Cards are dealt one by one from a well-shuffled pack until an ace appears. Show that the probability that exactly $n$ cards are dealt before the first ace appears.
My attempt: For an ace card to appear after $n$ cards $$\text{No. of ways }= \, ^{48}C_n \times \, ^{4}C_1$$ $\text{Sample space }= \, ^{52}C_{n+1}$ or should I take $\, ^{52}C_n \times \, ^{52-n}C_1$ but why?
$$\Pr(E) = \, \dfrac{^{48}C_n \times \, ^{4}C_1}{\, ^{52}C_{n+1}}$$
$n$ takes the values $0, 1, \ldots , 48$
$$P(X=0)= \frac {4}{52} $$ $$P(X=1)= \frac {48}{52} \frac{4}{51} $$ $$P(X=2)= \frac {48}{52} \frac{47}{51} \frac{4}{50}$$ etc.
Thus
$$ P(X=n)= \frac {\frac {4(48!)}{(48-n)!}}{\frac{52!}{(52-n-1)!}} =\frac{{48\choose n}{4\choose 1}}{{52\choose n}{52-n\choose 1}}$$