Question

Two persons \(A\) and \(B\) are alternately throwing two dice indefinitely. If \(A\) starts the game and the person who gets a prime number on one die and a composite number on the other for the first time wins the game, then the probability that \(B\) wins the game is

• A. \(\dfrac35\)
• B. \(\dfrac25\)
• C. \(\dfrac56\)
• D. \(\dfrac16\) \bigskip

Hint:

In alternating-turn games, if the probability of success in one turn is \(p\) and failure is \(q\), then the probability that the second player wins is \[ qp+q^3p+q^5p+\cdots = \frac{qp}{1-q^2}. \] This is a standard geometric series result.

Answer 1

The Correct Option is B

Concept: Model the game as a sequence of independent trials. B wins if A fails, then B succeeds; or both fail twice, then B succeeds; etc. Sum the infinite geometric series.

Step 1: Single trial success probability. Prime={2,3,5}, Composite={4,6}. Favorable: (P,C) or (C,P) = \(3\cdot2 + 2\cdot3 = 12\). Total = 36. \(p=\frac13, q=\frac23\).

Step 2: Probability B wins. \(P(B) = qp + q^3p + q^5p + \cdots = qp(1 + q^2 + q^4 + \cdots) = \frac{qp}{1-q^2}\).

Step 3: Substitute values. \(\frac{\frac23\cdot\frac13}{1-\frac49} = \frac{\frac29}{\frac59} = \frac25\).

Step 4: Write the final answer. \(\boxed{\frac25}\)