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Unformatted text preview: Solutions to Homework Set No. 1 Probability Theory (235A), Fall 2011 1. (a) If ( , F , P ) is a probability space and A,B F are events such that P ( B ) 6 = 0, the conditional probability of A given B is denoted P ( A  B ) and defined by P ( A  B ) = P ( A B ) P ( B ) . Prove the total probability formula: if A,B 1 ,B 2 ,...,B k F such that is the disjoint union of B 1 ,...,B k and P ( B i ) 6 = 0 for 1 i k , then P ( A ) = k X i =1 P ( B i ) P ( A  B i ) . (TPF) Solution. P ( A ) = P k [ i =1 ( A B i ) = k X i =1 P ( A B i ) = k X i =1 P ( A  B i ) P ( B i ) (b) An urn initially contains one white ball and one black ball. At each step of the experi ment, a ball is drawn at random from the urn, then put back and another ball of the same color is added. Prove that the number of white balls that are in the urn after N steps is a uniform random number in { 1 , 2 ,...,N + 1 } . That is, the event that the number of white balls after step N is equal to k has probability 1 / ( N + 1) for each 1 k N + 1. (Note: The idea is to use (TPF), but there is need to be too formal about constructing the relevant probability space  you can assume an intuitive notion of probabilities.) Solution. Let A N,k be the event that the number of white balls after the N th step is k. We will show that P ( A N,k ) = 1 N +1 for 1 k N + 1. By investigating some cases, we may conjecture that P ( A N,k ) = 1 / ( N + 1) for each 1 k N + 1. The proof follows from the mathematical induction on N . If N = 0, it is obvious that P ( A , 1 ) = 1. Suppose that P ( A N,k ) = 1 N +1 . Then by the total probability formula, if 1 < k < n + 2, P ( A N +1 ,k ) = P ( A N,k ) P ( A N +1 ,k  A N,k ) + P ( A N,k 1 ) P ( A N +1 ,k  A N,k 1 ) = 1 N + 1 N + 2 k N + 2 + 1 N + 1 k 1 N + 2 = 1 N + 2 ....
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 Fall '11
 DanRomik
 Conditional Probability, Probability

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