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Unformatted text preview: CS 2800 Fall 2008 Homework #7 Solution Section 6.1 Problem 38 [5pts] (a) p ( E 1 ) = 1 / 2. p ( E 2 ) = 1 / 2. E 1 E 2 is the event that the first coin comes up heads and the second coin comes up heads. p ( E 1 E 2 ) = 1 / 2 1 / 2 = 1 / 4 = p ( E 1 ) p ( E 2 ). Thus, the two events are independent. (b) p ( E 1 ) = 1 / 2. There are only 2 configurations in which exactly 2 heads can come up in a row HHT, THH. Thus, p ( E 2 ) = 1 / 8 + 1 / 8 = 1 / 4. E 1 E 2 is the event that the first coin comes up tails and two but not three heads come up in a row this corresponds to the configuration THH. Thus, p ( E 1 E 2 ) = 1 / 8 = 1 / 4 1 / 2 = p ( E 1 ) p ( E 2 ). The two events are therefore independent. (c) p ( E 1 ) = 1 / 2. As in the previous part, p ( E 2 ) = 1 / 4. E 1 E 2 is the event that the second coin comes up tails and two, but not three, heads come up a in a row this event is impossible and thus p ( E 1 E 2 ) = 0 6 = p ( E 1 ) P ( E 2 ). Therefore, E 1 and E 2 are not independent. Problem 40 [5pts] The chances of winning with the original selection are 1 / 4. The hosts action does not change this, just like in the original problem, since his actions are the same regardless of whether...
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 '07
 SELMAN

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