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Unformatted text preview: 1. Let A denotes the event that two dice land on different numbers, and B denotes the event that one lands on 6. So, what we need to compute is p ( B  A ). By formula, p ( B  A ) = p ( AB ) p ( A ) . p ( AB ) = 1 * 5 6 * 6 + 1 * 5 6 * 6 , and p ( A ) = 6 * 5 6 * 6 . Hence, p ( B  A ) = 1 * 5 6 * 6 + 1 * 5 6 * 6 6 * 5 6 * 6 = 1 3 . 2. Let A denotes the event that the first dice lands on 6. Let E i denotes the event that the sum of the two dice is i . We need to compute p ( A  E i ) for i ∈ [2 , 12] . By formula, p ( A  E i ) = p ( AE i ) p ( E i ) . When i ≤ 6, p ( AE i ) = 0 because it is impossible for the sum of the two dice is less than 6 when the first one already achieves 6; When i = 7, then the possible cases to get the sum of the two dice equals 7 is { (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) } . Then, p ( AE 7 ) = 1 * 1 6 * 6 , and p ( E 7 ) = 6 * 1 * 1 6 * 6 . Finally, p ( A  E 7 ) = 1 6 ; When i = 8, then the possible cases to get the sum of the two dice equals 8 is { (2,6),(3,5),(4,4),(5,3),(6,2) } . Then, p ( AE 8 ) = 1 * 1 6 * 6 , and p ( E 8 ) = 5 * 1 * 1 6 * 6 ....
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 Fall '08
 G.JOGESHBABU
 Probability, Summation, possible cases, E2 E3

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