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 and B winning.

Answer 1

P(A wins) = 5/11, P(B wins) = 6/11

Answer 2

Let $S_6$ be the event of rolling a sum of 6, and $S_7$ be the event of rolling a sum of 7. The probability of rolling a sum of 6 is $P(S_6) = 5/36$. The probability of rolling a sum of 7 is $P(S_7) = 6/36$.

A wins if a 6 is rolled before a 7. B wins if a 7 is rolled before a 6. The game ends when either a 6 or a 7 is rolled. The probability of A winning is the probability of rolling a 6 given that either a 6 or a 7 is rolled: $P(\text{A wins}) = \frac{P(S_6)}{P(S_6) + P(S_7)} = \frac{5/36}{5/36 + 6/36} = \frac{5/36}{11/36} = \frac{5}{11}$ The probability of B winning is the probability of rolling a 7 given that either a 6 or a 7 is rolled: $P(\text{B wins}) = \frac{P(S_7)}{P(S_6) + P(S_7)} = \frac{6/36}{5/36 + 6/36} = \frac{6/36}{11/36} = \frac{6}{11}$