Sonal and Ananya are playing a game by throwing a die alternatively till one of them gets a '1' and wins the game. Find their respective probabilities of winning if Sonal starts first.

Sonal: $\frac{6}{11}$, Ananya: $\frac{5}{11}$

The correct answer is Option (2) → Sonal: $\frac{6}{11}$, Ananya: $\frac{5}{11}$

When a die is thrown, then sample space $S = \{1, 2, 3, 4, 5, 6\}$.

It has 6 equally likely outcomes.

Let E be the event of throwing '1', then

$p = P(E)=\frac{1}{6}$, so $q = 1-\frac{1}{6}=\frac{5}{6}$.

As Sonal starts the game by throwing the first die

P(Sonal winning in the first throw) = $\frac{1}{6}$

Sonal gets second chance only if Sonal and Ananya both fail in the first round.

∴ P(Sonal winning in the 2nd round i.e. third throw) = $\frac{5}{6}×\frac{5}{6}×\frac{1}{6}$.

Similarly, P(Sonal winning in the 3rd round i.e. fifth throw) = $\frac{5}{6}×\frac{5}{6}×\frac{5}{6}×\frac{5}{6}×\frac{1}{6}$ and so on.

∴ The probability of Sonal winning the game is the sum of the series

$\frac{1}{6}+(\frac{5}{6})^2.\frac{1}{6}+(\frac{5}{6})^4+....$

which is an infinite geometrical series with $a =\frac{1}{6},r=(\frac{5}{6})^2<1$

$=\frac{\frac{1}{6}}{1-(\frac{5}{6})^2}=\frac{6}{11}$   $(∵S_∞=\frac{a}{1-r})$

Hence, P(Sonal winning) = $\frac{6}{11}$

Now, P(Sonal winning) + P(Ananya winning) = 1

⇒ P(Ananya winning) = $1-\frac{6}{11}=\frac{5}{11}$

Hence, P(Ananya winning) = $\frac{5}{11}$