Two persons A and B throw a die alternately till one of them gets a 'three' and wins the game. The probability of A's winning if A starts first is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{6}{11}\)
On each throw, P(three) \(= \frac{1}{6}\) and P(not three) \(= \frac{5}{6}\).
A throws 1st, 3rd, 5th, …. A wins on the 1st throw, or on the 3rd if both miss once, and so on:
\( \displaystyle P(A) = \frac{1}{6} + \left(\frac{5}{6}\right)^2\frac{1}{6} + \left(\frac{5}{6}\right)^4\frac{1}{6} + \cdots \)
This is an infinite geometric series with first term \(a = \frac{1}{6}\) and common ratio \(r = \frac{25}{36}\):
\( \displaystyle P(A) = \frac{a}{1 - r} = \frac{1/6}{11/36} \)
\( \displaystyle P(A) = \frac{36}{66} = \frac{6}{11} \)