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A man draws a card from a pack of 52 playing cards, replaces it and shuffles the pack. He continues this processes until he gets a card of spade. The probability that he will fail the first two times is:

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$\frac{1}{16}$

$\frac{9}{16}$

$\frac{9}{64}$

None of these

answer is C.

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Detailed Solution

The required probability is given by $\frac{{}^{39}C_{1}}{{}^{52}C_{1}} \times \frac{{}^{39}C_{1}}{{}^{52}C_{1}} \times \frac{{}^{13}C_{1}}{{}^{52}C_{1}} = \frac{3}{4} \times \frac{3}{4} \times \frac{1}{4} = \frac{9}{64}$

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