Three persons A,B,C in order cut a pack of cards replacing them after each cut. The person who first cuts a club shall win a prize. The probabilities of their winning are

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$\frac{1}{7}, \frac{3}{7}, \frac{3}{7}$

$\frac{4}{7}, \frac{2}{7}, \frac{1}{7}$

$\frac{16}{37}, \frac{12}{37}, \frac{9}{37}$

$\frac{4}{77}, \frac{3}{77}, \frac{70}{77}$

answer is C.

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Detailed Solution

$p = \frac{13}{52} = \frac{1}{4} and q = 1 - p = 1 - \frac{1}{4} = \frac{3}{4}$
$∴$ Probability of A’s winning $= \frac{P}{1 - q^{3}} = \frac{\frac{1}{4}}{1 - {(\frac{3}{4})}^{3}} = \frac{16}{64 - 27} = \frac{16}{37}$
Probability of B’ s winning $= \frac{qp}{1 - q^{3}} = \frac{(\frac{3}{4}) (\frac{1}{4})}{- {(\frac{3}{4})}^{3}} = \frac{12}{64 - 27} = \frac{12}{37}$
Probability of C’s winning $= \frac{q^{2} p}{1 - q^{2}} = \frac{{(\frac{3}{4})}^{2} (\frac{1}{4})}{1 - {(\frac{3}{4})}^{3}} = \frac{9}{64 - 27} = \frac{9}{37}$