Question

Two persons A and B throw a die alternatively. Person who first throws 6 is the winner. If A starts throwing. Then the probability of A’s winning is

• A. 5/11
• B. 6/11
• C. 2/11
• D. 9/11

Answer 1

Answer: (B)

6/11

Answer 2

Let P_k be the probability of A's winning in the K^th throw.

P(1) = P₁ + P₃ + P₅ + ......∞

= $\frac { 1 } { 6 } + \left( \frac { 5 } { 6 } \right) ^ { 2 } \frac { 1 } { 6 } + \left( \frac { 5 } { 6 } \right) ^ { 4 } \cdot \frac { 1 } { 6 } + \ldots \ldots \ldots \ldots$

= $\frac { \frac { 1 } { 6 } } { 1 - \frac { 25 } { 36 } } = \frac { 6 } { 11 }$