mont4e_sm_ch02_sec06

a 10 p a 1 p a 2 p a 10 1 10 0000976 2

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: - P(B). From part (a), P(B') = 1 - 0.59 = 0.41. 2-108. 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 2 ' ' ' 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⎠ () 2-109. (a) 3(0.2 ) =0.0048 (b) 3(4 * 0.2 * 0.8) =0.0768 3 4 2-110. (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 2-111. (a) The probability that one technician obtains equivalence at 100 mL is 0.1. So the probability that both technicia...
View Full Document

Ask a homework question - tutors are online