MHTCET 20 Apr 2026 Shift 2 Paper

Concept:Use a Markov-chain recursion based on the previous roll type.Explanation:An $A$-type roll means 3 or 5, so $P(A\text{-type}) = \frac{2}{6} = \frac{1}{3}$.A $B$-type roll means 1, 2, 4, or 6, so $P(B\text{-type}) = \frac{4}{6} = \frac{2}{3}$.Let $x$ be the probability that A eventually wins when the previous roll was $A$-type.Let $y$ be the probability that A eventually wins when the previous roll was $B$-type.If the previous roll was $A$-type: with probability $\frac{1}{3}$, the next roll is $A$-type, so A wins immediately; with probability $\frac{2}{3}$, the next roll is $B$-type, and the state becomes $y$.Thus, $x = \frac{1}{3} + \frac{2}{3}y$.If the previous roll was $B$-type: with probability $\frac{1}{3}$, the next roll is $A$-type, moving to state $x$; with probability $\frac{2}{3}$, the next roll is $B$-type, so B wins and A gets probability $0$.Thus, $y = \frac{1}{3}x$.Substitute $y$ into the first equation:$$x = \frac{1}{3} + \frac{2}{3} \cdot \frac{1}{3}x = \frac{1}{3} + \frac{2}{9}x.$$So, $\frac{7}{9}x = \frac{1}{3}$, giving $x = \frac{3}{7}$.Then, $y = \frac{1}{3} \cdot \frac{3}{7} = \frac{1}{7}$.Before the first roll, there is no previous roll: the first roll is $A$-type with probability $\frac{1}{3}$, contributing $x$; it is $B$-type with probability $\frac{2}{3}$, contributing $y$.Therefore, required probability is$$\frac{1}{3} \cdot \frac{3}{7} + \frac{2}{3} \cdot \frac{1}{7} = \frac{1}{7} + \frac{2}{21} = \frac{5}{21}.$$Answer:$\frac{5}{21}$, i.e. Option B.