S2 - Math 4653 Elementary Probability Spring 2007 Homework#2 Problems and Solutions(corrected 1 Sec 1.5#2 Polyas urn scheme An urn contains 4 white

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Unformatted text preview: Math 4653: Elementary Probability: Spring 2007 Homework #2. Problems and Solutions (corrected) 1. Sec. 1.5: #2: Polyas urn scheme. An urn contains 4 white balls and 6 black balls. A ball is chosen at random, and its color noted. The ball is then replaced, along with 3 more balls of the same color (so that there are now 13 balls in the urn). Then another ball is drawn at random from the urn. a) Find the chance that the second ball drawn is white. b) Given that the second ball drawn is white, what is the probability that the first ball drawn is black? c) Suppose the original contents of the urn are w white and b black balls, and that after a ball is drawn from the urn, it is replaced along with d more balls of the same color. In part a), w was 4, b was 6, and d was 3. Show that the chance that the second ball drawn is white is w w + b . [Note that the probability above does not depend on the value of d .] Solution. Let B = { the first ball drawn is white } , A = { the second ball drawn is white } . Note that B c = { the first ball drawn is black } , A c = { the second ball drawn is black } , and that P ( B ) = 4 10 = 2 5 , P ( B c ) = 6 10 = 3 5 , P ( A | B ) = 7 13 , P ( A | B c ) = 4 13 . a) P ( A ) = P ( A | B ) P ( B ) + P ( A | B c ) P ( B c ) = 7 13 · 2 5 + 4 13 · 3 5 = 2 5 = 0 . 4 . b) P ( B c | A ) = P ( A | B c ) P ( B c ) P ( A ) = 4 13 · 3 5 2 5 = 6 13 ≈ . 4615 . c) This time we have P ( B ) = w w + b , P ( B c ) = b w + b , P ( A | B ) = w + d w + d + b , P ( A | B c ) = w w + d + b . Therefore, P ( A ) = P ( A | B ) P ( B ) + P ( A | B c ) P ( B c ) = w + d w + d + b · w w + b + w w + d + b · b w + b = w ( w + d + b ) ( w + d + b )( w + b ) = w w + b . 2a. Sec. 1.5: #3: A manufacturing process produces integrated circuit chips. Over the long run the fraction of bad chips produced by the process is around 20%. Thoroughly testing a chip to determine whether it is good or bad is rather expensive, so a cheap test is tried. All good chips will pass the cheap test, but so will 10% of bad chips. a) Given a chip passes the cheap test, what is the probability that it is a good chip? b) If a company using this manufacturing process sells all chips which pass the cheap test, over the long run what percentage of chips sold will be bad? 1 Solution. Let B = { the chip is good } , A = { the chip passes the cheap test } . Note that B c = { the chip is bad } , A c = { the chip fails the cheap test } , and that P ( B ) = 0 . 8 , P ( B c ) = 0 . 2 , P ( A | B ) = 1 , P ( A | B c ) = 0 . 1 . a) P ( B | A ) = P ( A | B ) P ( B ) P ( A | B ) P ( B ) + P ( A | B c ) P ( B c ) = 1 · . 8 1 · . 8 + 0 . 1 · . 2 = 40 41 ≈ . 9756 ....
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This note was uploaded on 09/14/2009 for the course STAT 134 taught by Professor Aldous during the Fall '03 term at University of California, Berkeley.

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S2 - Math 4653 Elementary Probability Spring 2007 Homework#2 Problems and Solutions(corrected 1 Sec 1.5#2 Polyas urn scheme An urn contains 4 white

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