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Cards are drawn one by one without replacement until two aces are drawn. Let P(m) be the probability that the event occurs in exactly m trials, then P(m) must be zero at

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a

m = 2

b

m = 50

c

m = 51

d

m = 52

answer is C, D.

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Detailed Solution

$\begin{array}{l} P = \frac{^{4} C_{2} \times 2 (49)!}{52!} = \frac{12}{52 \times 51 \times 50} \\ q = \frac{49_{C_{4}} \times 4 \times 48!}{52!} \\ = \frac{49 \times 48 \times 47 \times 46}{52 \times 51 \times 50 \times 49} = \frac{48 \times 47 \times 46}{52 \times 51 \times 50} \end{array}$

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