Probability Ques 83
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)
Show Answer
Solution:
Formula:
- $P(N$ th draw gives 2nd 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} $
BETA
