Question

A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability that A wins if A makes the first throw is:

• A. \( \frac{9}{17} \)
• B. \( \frac{9}{19} \)
• C. \( \frac{8}{17} \)
• D. \( \frac{8}{19} \)

Hint:

To solve probability problems involving alternating events, use recursive equations to account for future possibilities. In this case, calculate the probability of each player's win on their turn and then set up an equation based on conditional probabilities.

Answer 1

The Correct Option is B

Given the following sums:

For sum '5': 

The possible pairs are \( (1,4), (2,3), (3,2) \), leading to:

\[ P(A) = \frac{4}{36}. \]

For sum '8':

The possible pairs are \( (2,6), (3,5), (4,4), (5,3), (6,2) \), leading to:

\[ P(B) = \frac{5}{36}. \]

Probability of \( A \) winning:

The probability of \( A \) winning is given by the series: \[ P(A \text{ wins}) = P(A) + P(A^C)P(B)P(A) + P(A^C)P(B)P(A^C)P(B) + \dots \] This series simplifies to: \[ P(A \text{ wins}) = \frac{9}{19}. \]

Final Answer:

The probability of \( A \) winning is \( \frac{9}{19} \).