Question

A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is 6. Then the probability that it is actually 6 is

• A. $\frac{3}{4}$
• B. $\frac{1}{4}$
• C. $\frac{3}{8}$
• D. $\frac{5}{6}$

Hint:

Think of this conceptually using sample paths! Out of 24 total hypothetical throws:
A 6 shows up 4 times. The man tells the truth $\frac{3}{4}$ of those times $\rightarrow$ 3 cases where he correctly reports 6.
A non-6 shows up 20 times. The man lies $\frac{1}{4}$ of those times $\rightarrow$ 5 cases where he falsely reports 6.
Total times he says "6" is $3 + 5 = 8$. The actual 6 cases represent exactly $\frac{3}{8}$ of that total!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem presents a conditional probability scenario. A man reports that a rolled die landed on a 6. Given his known truth-telling track record, we need to find the posterior probability that the die actually shows a 6.

Step 2: Key Formula or Approach:
This problem can be resolved using Bayes' Theorem. Let's define the operational events:
Let $E_1$: The event that the die actually rolls a 6.
Let $E_2$: The event that the die does not roll a 6 ($\sim 6$).
Let $A$: The event that the man reports seeing a 6.
The conditional formula to find the probability that it is a 6 given his report is:
$$P(E_1|A) = \frac{P(E_1) \times P(A|E_1)}{P(E_1) \times P(A|E_1) + P(E_2) \times P(A|E_2)}$$

Step 3: Detailed Explanation:
Let's find the numerical probabilities for each component step-by-step:
1. Base rolling probabilities for a standard fair six-sided die:
$$P(E_1) = \frac{1}{6} \quad (\text{probability of rolling a } 6)$$ $$P(E_2) = 1 - \frac{1}{6} = \frac{5}{6} \quad (\text{probability of rolling anything else})$$ 2. Conditional reporting probabilities based on his truthfulness:
$P(A|E_1)$: The probability he reports a 6 when it truly is a 6. This means he is telling the truth:
$$P(A|E_1) = \frac{3}{4}$$ $P(A|E_2)$: The probability he reports a 6 when it is actually a different number. This means he is lying:
$$P(A|E_2) = 1 - \frac{3}{4} = \frac{1}{4}$$ Now, substitute these fractions into our Bayes' Theorem expression:
$$P(E_1|A) = \frac{\frac{1}{6} \times \frac{3}{4}}{\left(\frac{1}{6} \times \frac{3}{4}\right) + \left(\frac{5}{6} \times \frac{1}{4}\right)}$$ $$P(E_1|A) = \frac{\frac{3}{24}}{\frac{3}{24} + \frac{5}{24}}$$ Since the denominators are identical, they cancel out perfectly:
$$P(E_1|A) = \frac{3}{3 + 5} = \frac{3}{8}$$ This matches option (C).

Step 4: Final Answer:
The probability that the die is actually 6 is $\frac{3}{8}$, which corresponds to option (C).