Unformatted text preview:  P(B). From part (a), P(B') = 1  0.59 = 0.41. 2108. Let A i denote the event that the ith bit is a one. a) By independence P( A 1 ∩ A 2 ∩...∩ A 10 ) = P( A 1 )P( A 2 )...P( A 10 ) = ( 1 )10 = 0.000976
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' ' ' c b) By independence, P ( A 1 ∩ A '2 ∩... ∩ A 10 ) = P ( A 1 ) P ( A '2 )... P ( A 10 ) = ( 1 ) 10 = 0. 000976 2 c) The probability of the following sequence is
1 ' ' ' ' ' P( A 1 ∩ A 2 ∩ A 3 ∩ A 4 ∩ A 5 ∩ A 6 ∩ A 7 ∩ A 8 ∩ A 9 ∩ A10 ) = ( )10 , by independence. The number of 2
10 10 ! sequences consisting of five "1"'s, and five "0"'s is 10 = = 252 . The answer is 252⎛ 1 ⎞ = 0.246 ⎜⎟ 5 5! 5! ⎝ 2⎠ () 2109. (a) 3(0.2 ) =0.0048 (b) 3(4 * 0.2 * 0.8) =0.0768
3 4 2110. (a) P = (0.8) = 0.4096
4 (b) P = 1 − 0.2 − 0.8 × 0.2 = 0.64 (c) Probability defeats all four in a game = 0.84 = 0.4096. Probability defeats all four at least once = 1 – (1 – 0.4096)3 = 0.7942 2111. (a) The probability that one technician obtains equivalence at 100 mL is 0.1. So the probability that both technicia...
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This note was uploaded on 02/17/2010 for the course IE 33214 taught by Professor Feng during the Spring '10 term at University of Illinois at Urbana–Champaign.
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