Question

Two persons A and B throw a pair of dice alternately until one of them gets the sum of the numbers appeared on the dice as 4 and the person who gets this result first is declared as the winner. If A starts the game, then the probability that B wins the game is:

• A. \( \frac{11}{23} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{5}{11} \)
• D. \( \frac{8}{17} \)

Hint:

For solving problems involving alternating events, break the problem into stages, find the probability of each event happening, and use geometric series when the events repeat in cycles.

Answer 1

The Correct Option is A

Step 1: Find the probability of getting a sum of 4. The possible outcomes for the sum of two dice to be 4 are: \[ (1, 3), (2, 2), (3, 1). \] Thus, the probability of getting a sum of 4 in one roll is: \[ P(\text{sum} = 4) = \frac{3}{36} = \frac{1}{12}. \] Step 2: Calculate the probability of B winning. The probability that B wins is the sum of the probabilities for B winning on subsequent turns. This is a geometric series, where: \[ P(\text{B wins}) = \frac{11}{12} \times \frac{1}{12} + \left( \frac{11}{12} \right)^2 \times \frac{1}{12} + \left( \frac{11}{12} \right)^3 \times \frac{1}{12} + \cdots \] Step 3: Sum the geometric series. Using the formula for the sum of an infinite geometric series, we get: \[ P(\text{B wins}) = \frac{\frac{11}{144}}{1 - \frac{11}{12}} = \frac{11}{23}. \]