Probability 3 Question 42
42. Cards are drawn one by one at random from a well shuffled full pack of 52 playing cards until 2 aces are obtained for the first time. If $N$ is the number of cards required to be drawn, then show that
$$ P _r{N=n}=\frac{(n-1)(52-n)(51-n)}{50 \times 49 \times 17 \times 13} $$
where, $2<n \leq 50$.
(1983, 3M)
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Answer:
Correct Answer: 42. 0.6976
Solution:
- $P(N$ th draw gives 2 nd ace $)$
$=P{1$ ace and $(n-2)$ other cards are drawn in $(N-1)$ draws $} \times P{N$ th draw is 2 nd ace $}$
$$ \begin{aligned} & =\frac{4 \cdot(48) ! \cdot(n-1) !(52-n) !}{(52) ! \cdot(n-2) !(50-n) !} \cdot \frac{3}{(53-n)} \\ & =\frac{4(n-1)(52-n)(51-n) \cdot 3}{52 \cdot 51 \cdot 50 \cdot 49} \\ & =\frac{(n-1)(52-n)(51-n)}{50 \cdot 49 \cdot 17 \cdot 13} \end{aligned} $$
