A researcher says that there is a 80% chance a polygraph test (lie detector test

Question

A researcher says that there is a 80% chance a polygraph test (lie detector test) will catch a person who is, in fact, lying. Furthermore, there is approximately a 9% chance that the polygraph will falsely accuse someone of lying. (a) If the polygraph indicated that 30% of the questions were answered with lies, what would you estimate for the actual percentage of lies in the answers? Hint: Let B = event detector indicates a lie. We are given P(B) = 0.3. Let A = event person is lying, so Ac = event person is not lying. Then

P(B) = P(A and B) + P(Ac and B)
P(B) = P(A)P(B | A) + P(Ac)P(B | Ac)

Replacing P(Ac) by 1 P(A) gives

P(B) = P(A) · P(B | A) + [1 P(A)] · P(B | Ac)

Substitute known values for P(B), P(B | A), and P(B | Ac) into the last equation and solve for P(A). (Round your answer to two decimal places.)

a.) P(A) =

(b) If the polygraph indicated that 70% of the questions were answered with lies, what would you estimate for the actual percentage of lies? (Round your answer to one decimal place.)

Explanation / Answer

a)

From the given information we have

P(B) = 0.3, P(B|A) = 0.80, P(B|Ac) = 0.09

Putting values in the equation

P(B) = P(A) · P(B | A) + [1 P(A)] · P(B | Ac) gives

0.3 = P(A) * 0.80 + [1-P(A)] * 0.09

0.3 = 0.8 P(A) +0.09 - 0.09P(A)

0.21 = 0.71 P(A)

P(A) = 0.296

So requried percentage is 29.6%

b)

From the given information we have

P(B) = 0.7, P(B|A) = 0.80, P(B|Ac) = 0.09

Putting values in the equation

P(B) = P(A) · P(B | A) + [1 P(A)] · P(B | Ac) gives

0.7 = P(A) * 0.80 + [1-P(A)] * 0.09

0.7 = 0.8 P(A) +0.09 - 0.09P(A)

0.61 = 0.71 P(A)

P(A) = 0.859

So required percentage is 85.9%.

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