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A policy requiring all hospital employees to take lie detector tests may reduce losses due to theft, but some employees regard such tests as a violation of their rights. To gain some insight into the risks that employees face when taking a lie detector test, suppose that the probability is 0.06 that a lie detector concludes that a person is lying who, in fact, is telling the truth and suppose that any pair of tests are independent.

What is the probability that a machine will conclude that each of three employees is lying when all are telling the truth?

For this one I did (0.06)^3

What is the probability that the machine will conclude that none of the employees is lying when all are telling the truth?

For this one I did (0.94)^3

What is the probability that a machine will conclude that at least one of the three employees is lying when all are telling the truth?

For this one I did (0.94)^2(0.6)

I am not 100% sure if I am doing this right. Also, I am pretty bad with probability.

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  • $\begingroup$ For the last question, easiest to find $1-(0.94)^3$. Your answer is not correct, for your way we would need to add up the probabilities it will conclude $1$ is lying, $2$ are lying, $3$ are lying. The probability it will conclude exactly one is lying is $(3)(0.94)^2(0.06)$. For $2$ lying it is $(3)(0.94)(0.06)^2$, and you already did $3$. Now add up. But that's the hard way, the easy way is $1-(0.94)^3$. $\endgroup$ – André Nicolas Feb 15 '14 at 20:03
  • $\begingroup$ @AndréNicolas Ahhh I see that makes so much sense thanks!!! $\endgroup$ – user125627 Feb 15 '14 at 20:08
  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Feb 15 '14 at 20:09
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For the first one, the probability is $$ \frac{P(\text{detect lying, telling truth})^3}{P(\text{telling truth})^3}=0.06^3 $$ Similarly the second one is $$ \frac{P(\text{detect telling truth, telling truth})^3}{P(\text{telling truth})^3}=0.94^3 $$ So you got the first two right. But for the third one, it should be $$ 1-P(\text{detect no on is lying|all are telling the truth})=1-0.94^3 $$

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Your first two answers look good. However, for the last one, you're calculating the probability that exactly one person — and a specific person at that — is erroneously said to be lying and the other two are telling the truth.

Instead, note the relationship between the third question and the one that immediately precedes it. If it's not the case that the polygraph concluded that all employees were telling the truth, then what did it conclude? In terms of probability, what's the relationship between the event described in the third question and the one in the second?

Let me know if you need another hint.

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Instead of trying to evaluate for exactly one liar, just assume they're all being truthful and take the complement of that. That way you'll know how likely it is that even one of them lied. In probability you can usually easily what find what you don't know based on what you do know.

let A = no lies detected let B = telling truth let C = at least one employee truthful

P(C) = 1 - P(A | B) = 1 - (0.94)^3 with "1' assuming full probability

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