Question

A and B alternately cut a card each from a well-shuffled pack of $52$ cards with replacement and the pack is shuffled after each cut. If A starts the game and the game is continued till one cuts a spade or club, the respective probabilities of And B cutting a spade or club card are

• A. $\frac{1}{3},\frac{2}{3}$
• B. $\frac{2}{3},\frac{1}{3}$
• C. $\frac{4}{7},\frac{3}{7}$
• D. $\frac{3}{7},\frac{4}{7}$

Answer 1

The correct option is B $\frac{2}{3},\frac{1}{3}$

Let WA denotes the event that in particular chance A wins and NA be the A not wins, similarly WB & NB denotes B wins and not wins respectively.
Therefore probability A winning first,$P\left(A\right)=P\left(WA\right)+P\left(NANBWA\right)+P\left(NANBNANBWA\right)+...$
$=1/2+\left(1/2\right)\left(1/2\right)\left(1/2\right)+\left(1/2\right)\left(1/2\right)\left(1/2\right)\left(1/2\right)\left(1/2\right)+..$
$=\left(1/2\right)/(1-1/4)=2/3$
As the events A winning first & B winning first are mutually exclusive & $P\left(A\right)+P\left(B\right)=1$
Therefore $P\left(B\right)=1/3$