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. $\frac{(n-1)(52-n)(51-n)}{50\times 49\times 17\times 3}$
• B. $\frac{2(n-1)(52-n)(51-n)}{50\times 49\times 17\times 3}$
• C. $\frac{3(n-1)(52-n)(51-n)}{50\times 49\times 17\times 3}$
• D. $\frac{4(n-1)(52-n)(51-n)}{50\times 49\times 17\times 3}$

Answer 1

The correct option is A $\frac{(n-1)(52-n)(51-n)}{50\times 49\times 17\times 3}$

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
$=\frac{{}^4{C}_1{\times }^{48}{C}_{n-2}}{{}^{52}{C}_{n-1}}\times \frac{3}{52-(n-1)}$
$=\frac{4\times \left(48\right)!}{(n-2)!(48-n+2)!}\times \frac{(n-1)!(52-n+1)!}{\left(52\right)!}\times \frac{3}{52-n+1}$
$=\frac{(n-1)(52-n)(51-n)}{50\times 49\times 17\times 3}$ (on simplification).