Question

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

• A. \(\frac{6}{11},\frac{5}{11}\)
• B. \(\frac{5}{11},\frac{6}{11}\)
• C. \(\frac{3}{11},\frac{8}{11}\)
• D. \(\frac{8}{11},\frac{3}{11}\)

Answer 1

The Correct Option is A

To solve the problem, consider the nature of the game where Ram and Shyam take turns throwing a die to get a '1'. Ram starts first. We aim to find the probabilities of Ram and Shyam winning, respectively.

The sample space for a single die roll not resulting in '1' (i.e., the event of rolling a number from 2 to 6) has a probability:

\[ P(\text{Not } 1) = \frac{5}{6} \]

The probability of a '1' coming up is:

\[ P(1) = \frac{1}{6} \]

Let P be the probability that Ram wins the game. If Ram doesn't win on his first throw, Shyam will have the chance to throw next and their turns will continue alternately.

An expression for the probability P that Ram wins is:

\[ P = \frac{1}{6} + \frac{5}{6} \times P' \]

where \( P' \) is the probability that Ram wins given Shyam didn't roll a '1' (i.e., they jump back to the situation where it's Ram's turn again after a cycle of both playing, which has the probability \( P \)). Therefore:

\[ P' = P \]

Thus, the equation becomes:

\[ P = \frac{1}{6} + \frac{5}{6}P \]

Rearranging gives:

\[ P - \frac{5}{6}P = \frac{1}{6} \]

\[ \frac{1}{6}P = \frac{1}{6} \]

\[ P = 1 - \frac{5}{6} = \frac{6}{11} \]

Next, the probability that Shyam wins, denoted as \( P' \), is the complementary probability:

\[ P' = 1 - P = 1 - \frac{6}{11} = \frac{5}{11} \]

Thus, the probabilities of Ram and Shyam winning are \(\frac{6}{11}\) and \(\frac{5}{11}\) respectively, aligning with the given answer option \(\frac{6}{11},\frac{5}{11}\).