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) → \(\displaystyle\frac{6}{11},\frac{5}{11}\)
On each throw, \(P(\text{six}) = p = \frac{1}{6}\) and \(P(\text{not six}) = q = \frac{5}{6}\).
A throws 1st, 3rd, 5th, …. A wins on his k-th throw if the previous 2(k − 1) throws all fail and then A gets a six:
\( P(A) = p + q^2 p + q^4 p + \cdots \)
This is an infinite geometric series with first term p and common ratio \(q^2 = \frac{25}{36}\):
\( \displaystyle P(A) = \frac{p}{1 - q^2} = \frac{\frac{1}{6}}{\frac{11}{36}} = \frac{6}{11} \)
The game ends with probability 1, so
\( \displaystyle P(B) = 1 - \frac{6}{11} = \frac{5}{11} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{1}{6},\frac{1}{6}\) — Incorrect. \(\frac{1}{6}\) is only the chance of a six on one throw; the two winning probabilities must add up to 1.
Option (2) \(\displaystyle\frac{2}{3},\frac{1}{3}\) — Incorrect. The sum is 1, but \(P(A) = \frac{6}{11}\), not \(\frac{2}{3}\).
Option (4) \(\displaystyle\frac{5}{11},\frac{6}{11}\) — Incorrect. The values are swapped. A throws first, so A has the larger chance.
NCERT (2026-27 edition):
Probability
