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: Set up the two possibilities.
Let event $S$ be the die truly showing a 6, and $N$ be the die showing something other than 6. We want the chance it really is a 6 given the man reported a 6.
Step 2: Prior probabilities.
On a fair die, $P(S)=\dfrac{1}{6}$ and $P(N)=\dfrac{5}{6}$.
Step 3: Reporting reliability.
He tells the truth $\dfrac34$ of the time. If it really is a 6, he truthfully says 6: $P(\text{says }6\mid S)=\dfrac34$. If it is not a 6, he would have to lie and call it a 6: $P(\text{says }6\mid N)=\dfrac14$.
Step 4: Weight each branch.
True-6 branch: $\dfrac16\cdot\dfrac34=\dfrac{3}{24}$. False-report branch: $\dfrac56\cdot\dfrac14=\dfrac{5}{24}$.
Step 5: Apply Bayes' rule.
$P(S\mid\text{says }6)=\dfrac{\frac{3}{24}}{\frac{3}{24}+\frac{5}{24}}$. The common denominator 24 cancels.
Step 6: Simplify.
$P=\dfrac{3}{3+5}=\dfrac{3}{8}$, matching option (3).
\[ \boxed{\dfrac{3}{8}} \]