JEE Main 2026 — 8 April, Evening Shift — Question 25

Let $B$: event that candidate travels by bus, $S$: by scooter, $C$: by car. Given $P(B)=\frac{2}{5},\; P(S)=\frac{1}{5},\; P(C)=\frac{2}{5}$. Let $L$: event of reaching late. Given $P(L|B)=\frac{1}{5},\; P(L|S)=\frac{1}{3},\; P(L|C)=\frac{1}{4}$.

By Bayes' theorem, $P(B|L)=\frac{P(B)P(L|B)}{P(B)P(L|B)+P(S)P(L|S)+P(C)P(L|C)}$.

Numerator: $\frac{2}{5}\times\frac{1}{5}=\frac{2}{25}$.

Denominator: $\frac{2}{5}\times\frac{1}{5}+\frac{1}{5}\times\frac{1}{3}+\frac{2}{5}\times\frac{1}{4}=\frac{2}{25}+\frac{1}{15}+\frac{1}{10}$.

Convert to common denominator 150: $\frac{12}{150}+\frac{10}{150}+\frac{15}{150}=\frac{37}{150}$.

Thus $P(B|L)=\frac{2/25}{37/150}=\frac{2}{25}\times\frac{150}{37}=\frac{12}{37}$.