Question

A person, who speaks the truth 3 out of 4 times, throws a fair die with six faces and informs that the outcome is 5. The probability that the outcome is really 5 is \(\underline{\hspace{2cm}}\) (round off to three decimal places).

Hint:

For conditional probability, Bayes' Theorem is a useful tool to calculate the probability of an event based on prior knowledge and observed information.

Answer 1

Correct Answer: 0.375

This is a problem of conditional probability. We need to find the probability that the outcome is really 5, given that the person informed that the outcome is 5. Let the events be defined as:
- \( A \) : The outcome is 5.
- \( B \) : The person informs that the outcome is 5.
We need to find \( P(A|B) \), which is the probability that the outcome is 5, given that the person informed that it is 5. Using Bayes' Theorem, we have: \[ P(A|B) = \frac{P(B|A)P(A)}{P(B)} \] Where:
- \( P(B|A) \) is the probability that the person informs 5 when the outcome is actually 5. Since the person tells the truth 3 out of 4 times, \( P(B|A) = \frac{3}{4} \).
- \( P(A) \) is the probability that the outcome is 5, which is \( \frac{1}{6} \) (since it is a fair die).
- \( P(B) \) is the total probability that the person informs 5. This can be split into two cases: the person tells the truth or lies.
- The probability that the person informs 5 and the outcome is indeed 5 is \( P(B|A)P(A) = \frac{3}{4} \times \frac{1}{6} \).
- The probability that the person lies and informs 5 while the outcome is not 5 is \( P(B|A^c)P(A^c) = \frac{1}{4} \times \frac{5}{6} \).
So, \( P(B) \) is: \[ P(B) = \frac{3}{4} \times \frac{1}{6} + \frac{1}{4} \times \frac{5}{6} \] Now, substitute these values into Bayes' Theorem: \[ P(A|B) = \frac{\frac{3}{4} \times \frac{1}{6}}{\frac{3}{4} \times \frac{1}{6} + \frac{1}{4} \times \frac{5}{6}} \] Simplifying: \[ P(A|B) = \frac{\frac{3}{24}}{\frac{3}{24} + \frac{5}{24}} = \frac{3}{8} = 0.375 \] Thus, the probability that the outcome is really 5 is \( 0.375 \).