Question-50 A and B alternately cut a card each from a pack of cards with replacement and pack is shuffled after each cut. If A starts the game and the game is continued till one cuts a spade, the respective probabilities of A and B cutting a spade are
A)
\frac{1}{3}, \frac{2}{3}B)
\frac{3}{4}, \frac{1}{4}C)
\frac{4}{7}, \frac{3}{7}D)
\frac{3}{7}, \frac{4}{7}Answer: C)
Explanation
We have,
\begin{aligned} & P(\text { spade })=p=\frac{13}{52}=\frac{1}{4} \\ & P(\text { non-spade })=q=1-\frac{1}{4}=\frac{3}{4} \end{aligned}Now,
\begin{aligned} & P(A \text { win })=\frac{1}{4}+\left(\frac{3}{4} \times \frac{3}{4} \times \frac{1}{4}\right)+\left(\frac{3}{4} \times \frac{3}{4} \times \frac{3}{4} \times \frac{3}{4} \times \frac{1}{4}\right)+\ldots \infty \\ & \Rightarrow P(A \text { win })=\frac{\frac{1}{4}}{1-\frac{9}{16}}=\frac{4}{7} \end{aligned}Hence,
P(B \text { win })=1-\frac{4}{7}=\frac{3}{7}