mont4e_sm_ch02_sec06

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

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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...
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