Question

Cards are dealt one by one from a well shuffled pack until an ace appears. the probability that exactly n cards are dealt befor the first ace appears is

• A. $\frac{4(51-n)(50-n)(49-n)}{\mathrm{52.51.50.49}}$
• B. $\frac{4(52-n)(51-n)(49-n)}{\mathrm{52.51.50.49}}$
• C. $\frac{4(52-n)(51-n)(49-n)}{\mathrm{51.50.49.48}}$
• D. $\frac{4(51-n)(50-n)(49-n)}{\mathrm{51.50.49.48}}$

Answer 1

The correct option is A $\frac{4(51-n)(50-n)(49-n)}{\mathrm{52.51.50.49}}$

A: number of ace is drawn in the first n draw
B: an ace appear in the ${(n+1)}^{th}$ draw
Hence the probability that exactly n cards are dealt before the first ace appear is equal to $P(A\cap B)$
$P\left(A\right)=\frac{{}^{48}{C}_n}{{}^{52}{C}_n},P\left(\frac{B}{A}\right)=\frac{4}{52-n}$
$\therefore P(A\cap B)=P\left(A\right).P\left(\frac{B}{A}\right)=\frac{{}^{48}{C}_n}{{}^{52}{C}_n}.\frac{4}{52-n}$
$=\frac{48!}{(48-n)!n!}\times \frac{(52-n)!n!}{52!}\times \frac{4}{52-n}$
$=\frac{4(51-n)(50-n)(49-n)}{52\times 51\times 50\times 49}$