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Question:

Saransh, Akhil and Ronak start a game of dice. They throw the dice by turns, first Saransh, Akhil then Ronak, then once again Saransh, Akhil and Ronak and so on until one of them wins. Saransh is considered to have won if he throws 1 or 2 or 3, Akhil if he throws a prime number and Ronak if he throws 4 or 5 or 6. What is the probability that Saransh wins?

Explanation:

P(S) = P(Saransh getting 1 or 2 or 3) = $\frac{3}{6}$ = $\frac{1}{2}$
P(A) = P(Akhil getting 2 or 3 or 5) = $\frac{3}{6}$ = $\frac{1}{2}$
P(R) = P(Ronak getting 4 or 5 or 6) = $\frac{3}{6}$ = $\frac{1}{2}$

P(Saransh wins) =
P(Saransh gets 1 or 2 or 3 in his first turn) +
P(Saransh gets 1 or 2 or 3 in his second turn) +
P(Saransh gets 1 or 2 or 3 in his third turn) + …

For Saransh to win in his second turn, Akhil and Ronak must not win in their first turn. Same applies for Saransh winning in his further turns.

⇒ P(Saransh wins) = P(S) + P(S’) × P(A’) × P(R’) × P(S) + P(S’) × P(A’) × P(R’) × P(S’) × P(A’) × P(R’) × P(S) + …

= $\frac{1}{2}$ + $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$ + $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$ +⋯

= $\frac{1}{2}$ + ${(\frac{1}{2})}^{4}$ + ${(\frac{1}{2})}^{7}$ + ⋯ [Sum of an infinite GP]

= $\frac{\frac{1}{2}}{1 - \frac{1}{8}}$ = $\frac{4}{7}$ .

Hence, option (b).

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