Question

The probability of a person telling the truth is $\dfrac{4}{6}$. An unbiased die is thrown by the same person twice and the person reports that the numbers appeared in both the throws are the same. Then the probability that actually the numbers appeared in both the throws are same is ______________ (rounded off to 2 decimal places).

Hint:

Model “truth vs. lie” explicitly and apply Bayes’ rule: $P(A|R)=\dfrac{P(R|A)P(A)}{P(R|A)P(A)+P(R|A^c)P(A^c)}$ with $P(R|A)=P(T)$ and $P(R|A^c)=P(L)$ here.

Answer 1

Correct Answer: 0.26

Step 1: Define events and priors.
Let $A$ = “actual outcomes are the same”, $R$ = “reports same”. Let $T$ = “tells truth”, $L$ = “lies”. Given $P(T)=\tfrac{4}{6}=\tfrac{2}{3}$, $P(L)=\tfrac{1}{3}$. For two fair throws, $P(A)=\tfrac{6}{36}=\tfrac{1}{6}$, $P(A^c)=\tfrac{5}{6}$.
Step 2: Reporting model.
If truthful, the report equals the reality; if lying, the report is the opposite. Hence \[ P(R|A,T)=1,\; P(R|A^c,T)=0,\qquad P(R|A,L)=0,\; P(R|A^c,L)=1. \]
Step 3: Compute $P(R)$ and $P(A\cap R)$.
\[ P(R)=P(T)P(R|A,T)P(A)+P(L)P(R|A^c,L)P(A^c) =\frac{2}{3} . 1 . \frac{1}{6}+\frac{1}{3} . 1 . \frac{5}{6} =\frac{7}{18}. \] \[ P(A\cap R)=P(T)P(A)P(R|A,T)=\frac{2}{3} . \frac{1}{6} . 1=\frac{1}{9}. \]
Step 4: Bayes’ rule for the required probability.
\[ P(A\,|\,R)=\frac{P(A\cap R)}{P(R)}=\frac{\tfrac{1}{9}}{\tfrac{7}{18}}=\frac{2}{7}\approx 0.285714\ldots \] Rounded to two decimals: $0.29$. Final Answer: \[ \boxed{0.29} \]