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 (  NE [ 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 (  NE [ 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.
 Spring '11
 MichealAlain
 Math

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