hw1_solution - 10-701, Fall 2008, Homework 1 Solution...

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Unformatted text preview: 10-701, Fall 2008, Homework 1 Solution September 25, 2008 1 Probability [20 Points, Jerry] 1. (8 pts) Andy has an old friend who has two children, but he don’t know their genders. (a) What is the probability that they are a girl and a boy? (Hint: For this whole problem, assume that the probabilities of having boys and girls are equal.) Solution: P (one girl one boy) = P ( C 1 = g, C 2 = b ) + P ( C 1 = b, C 2 = g ) = P ( C 1 = g ) P ( C 2 = b ) + P ( C 1 = b ) P ( C 2 = g ) = 0 . 5 ∗ . 5 + 0 . 5 ∗ . 5 = 0 . 5 (b) If Andy knows that at least one of them is a girl, then what is the prob- ability that the other child is a boy? Solution: Use Bayes’ Theorem P (one boy | at least one girl) = P (a girl and a boy) P (at least one girl) = P (a girl and a boy) P (a girl and a boy) + P (two girls) = . 5 . 5 + 0 . 5 ∗ . 5 = 2 3 (c) Assume again that Andy don’t know their gender, but he called his friend’s house and a girl answered the phone. Then, what is the probability that the other child is a boy? (Hint: You can assume that only the two children were at home and they had equal probabilities to answer the phone.) 1 Solution: Use Bayes’ Theorem again P (one boy | a girl answered phone) = P (a girl and a boy, a girl answered phone) P (a girl answered phone) = P (a girl and a boy, a girl answered phone) P (a girl and a boy, a girl answered phone) + P (two girls, a girl answered phone) = . 5 ∗ . 5 . 5 ∗ . 5 + 0 . 25 ∗ 1 = 0 . 5 PS: Here is an easier way to solve this question. The gender of the child who didn’t answer the phone is independent of the gender of the child who answered the phone, therefore the probability of the child who didn’t answer the phone being a boy should be 0.5. 2. (5 pts) Andy plays a card game with Mary. The game uses a deck of 26 black cards and 26 red cards. First, Mary randomly deals the cards and places the deck in her hand all facing down. Then, she flips the card one by one. Before the flip of each card, Andy can choose whether or not he wants to take the card. To win the game, Andy has to pick a Black card, but he can only pick one card. Andy wants to play it wisely. As his good friend, what is the best strategy you can tell him and what is the winning probability? (Hint: If Andy always pick the first card, he has a 50-percent chance to win. Therefore, your answer should be at least as good as 0.5.) Solution: : Any strategy will lead to a 50-percent winning probability. We are going to prove a more general statement: for any game with m black cards and n red cards ( m + n > 0, m ≥ 0, n ≥ 0), the winning probability of any strategy is m/ ( m + n ). Proof: For cases where m = 0 or n = 0, the statement is obviously true. For the rest of cases, we prove by induction on m + n ....
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This note was uploaded on 01/26/2010 for the course MACHINE LE 10701 taught by Professor Ericp.xing during the Fall '08 term at Carnegie Mellon.

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hw1_solution - 10-701, Fall 2008, Homework 1 Solution...

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