Question

A ans B throw a pair of dice. If sum 6 comes to A before 7 comes for A is winner. Find probability of A winning.

• A. 5/11
• B. 6/11
• C. 5/36
• D. 6/36

Answer 1

Answer: (A)

5/11

Answer 2

Let W be the event that A rolls a sum of 6. Let L be the event that A rolls a sum of 7. The probability of rolling a sum of 6 is $P(W) = 5/36$. The probability of rolling a sum of 7 is $P(L) = 6/36$. The probability of rolling neither a 6 nor a 7 is $P(C) = 1 - P(W) - P(L) = 1 - 5/36 - 6/36 = 25/36$.

A wins if a 6 is rolled before a 7. This can be modeled as a geometric distribution. The probability of A winning is the probability of rolling a 6 given that either a 6 or a 7 is rolled: $P(\text{A wins}) = \frac{P(W)}{P(W) + P(L)} = \frac{5/36}{5/36 + 6/36} = \frac{5/36}{11/36} = \frac{5}{11}$.