solution05 - Solutions to Homework 05 1. a. P [ace in first...

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Unformatted text preview: Solutions to Homework 05 1. a. P [ace in first draw] = 4 52 . b. Suppose the first draw is seen to be an ace. Then, we now have 3 aces remaining in the 51 cards. Thus, P [ace in second draw | ace in first draw] = 3 51 If the the first draw is not an ace, we will have 4 aces remaining in the 51 cards. Thus P [ace in second draw | no ace in first draw] = 4 51 If we do not look at the first draw, then, using the theorem on total probability P [ace in second draw] = P [ace in second draw | ace in first draw] P [ace in first draw] + P [ace in second draw | no ace in first draw] P [no ace in first draw] = 3 51 4 52 + 4 51 48 52 = 12 + 192 52 51 = 4 52 = P [ace in first draw] Thus the probability of an ace depends on the observations available. c. The total number of way to choose 7 cards out of a deck of 52 cards is given by parenleftBigg 52 7 parenrightBigg . We need 3 aces in these 7 cards out of a total of 4 possible aces. The number of ways to draw 3 aces out of 4 is given by parenleftBigg 4 3 parenrightBigg . The remaining 4 cards of the 7 are drawn from the 48 non-ace card which results in parenleftBigg 48 4 parenrightBigg possibilities. Then, P [3 aces in 7 cards] =...
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This note was uploaded on 09/03/2008 for the course ECSE 4510 taught by Professor Woods during the Spring '03 term at Rensselaer Polytechnic Institute.

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solution05 - Solutions to Homework 05 1. a. P [ace in first...

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