Question

A and B alternately cut a pack of cards which is shuffled after each cut. The game is started by A and continuous until one of the players cuts a club. The probability that B wins the game is..........

• A. $\frac{3}{7}$
• B. $\frac{4}{7}$
• C. $\frac{5}{7}$
• D. $\frac{6}{7}$

Answer 1

The correct option is A $\frac{3}{7}$

The probable ways $B$ can win the game are as follows :-
$A$ does not get a club ans $B$ gets a club.
$A,B,A$ in that order A do not get club and $B$ gets a club.
$A,B,A,B,A$ in that order A do not get a club and $B$ gets a club, so on......
Let $E$ be the event of getting a club.
$P\left(E\right)=\frac{13}{52}=\frac{1}{4},P\left(E^{\prime }\right)=\frac{3}{4}$
Probability that $B$ wins
$=P\left(E^{\prime }E\right)+P\left(E^{\prime }{E}^{\prime }{E}^{\prime }E\right)+P\left(E^{\prime }{E}^{\prime }{E}^{\prime }{E}^{\prime }{E}^{\prime }E\right)+...$
$=P\left(E\right)P\left(E^{\prime }\right)+P^3\left(E^{\prime }\right)P\left(E\right)+P^5\left(E^{\prime }\right)P\left(E\right)+...$
$=P\left(E\right)P\left(E^{\prime }\right)\{1+P^2\left(E^{\prime }\right)+P^4\left(E^{\prime }\right)+...\}$
$=P\left(E\right)P\left(E^{\prime }\right)\frac{1}{1-P^2\left(E^{\prime }\right)}$
$=\frac{1}{4}.\frac{3}{4}\frac{1}{1-{\left(\frac{3}{4}\right)}^2}=\frac{3}{7}$