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 $P_r\left\{N=n\right\}$, where $2\le n\le 50$, is

• A. $\frac{\left[\right(n-1\left)\right(52-n\left)\right(51-n\left)\right]}{[50.49.17.13]}$
• B. $\frac{\left[2\right(n-1\left)\right(52-n\left)\right(51-n\left)\right]}{[50.49.17.13]}$
• C. $\frac{\left[3\right(n-1\left)\right(52-n\left)\right(51-n\left)\right]}{[50.49.17.13]}$
• D. $\frac{\left[4\right(n-1\left)\right(52-n\left)\right(51-n\left)\right]}{[50.49.17.13]}$

Answer 1

The correct option is A

$\frac{\left[\right(n-1\left)\right(52-n\left)\right(51-n\left)\right]}{[50.49.17.13]}$

Explanation for the correct option:

We must get one ace in $n-1$ attempts and one ace in the $nth$ attempt.

Step 1. The probability of getting one ace in first $n-1$ attempts is
$=\frac{{}^{4}C_1\times {}^{48}C_{n-2}}{{}^{52}C_n}$

Sep 2. Probability of getting one ace in the nth attempt is
$=\frac{{}^{3}C1}{[52-(n-1\left)\right]}=\frac{3}{53-n}$

Therefore, the required probability is

$\frac{4\times 48!}{(n-2)!(50-n)!}\times \frac{(n-1)!(53-n)!}{52!}\times \frac{3}{53-n}$

$=\frac{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{52}\mathbf{-}\mathbf{n}\mathbf{)}\mathbf{(}\mathbf{51}\mathbf{-}\mathbf{n}\mathbf{)}}{\mathbf{50}\mathbf{\times }\mathbf{49}\mathbf{\times }\mathbf{17}\mathbf{\times }\mathbf{13}}$

Hence, Option ‘A’ is Correct.