mont4e_sm_ch02_sec06

# 999 0002 2 107 it is useful to work one of these

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

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . 2-18 2-106. (a) P = (0.001) = 10 2 2 −6 (b) P = 1 − (0.999) = 0.002 2-107. It is useful to work one of these exercises with care to illustrate the laws of probability. Let Hi denote the event that the ith sample contains high levels of contamination. ' ' ' ' ' ' ' ' ' ' a) P(H1 ∩ H2 ∩ H3 ∩ H4 ∩ H5 ) = P(H1)P(H2 )P(H3 )P(H4 )P(H5 ) by independence. Also, P(Hi' ) = 0.9 . Therefore, the answer is 0.9 5 = 0.59 ' ' b) A1 = (H1 ∩ H'2 ∩ H'3 ∩ H4 ∩ H5 ) ' A 2 = (H1 ∩ H2 ∩ H'3 ∩ H'4 ∩ H'5 ) ' ' ' A 3 = (H1 ∩ H2 ∩ H3 ∩ H4 ∩ H'5 ) ' ' ' A 4 = (H1 ∩ H2 ∩ H3 ∩ H4 ∩ H'5 ) ' ' ' ' A 5 = (H1 ∩ H2 ∩ H3 ∩ H4 ∩ H5 ) The requested probability is the probability of the union A 1 ∪ A 2 ∪ A 3 ∪ A 4 ∪ A 5 and these events are mutually exclusive. Also, by independence P( A i ) = 0.9 4 (0.1) = 0.0656 . Therefore, the answer is 5(0.0656) = 0.328. c) Let B denote the event that no sample contains high levels of contamination. The requested probability is P(B') = 1...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online