Question

Two players A and B are alternately throwing a coin and a die together. A player who first throws a head and a 6 wins the game. If A starts the game, then the probability that B wins the game is:

• A. \( \frac{12}{23} \)
• B. \( \frac{11}{23} \)
• C. \( \frac{5}{119} \)
• D. \( \frac{12}{119} \)

Hint:

In problems involving alternating turns and infinite sequences, identify the probabilities of the sequence of events and use the sum of an infinite geometric series to solve for the desired outcome.

Answer 1

The Correct Option is B

To solve this problem, let's first calculate the probability that both a head on the coin and a 6 on the die occur in one throw. The probability is: \[ P(\text{Head and 6}) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}. \] Next, we determine the probability of Player B winning the game. Player B wins if A does not win on the first, third, fifth, etc., throws, and then Player B wins on the subsequent throw. Hence, we need to consider the sequence of events where Player A does not win initially, followed by Player B winning. Step 1: Calculate the probability of B winning on the second, fourth, sixth throws, etc. The probability of A not winning on the first throw is: \[ P(\text{A does not win in 1st throw}) = \frac{11}{12}. \] The probability of B winning on the second throw is: \[ P(\text{B wins in 2nd throw}) = \frac{1}{12}. \] Thus, the probability that B wins on the second throw is: \[ P(\text{B wins in 2nd throw}) = \frac{11}{12} \times \frac{1}{12}. \] Similarly, for B to win on the fourth throw: \[ P(\text{B wins in 4th throw}) = \frac{11}{12} \times \frac{11}{12} \times \frac{1}{12}. \] Step 2: Set up the infinite series. We see a pattern here where the probability of B winning on the 2nd, 4th, 6th, etc., throws forms an infinite geometric series: \[ P(\text{B wins}) = \frac{11}{12} \times \frac{1}{12} + \frac{11}{12}^2 \times \frac{1}{12} + \frac{11}{12}^3 \times \frac{1}{12} + \dots \] The sum of this infinite geometric series is given by: \[ P(\text{B wins}) = \frac{1}{12} \cdot \frac{\frac{11}{12}}{1 - \left(\frac{11}{12}\right)^2}. \] Step 3: Simplify the expression. Simplifying the above expression: \[ P(\text{B wins}) = \frac{1}{12} \cdot \frac{\frac{11}{12}}{1 - \frac{121}{144}} = \frac{1}{12} \cdot \frac{\frac{11}{12}}{\frac{23}{144}}. \] \[ P(\text{B wins}) = \frac{1}{12} \times \frac{11}{12} \times \frac{144}{23} = \frac{11}{23}. \] Final Answer: \[ \boxed{\frac{11}{23}}. \]