2.10.6 - Modification of the Example on Slide 6 of Section...

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Modification of the Example on Slide 6 of Section 2.10 We let: A = event the toss resulted in heads B 1 = event the coin was fair (call this a type 1 coin) B 2 = event the coin was biased (call this a type 2 coin) B 3 = event the coin was two-headed (call this a type 3 coin). From the given information we were able to directly (and easily) calculate the probabilities on the right above, and then we could use the Law of Total Probability to find the desired P ( A ): P ( A ) = P ( B 1 ) P ( A | B 1 ) + P ( B 2 ) P ( A | B 2 ) + P ( B 3 ) P ( A | B 3 ) = (5/12)(1/2) + (4/12)(1/3) + (3/12)(1) = (5/24) + (4/36) + (3/12) = 41/72 Consider the following modification to the problem. Suppose that two coins are selected from the bag without replacement and tossed. What is the probability that both tosses were heads? There are (at least) two approaches that we can take. The first approach is to let S be the sample space of all possible outcomes from tossing two coins, keeping track of the type of coin selected on each draw and the result of its toss. Then partition the sample space S into 9 parts, B ij ( i = 1, 2, 3; j = 1, 2, 3), where event B ij occurs if the first coin was type i
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This note was uploaded on 02/23/2011 for the course MATH 444 taught by Professor Any during the Fall '10 term at Roosevelt.

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2.10.6 - Modification of the Example on Slide 6 of Section...

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