Question

Cards are dealt one by one from well shuffled pack of 52 playing cards until $r(1\le r\le 4)$ aces are obtained. If $p_r$ denotes the probability of drawing r aces for the first time at the nth draw (with $n\ge 4)$, and $p_r={(}^{52}{C}_4\left)p\right)r$, then

Answer 1

We must draw (r-1) aces in the (n-1) draws. The total number of ways of drawing (n-1) cards out of 52 is ${}^{52}{C}_{n-1}$.
The number of ways of drawing (r-1) aces and n-r other cards is ${(}^{48}{C}_{n-r}\left){(}^4{C}_{r-1}\right)$.
The probability of getting an ace at the rth draw is $(5-r)/(53-n)$. Thus,
$p_r=\frac{{(}^{48}{C}_{n-r}\left){(}^4{C}_{r-1}\right)}{{}^{52}{C}_{n-1}}\times \frac{5-r}{53-n}=\frac{48!4!(52-n)!(n-1)!}{(n-r)!(48+r-n)!(r-1)!(4-r)!52!}$
$=\frac{1}{{}^{52}{C}_4}\left[{(}^{n-1}{C}_{r-1}\right){(}^{52-n}{C}_{48+r-n}\left)\right]$
$\Rightarrow P_r{=}^{52}{C}_4{p}_r={(}^{n-1}{C}_{r-1}\left){(}^{52-n}{C}_{4-r}\right).$
A) $P_1{=}^{52}{C}_4{p}_1={(}^{n-1}{C}_0\left){(}^{52-n}{C}_{4-1}\right){=}^{52-n}{C}_{4-3}$
B) $P_2{=}^{52}{C}_4{p}_2={(}^{n-1}{C}_1\left){(}^{52-n}{C}_2\right)={(n-1)}^{52-n}{C}_2$
C) $P_3{=}^{52}{C}_4{p}_3={(}^{n-1}{C}_2\left){(}^{52-n}{C}_1\right)$
D) $P_4{=}^{52}{C}_4{p}_4={(}^{n-1}{C}_3\left){(}^{52-n}{C}_0\right){=}^{n-1}{C}_3$