
Robert B. answered 09/10/17
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This seems to be a poorly constructed problem since the probability of being selected once is 5(.86)4(.14)≈0.382905712. This is the probability of not getting selected 4 times times the probability of getting selected once times the number of combinations which include getting selected once.
Based on the answers that are provided my guess is that the problem means to ask what the probability of getting selected at least once in 5 consecutive random drug tests. This is most easily computed by using the identity P(A)=1 − P(not A) since the probability of not getting selected is (1 − .14)5 = .865. So the probability of getting selected at least once is 1 − .865 ≈ 0.5295729824.
Based on the answers that are provided my guess is that the problem means to ask what the probability of getting selected at least once in 5 consecutive random drug tests. This is most easily computed by using the identity P(A)=1 − P(not A) since the probability of not getting selected is (1 − .14)5 = .865. So the probability of getting selected at least once is 1 − .865 ≈ 0.5295729824.