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

Step 1: First, calculate the probability of A throwing a sum of 5 and the probability of B throwing a sum of 8 with a pair of dice. 
The total possible outcomes when rolling two dice is 36. The number of outcomes that result in a sum of 5 is 4 (i.e., (1, 4), (2, 3), (3, 2), (4, 1)). 
So, the probability that A rolls a 5 is: \[ P(A_{{wins}}) = \frac{4}{36} = \frac{1}{9}. \] The number of outcomes that result in a sum of 8 is 5 (i.e., (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)). 
So, the probability that B rolls an 8 is: \[ P(B_{{wins}}) = \frac{5}{36}. \] 
Step 2: Now, define \( p \) as the probability that A wins given that A makes the first throw. 
The probability of A winning can be broken into two parts: 
- If A throws a 5 on his first turn, A wins immediately with a probability of \( \frac{1}{9} \). 
- If A does not throw a 5, it is B’s turn. The probability that B does not throw an 8 is \( \frac{31}{36} \). 
Therefore, the equation for the total probability of A winning is: \[ p = \frac{1}{9} + \left( \frac{31}{36} \right) \left( \frac{1}{9} + \left( \frac{31}{36} \right) p \right). \] Solving this equation, we find: \[ p = \frac{9}{19}. \] Thus, the probability that A wins if A makes the first throw is \( \frac{9}{19} \).