Ram and Shyam are playing a game by throwing a die alternatively till one of them gets a ‘1’ and wins the game. The probabilities of winning by Ram and Shyam respectively if Ram starts first, is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{6}{11}, \frac{5}{11}\)
On each throw, P(getting 1) = \(\frac{1}{6}\) and P(not getting 1) = \(\frac{5}{6}\).
Ram wins on his 1st, 2nd, 3rd, … turn, that is on throw 1, 3, 5, …. Before each of his later turns, both players must have failed once more, which multiplies the probability by \(\left(\frac{5}{6}\right)^2 = \frac{25}{36}\):
\( \displaystyle P(\text{Ram wins}) = \frac{1}{6} + \frac{25}{36} \times \frac{1}{6} + \left(\frac{25}{36}\right)^{2} \times \frac{1}{6} + \cdots \)
This is an infinite geometric series with first term \(\frac{1}{6}\) and common ratio \(\frac{25}{36}\):
\( \displaystyle P(\text{Ram wins}) = \frac{\frac{1}{6}}{1 - \frac{25}{36}} = \frac{1}{6} \times \frac{36}{11} \)
\( \displaystyle = \frac{6}{11} \)
The game ends with a winner, so
\( \displaystyle P(\text{Shyam wins}) = 1 - \frac{6}{11} = \frac{5}{11} \)
Tip: The player who throws first always has the better chance, so Options (2) and (3), which give Ram the smaller probability, are ruled out at once.
CBSE Applied Mathematics (Class XII):
Binomial Distribution
