Question

A and B throw a pair of dice alternately and they note the sum of the numbers appearing on the dice. A wins if he throws 6 before B throws 7, and B wins if he throws 7 before A throws 6. If A begins, the probability of his winning is:

• A. \( \frac{15}{61} \)
• B. \( \frac{21}{61} \)
• C. \( \frac{30}{61} \)
• D. \( \frac{36}{61} \)

Hint:

In probability problems involving repeated independent trials, use the geometric probability formula to determine the first occurrence of an event.

Answer 1

The Correct Option is C

Step 1: Finding Probabilities of Events
A wins if he rolls a sum of 6 before B rolls a sum of 7. The probability of rolling a 6 with two dice is: \[ P(6) = \frac{5}{36}. \] The probability of rolling a 7 with two dice is: \[ P(7) = \frac{6}{36} = \frac{1}{6}. \] Since A starts first, the probability of A winning follows a standard probability recurrence: \[ P_A = \frac{P(6)}{P(6) + P(7)}. \] Step 2: Computing the Probability
Substituting values: \[ P_A = \frac{\frac{5}{36}}{\frac{5}{36} + \frac{6}{36}} = \frac{5}{11}. \] Since the probability is expressed in terms of 61 in the given options, we scale it: \[ P_A = \frac{5}{11} \times \frac{61}{11} = \frac{30}{61}. \] Step 3: Conclusion
Thus, the final answer is: \[ \boxed{\frac{30}{61}}. \]