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Problem 6 Among a certain group of people 5%5 \% are (professional) liars. A li...
Sep 4, 2024
Solution by Steps
step 1
Let P(L) P(L) be the probability of being a liar, which is 0.05 0.05 (5%). Let P(T+∣L) P(T^+|L) be the probability of a positive test given that the person is a liar, which is 0.94 0.94 . Let P(T+∣N) P(T^+|N) be the probability of a positive test given that the person is not a liar, which is 0.08 0.08 . We need to find P(L∣T+) P(L|T^+) , the probability that a person is a liar given that the test is positive. We will use Bayes' theorem: P(L∣T+)=P(T+∣L)⋅P(L)P(T+) P(L|T^+) = \frac{P(T^+|L) \cdot P(L)}{P(T^+)}
step 2
To find P(T+) P(T^+) , we use the law of total probability: P(T+)=P(T+∣L)⋅P(L)+P(T+∣N)⋅P(N) P(T^+) = P(T^+|L) \cdot P(L) + P(T^+|N) \cdot P(N) where P(N)=1−P(L)=0.95 P(N) = 1 - P(L) = 0.95 . Thus, P(T+)=(0.94⋅0.05)+(0.08⋅0.95) P(T^+) = (0.94 \cdot 0.05) + (0.08 \cdot 0.95)
step 3
Calculating P(T+) P(T^+) : P(T+)=(0.94⋅0.05)+(0.08⋅0.95)=0.047+0.076=0.123 P(T^+) = (0.94 \cdot 0.05) + (0.08 \cdot 0.95) = 0.047 + 0.076 = 0.123
step 4
Now substituting back into Bayes' theorem: P(L∣T+)=0.94⋅0.050.123=0.0470.123≈0.382 P(L|T^+) = \frac{0.94 \cdot 0.05}{0.123} = \frac{0.047}{0.123} \approx 0.382
Answer
The probability that a randomly picked person is a liar given that the test is positive is approximately 0.382 0.382 or 38.2% 38.2\% .
Key Concept
Bayes' theorem is used to update the probability of a hypothesis based on new evidence.
Explanation
The answer shows how to calculate the probability of being a liar given a positive test result, taking into account the rates of false positives and the actual prevalence of liars in the population.
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