Question

Five cards are drawn from a pack of 52 cards.
What is the chance that these 5 will contain :
(i) just one ace

(ii) at least one ace?

Answer 1

Since five cards are drawn from a pack to 52 cards =${}^{52}{C}_5$

Let E be the event that those five cards contain exactly one ace.

$\therefore n\left(E\right){=}^4{C}_1{\times }^{48}{C}_4$

$\therefore P\left(E\right)=\frac{{}^4{C}_1{\times }^{48}{C}_4}{{}^{52}{C}_5}=\frac{\frac{4!}{1!\times 3!}\times \frac{48}{4!\times 44!}}{\frac{52!}{5!\times 47!}}=\frac{4!\times 48!5!\times 47!}{4!\times 4!\times 52!}=\frac{4\times 48\times 47\times 46\times 45}{\frac{52\times 51\times 50\times 49\times 48}{5}}=\frac{3243}{10829}$

(ii) Let E be the event that five cards contain atleast one ace

$\therefore$ E={1 or 2 or 3 or 4}

$\therefore n\left(E\right)=\frac{{}^4{C}_1{\times }^{48}{C}_4{+}^4{C}_2{\times }^{48}{C}_3{+}^4{C}_3{\times }^{48}{C}_2{+}^4{C}_4{\times }^{48}{C}_1}{{}^{52}{C}_5}$

$=\frac{4\times \frac{48\times 47\times 46\times 45}{4\times 3\times 2\times 1}+\frac{4\times 3}{2}\times \frac{48\times 47\times 46}{3\times 2\times 1}+4\times \frac{48\times 47}{2}+48}{\frac{52\times 51\times 50\times 49\times 48}{5\times 4\times 3\times 2\times 1}}=\frac{18472}{54145}$