# 55PS8 - | S 1 = 1 p R 1 | S 1 = 9 Moreover we know p S 0 =...

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SOLUTION SET 8—DUE 3/12/2008 This solution set is currently only partially complete, to allow a post before the midterm. Since I appear to have left my copy of the text in a locked oﬃce in a locked building 35 miles away, I’ll write solutions from memory. The problems may be a little oﬀ, but should be broadly right. Please report any errors in this document to Ian Sammis ( [email protected] ). Problem 1 (#6.3.????) . A Neptune probe sends 0 or 1: 1 with probability 1 3 , and 1 with probability 2 3 . If a 1 is sent, a 1 is received with probability .9 and a 0 with probability .1; if a 0 is sent, a 0 is recieved with probability 0.8 and a 1 with probability .2. Given that a 1 is received, what’s the probability that a 1 was sent? Solution. Let S 0 ,S 1 be the events that a 0 (resp.1 ) was sent, and let R 0 ,R 1 be the events that a 0 (resp. 1) was received. We have p ( R 0 | S 0) = . 8; p ( R 1 | S 0) = . 2; p ( R
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Unformatted text preview: | S 1) = . 1; p ( R 1 | S 1) = . 9 . Moreover, we know p ( S 0) = 2 3 and p ( S 1) = 1 3 . Thus p ( S 1 | R 1) = p ( R 1 | S 1) p ( S 1) p ( R 1 | S 1) p ( S 1) + p ( R 1 | S 0) p ( S 0) = . 9( 1 3 ) . 9( 1 3 ) + . 2( 2 3 ) ≈ . 69 Problem 2 (#6.4.????b) . A factory produces cans of soda. The expected number of cans is 10,000; the variance is 1000. Use Chebyshev’s inequality to write a lower bound on the probability that the number of cans produced is between 9000 and 11000. Solution. Chebyshev’s inequality states that P ( | N-E [ N ] | ≥ r ) ≤ V [ N ] /r 2 . In this case we are interested in r = 1000 (properly speaking, 1001, but let’s make the math simple, so P ( | N-E [ N ] | ≥ r ) ≤ 1 / 1000. Thus the probability of the number of cans produced being between 9000 and 11000 is at least 0.999. 1...
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## This note was uploaded on 12/07/2011 for the course MATH MATH192 taught by Professor Michealalain during the Spring '11 term at Université Joseph Fourier Grenoble I.

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