease? What changes if it is known that the first child has the disease? Solution: Let D be the event that the child has the disease, A i be the event that the parent i has only genotypes A a i the event that a parent has the genotype (A, a), i = 1, 2. We know P(D A 1 A 2 ) = P(D A 1 a 2 ) = P(D a 1 A 2 ) = 0 P(D a 1 a 2 ) = 1 4. Since each parent can be of type (A, A) with probability p of type (a, A) with probability 1 p, independently of each other, we also get P(a 1 a 2 ) = (1 p) 2. Therefore P(D) = (1 p)2. 4

5 Let E denote the event that the eldest child has the disease. Let P E denote the probability after knowing that the eldest child has the disease. Then P(a 1 a 2 E) = P E (a 1 a 2 ) = 1 P E (D) = P E (D a 1 a 2 )P E (a 1 a 2 ) + P E (D a c 1 ac 2 )P E(a c 1 ac 2 ) = 1 4. Note that P E (D a 1 a 2 ) = P(D a 1 a 2 ) is unchanged by independence. Problem 10: A quiz has 5 multiple choice questions. Problem i has i+3 possible answers (i = 1, 2, 3, 4, 5). A student decides to choose an answer to each problem at rom independently. What is the probability to pass the quiz (i.e. to have two answers correct)? Solution: The probability of choosing the correct answer in the i-th question is 1 i+3. Therefore, by independence, splitting the event that 2 answers are correct into the 10 disjoint subevents that questions i < j are correct but all other answers to questions k < j are not correct, we get P(2 correct) = 1 4 ( ( ) ( ) = )