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Unformatted text preview: i th missile (with i ∈ { 1 , 2 , 3 } ). Then, by independence, the probability of a hit is P ( H 1 ∪ H 2 ∪ H 3 ) = P ( H 1 ) + P ( H 2 ) + P ( H 3 ) − P ( H 1 H 2 ) − P ( H 1 H 3 ) − P ( H 2 H 3 ) + P ( H 1 H 2 H 3 ) = . 7 + . 8 + . 9 − P ( H 1 ) P ( H 2 ) − P ( H 1 ) P ( H 3 ) − P ( H 2 ) P ( H 3 ) + P ( H 1 ) P ( H 2 ) P ( H 3 ) = 2 . 4 − . 56 − . 63 − . 72 + . 504 = . 994 . 29. Let E i be the event that the i th switch is closed. Then, for a signal to get through, we need one of the following events to occur: E 1 E 2 E 4 E 6 or E 1 E 3 E 5 E 6 . By the independence of the switches we have P ( E 1 E 2 E 4 E 6 ∪ E 1 E 3 E 5 E 6 ) = P ( E 1 E 2 E 4 E 6 ) + P ( E 1 E 3 E 5 E 6 ) − P ( E 1 ··· E 6 ) = 2 p 4 − p 6 . 1...
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This note was uploaded on 03/31/2008 for the course MATH 331 taught by Professor Anderson during the Spring '08 term at Wisconsin.
 Spring '08
 Anderson
 Math, Probability

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