Question

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

Answer

Short answer

The probability that both selected cards are ace is$\frac{1}{221}$, the probability that both the selected cards are aces when at least one is an ace is$\frac{1}{33}$and the probability that both are aces when the first card is an ace is$\frac{1}{17}$.

Step-by-step solution

Given Information 

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

Definition of Independent Event

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

Finding the probability that both the selected cards are aces 

There are 4 Aces and out of which 2 are to be selected and can be done 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$\frac{1}{221}$

Finding the probability that both the selected cards are aces when at least one is an ace 

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$\frac{1}{33}$

Finding the probability that both the selected cards are aces when it is known that one is the ace of spades

When first card is a spade, this implies there are 3 aces left in the deck of 51, this the probability of getting a second space $\frac{3}{51}$is or$\frac{1}{17}$