Question

Cards are drawn one by one at random from a well shuffled full pack of 52 cards until two aces are obtained for the first time. If N is the number of cards required to be drawn, then $P_{r}N=n$ where $2\le n\le 50,$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Here the least number of draws to obtain 2 aces are 2 and the maximum number is 50 thus n can take value from 2 to 50.
Since we have to make n draws for getting two aces, in (n – 1) draws, we get any one of the 4 aces and in the $n^{th}$ draw we get one ace. Hence the required probability
$=\frac{{}^{48}{C}_{n-2}{\times }^4{C}_1}{{}^{52}{C}_{n-1}}\times \frac{3}{52-(n-1)}$

The first factor is the probability of drawing (n - 2) non-aces and one ace in the first (n − 1) draws while the second factor is the probability of drawing an ace in the nth draw.
$=\frac{4\times \left(48\right)!}{(n-2)!(48-n+2)!}\times \frac{(n-1)!(52-n+1)!}{\left(52\right)!}\times \frac{3}{52-n+1}$
$=\frac{4\times \left(48\right)!}{(n-2)!(50-n)!}\times \frac{(n-1)(n-2)!(53-n)(52-n)(51-n)(50-n)!}{\left(52\right)\left(51\right)\left(50\right)\left(49\right)\left(48\right)!}\times \frac{3}{53-n}$
$=\frac{(n-1)(52-n)(51-n)}{50\times 49\times 17\times 3}$ (on simplification).