Two cards are drawn from a shuffled deck. What is the probability that both areaces? If you know that at least one is an ace, what is the probability that both areaces? If you know that one is the ace of spades, what is the probability that bothare aces?

Adeck of 52 cards is given out of which 2 cards are to be drawn.

When the order of arrangement is definite, the permutation is applied and when the order is not definite, combination is applied.

There are 4 Aces and out of which 2 are to be selected and can be do in $C\left(4,2\right)$ways. And 2 cards out of whole deck can be selected in$C\left(52,2\right)$ways.

This implies that the number of outcomes favourable are $C\left(4,2\right)$and total number of outcomes are$C\left(52,2\right)$.

Apply the formula for probability, that is$p=\frac{numberofoutcomesfavorabletoE}{totalnumberofoutcomes}$to get the probability that both the selected cards are aces.

$\begin{array}{c}P\left(BothAces\right)=\frac{C\left(4,2\right)}{C\left(52,2\right)}\\ =\frac{\frac{4!}{2!2!}}{\frac{52!}{2!50!}}\\ =\frac{1}{221}\end{array}$

Thus the required probability is$\frac{1}{221}$

There are 48 non-Aces and out of which 2 are to be selected and can be done in $C\left(48,2\right)$ways.

The probability of at least one ace can be obtained by subtracting probability of no ace from 1.

Find the probability of at least one ace.

$\begin{array}{c}P\left(Atleast1ace\right)=1-\frac{C\left(48,2\right)}{C\left(52,2\right)}\\ =1-\frac{\frac{48!}{2!46!}}{\frac{52!}{2!50!}}\\ =1-\frac{4\times 47}{13\times 17}\\ =\frac{33}{221}\end{array}$

Find the probability that both the selected cards are aces when at least one is an ace using the Bayes theorem.

$\begin{array}{c}P\left(BothAces|Atleastoneace\right)=\frac{\frac{1}{221}}{\frac{33}{221}}\\ =\frac{1}{33}\end{array}$

Thus the required probability is.