suppose that any pair of tests are independent. What is the probability that a machine will conclude that
a each of three employees is lying when all are telling the truth?
b at least one of the three employees is lying when all are telling the truth?
Answer:
a. The probability that each of three employees is lying when all are telling the truth is 0.000125
b. The probability that at least one of the three employees is lying when all are telling the truth is 0.1426
Step-by-step explanation:
Let the event be;
A= the lie detector concludes that a person is lying who, in fact, is telling the truth
B= the lie detector concludes that a person is lying who, in fact, is telling lie
Given their probabilities as:
P(A)= 0.05 and P(B)= (1 - 0.05)= 0.95
a. What is the probability that a machine will conclude that each of three employees is lying when all are telling the truth?
A binomial problem with n= 3 and P= 0.05
P(all are telling truth) = P^n= 0.05^3 = 0.000125
b. What is the probability that at least one of the three employees is lying when all are telling the truth?
P(at least one is telling) = P(one is telling lie) + P(two are telling lie) + P(three are telling lie)
Or
P(at least one is telling) = 1 - P(B) = 1 - (0.95)^3 = 0.1426
