Two persons A and B throw a die alternately till one of them gets a six and wins the game. If A begins, then the probabilities of winning of A and B respectively are
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\frac{6}{11},\frac{5}{11}$
Let the probability of getting a six on a single throw be $p=\frac{1}{6}$, and not getting a six be $q=\frac{5}{6}$.
A wins if:
He gets a six on the 1st, 3rd, 5th, ... trials.
Hence, probability that A wins:
$P(A) = p + (q^2)p + (q^2)^2p + \dots$
$P(A) = p[1 + q^2 + q^4 + \dots]$
This is an infinite geometric series with ratio $q^2 = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$.
$\Rightarrow P(A) = \frac{p}{1 - q^2} = \frac{\frac{1}{6}}{1 - \frac{25}{36}} = \frac{\frac{1}{6}}{\frac{11}{36}} = \frac{36}{66} = \frac{6}{11}$
Since one of them must win, $P(B) = 1 - P(A) = 1 - \frac{6}{11} = \frac{5}{11}$.
Therefore, $P(A) = \frac{6}{11}$ and $P(B) = \frac{5}{11}$.