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Unformatted text preview: Basic Probability: Problem Set One Summer 2011 1.3.1 We have A B B P ( A B ) P ( B ) = 1 3 . We also have from the inclusionexclusion principle that P ( A B ) = P ( A ) + P ( B ) P ( A B ) = 13 12 P ( A B ) 13 12 1 = 1 12 since P ( A B ) 1. For examples of attaining each bound, let = { 1 , 2 ,..., 12 } let F = P (), and let P ( ) = 1 12 for each . Further, let A = {  9 } . When B = {  9 } , P ( A B ) = 1 12 and for B = {  4 } , P ( A B ) = 1 3 . In each case, P ( A ) = 3 4 and P ( B ) = 1 3 . For comparable bounds on P ( A B ), we have A A B so 3 4 = P ( A ) P ( A B ). Also, P ( A B ) = 13 12 P ( A B ) 13 12 1 12 = 1 since P ( A B ) 1 12 . 1.4.3 Let 0, 1, and 2 be the events that the coin has 0, 1, and 2 heads, respectively. Let H L (resp. H U ) be the events that the lower (upper) side is heads. Note that H U H L = 2 (the event 2) and P ( H L ) = P ( H U ) P ( H L ) = P ( H L  0) P (0) + P ( H L  1) P (1) + P ( H L  2) P (2) = 0 1 5 + 1 2 2 5 + 1 2 5 = 3 5 P ( H L  H U ) = P ( H L H U ) P ( H U ) = P (2) P ( H L ) = 2 / 5 3 / 5 = 2 3 1 Let H U H L be the event that the first toss is heads up and the second is heads down. P ( H U H L  H U ) = P ( H U H L H U ) P ( H U ) = P ( H U H L ) 3 / 5 = 5 3 ( P ( H U H L  0) P (0) + P ( H U H L  1) P (1) + P ( H U H L  2) P (2)) = 5 3 1 5 + 1 4 2 5 + 1 2 5 = 5 6 P ( H U H L  H U H U ) = P ( H U H L H U H U ) P ( H U H U ) = P (2) P ( H U H U ) = P (2) P ( H U H L ) = 2 / 5 1 / 2 = 4 5 We do the final question in two parts. First, we condition on what the first coin was. First, suppose the first coin was 1 (it cannot have been 0). P ( H U H U H U  1 H U H U ) = P ( H U H U H U  10 H U H U ) P (10 H U H U ) + P ( H U H U H U  11 H U H U ) P (11 H U H U ) + P ( H U H U H U  12 H U H U ) P (12 H U H U ) = 0 1 4 + 1 2 1 4 + 1 1 2 = 5 8 Next, suppose it was 2. P ( H U H U H U  1 H U H U ) = P ( H U H U H U  20 H U H U ) P (20 H U H U ) + P ( H U H U H U  21 H U H U ) P (21 H U H U ) + P ( H U H U H U  22 H U H U ) P (22 H U H U ) = 0 1 4 + 1 2 1 2 + 1 1 4 = 1 2 2 We have P (2  H U H U ) = P ( H U H L  H U H U ) = 4 5 , and since P (0  H U H U ) = 0, we must have P (1  H U H U ) = 1 5 Finally, combining the results: P ( H U H U H U  1 H U H U ) = P ( H U H U H U  1 H U H U ) P (1  H U H U ) + P ( H U H U H U  2 H U H U ) P (2  H U...
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This note was uploaded on 10/28/2011 for the course MATH 795 taught by Professor Thompson during the Spring '11 term at CUNY Hunter.
 Spring '11
 Thompson
 Algebra, Algebraic Topology, Probability

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