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Unformatted text preview: ( D 1 âˆ© G 2 âˆ© D 3 ) P ( D 1 âˆ© G 2 ) = 5 40 Â· 35 39 Â· 4 38 5 40 Â· 35 39 = 4 / 38 = 2 / 19 . 2. The probability that a lab specimen contains high levels of contamination is 0 . 05. Four samples are checked, and the samples are independent. (a) What is the probability that none contains high levels of contamination? (b) What is the probability that exactly one contains high levels of contamination? (c) What is the probability that at least one contains high levels of contamination? Solution . Let H i denote the event that the i th sample contains high levels of con tamination for i = 1 , 2 , 3 , 4. (a) The event that none contains high levels of contamination is equivalent to H 1 âˆ© H 2 âˆ© H 3 âˆ© H 4 . So, by independence, the desired probability is P ( H 1 âˆ© H 2 âˆ© H 3 âˆ© H 4 ) = P ( H 1 ) P ( H 2 ) P ( H 3 ) P ( H 4 ) = (1 . 05) 4 = 0 . 8145 . (b) Let A 1 = ( H 1 âˆ© H 2 âˆ© H 3 âˆ© H 4 ), A 2 = ( H 1 âˆ© H 2 âˆ© H 3 âˆ© H 4 ), A 3 = ( H 1 âˆ© H 2 âˆ© H 3 âˆ© H 4 ), 1 A 4 = ( H 1 âˆ© H 2 âˆ© H 3 âˆ© H 4 ). Then, the requested probability is the probability of the union A 1 âˆª A 2 âˆª A 3 âˆª A 4 and these events are mutually exclusive. Also, by independence, P ( A i ) = (0 . 95) 3 (0 . 05) = . 0429 ,i = 1 , 2 , 3 , 4 . Therefore, the answer is 4(0 . 0429) = 0 . 1716 . (c). Let B be the event that no sample contains high levels of contamination. The event that at least one contains high levels of contamination is the complement of B , i.e. B . By part (a), it is known that P ( B ) = 0 . 8145. So, the requested probability is P ( B ) = 1 P ( B ) = 1 . 8145 = 0 . 1855. 2...
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 Summer '08
 STAFF
 Statistics, Conditional Probability, Probability, HMS H2, G protein coupled receptors, HMS H1

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