Question

Two cards are drawn at random from a standard deck of $52$ cards. What is the probability that both cards are aces?

Answer 1

Step 1: Find the probability that both cards are aces.

Given that a standard deck of $52$ cards,

Total number of cards$=52$

Total number of aces$=4$

Two cards are drawn at random.

The probability that both drawn cards are aces are:

$\mathrm{P}(\mathrm{Both}\mathrm{aces})=\frac{{}^4\mathrm{C}_2\times {}^{48}\mathrm{C}_0}{{}^{52}\mathrm{C}_2}$.

Where ${}^{4}C_2$ denotes the possibility of $2$ aces out of $4$ aces.

${}^{48}C_0$ denotes the remaining cards $48$ cards apart from $4$ aces from $52$ cards .

${}_{}{}^{52}C_2$ denoted the total number of cards out of the possibility of $2$ aces.

Step 2: Simplify the expression.

Apply the combination formula ${}_{}{}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{r})!\mathrm{r}!}$:

$\begin{array}{rcl}\mathrm{P}(\mathrm{Both}\mathrm{aces})& =& \left(\frac{\frac{4!}{(4-2)!2!}\times {}_{}{}^{48}\mathrm{C}_0}{{}_{}{}_{}{}^{52}{\mathrm{C}}_2}\right)\\ & =& \left(\frac{\frac{4!}{2!2!}\times \frac{48!}{(48+0)!0!}}{\frac{52!}{(52-2)!2!}}\right)\\ & =& \left(\frac{\frac{4\times 3\times 2!}{2!2!}\times \frac{48!}{\left(48\right)!0!}}{\frac{52!}{\left(50\right)!2!}}\right)\\ & =& \left(\frac{\frac{4\times 3}{2!}\times {\displaystyle \frac{1}{0!}}}{\frac{52\times 51\times 50!}{\left(50\right)!2!}}\right)[\mathrm{cancel}\mathrm{the}\mathrm{common}\mathrm{term}]\end{array}$

Simplify the expression again:

$\begin{array}{rcl}\mathrm{P}(\mathrm{Both}\mathrm{aces})& =& \left(\frac{\frac{4\times 3}{2!}\times {\displaystyle \frac{1}{0!}}}{\frac{52\times 51\times 50!}{\left(50\right)!2!}}\right)\\ & =& \left(\frac{\frac{12}{2\times 1!}\times {\displaystyle 1}}{\frac{52\times 51}{2\times 1!}}\right)\\ & =& \left(\frac{6\times {\displaystyle 1}}{\frac{52\times 51}{2\times 1!}}\right)\\ & =& \left(\frac{6}{\frac{2652}{2}}\right)\\ & =& \frac{6}{1326}\\ & =& \frac{1}{221}\end{array}$

Hence, the probability that both cards are aces are $\frac{\mathbf{1}}{\mathbf{221}}\mathbf{.}$