sol9 - CS 70 Fall 2006 - Solutions to Homework #9 November...

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CS 70 Fall 2006 — Solutions to Homework #9 November 30, 2006 1. Marbles. One approach is to explicitly construct the probability space with The sample space Ω = { ( A, b 1 ) , ( A, b 2 ) , ( A, w 1 ) , . . . , ( A, w 5 ) , ( B, b 1 ) , ( B, w 1 ) , . . . , ( B, w 4 ) } ; and The correct probability on each point—probabilities of 1 14 and 1 10 on each point with box A and B respectively. (the events corresponding to box A and B must have total probability 0.5 each, and each sample point within each event must have equal probability summing to total probability of the event). Then we can calculate the probabilities directly from Frst principles as follows. The probability that the marble is black: Involves deFning the event B = { ( A, b 1 ) , ( A, b 2 ) , ( B, b 1 ) } that the marble is black and noticing that Pr ( B ) = X ω B Pr ( ω ) = Pr (( A, b 1 )) + Pr (( A, b 2 )) + Pr (( B, b 1 )) = 1 14 + 1 14 + 1 10 = 17 70 . The probability that the marble came from A, given that it is white: Involves the deFnition of condi- tional probability: let A = { ( A, b 1 ) , ( A, b 2 ) , ( A, w 1 ) , . . . , ( A, w 5 ) } be the event that the marble came from box A and W = { ( A, w 1 ) , . . . , ( A, w 5 ) , ( B, w 1 ) , . . . , ( B, w 4 ) } be the event that the marble is white; then Pr ( A | W ) = Pr ( A W ) Pr ( W ) = ω A W Pr ( ω ) ω W Pr ( ω ) = Pr (( A, w 1 )) + . . . + Pr (( A, w 5 )) Pr (( A, w 1 )) + . . . + Pr (( A, w 5 )) + Pr (( B, w 1 )) + . . . + Pr (( B, w 4 )) = 5 / 14 (5 / 14) + (4 / 10) = 25 53 . 1
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Another approach is to acknowledge the existence of the underlying sample space but not bother with the probabilities on each sample point, rather write down the ‘interesting’ events and the various probabilities given to us directly: The sample space Ω = { ( A, b 1 ) , ( A, b 2 ) , ( A, w 1 ) , . . . , ( A, w 5 ) , ( B, b 1 ) , ( B, w 1 ) , . . . , ( B, w 4 ) } ; Let A = { ( A, b 1 ) , ( A, b 2 ) , ( A, w 1 ) , . . . , ( A, w 5 ) } be the event that the marble came from box A and let W = { ( A, w 1 ) , . . . , ( A, w 5 ) , ( B, w 1 ) , . . . , ( B, w 4 ) } be the event that the marble is white (NB: the complement ¬ A corresponds to box B and ¬ W corresponds to black marble); then We are told that Pr ( A ) = 1 2 , Pr ( W | A ) = 5 7 and Pr ( W A ) = 4 5 . We can then transform the probabilities we know to get those we want: The probability that the marble is black: Use the fact that Pr ( ¬ W ) = 1 - Pr ( W ) and that the sets A and ¬ A partition 1 Ω: Pr ( W ) = Pr ( W | A ) Pr ( A ) + Pr ( W A ) Pr ( ¬ A ) = 5 7 · 1 2 + 4 5 · 1 2 = 53 70 . The probability that the marble came from A, given that it is white: The same ideas apply here.
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This note was uploaded on 09/14/2009 for the course CMPSC 360 taught by Professor Haullgren during the Fall '08 term at Pennsylvania State University, University Park.

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sol9 - CS 70 Fall 2006 - Solutions to Homework #9 November...

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