GUJCET 2025 Math Paper

Concept:Use Bayes' theorem to reverse the conditional probability after the man reports a six.Explanation:Let $E$ be the event that the die actually shows six.Then $P(E)=\frac{1}{6}$ and $P(E')=\frac{5}{6}$.Let $A$ be the event that the man reports a six.He speaks truth $\frac{4}{5}$ of the time, so $P(A|E)=\frac{4}{5}$.When the die is not six, reporting six is a lie, so $P(A|E')=1-\frac{4}{5}=\frac{1}{5}$.Using Bayes' theorem:$$P(E|A)=\frac{P(A|E)P(E)}{P(A|E)P(E)+P(A|E')P(E')}$$Substitute the values:$$P(E|A)=\frac{\frac{4}{5}\cdot\frac{1}{6}}{\frac{4}{5}\cdot\frac{1}{6}+\frac{1}{5}\cdot\frac{5}{6}}$$Simplify:$$P(E|A)=\frac{\frac{4}{30}}{\frac{4}{30}+\frac{5}{30}}=\frac{4}{9}$$Answer:$\frac{4}{9}$So option B is correct.