A friend randomly chooses two cards, without replacement, from an ordinary deck of $52$playing cards. In each of the following situations, determine the conditional probability that both cards are aces. $110$Chapter $3$Conditional Probability and Independence.

(a) You ask your friend if one of the cards is the ace of spades, and your friend answers in the affirmative.

(b) You ask your friend if the first card selected is an ace, and your friend answers in the affirmative.

(c) You ask your friend if the second card selected is an ace, and your friend answers in the affirmative.

(d) You ask your friend if either of the cards selected is an ace, and your friend answers in the affirmative.

Two cards are selected, in an order from a deck of 52 cards

Events:

$B$ - both cards are aces.

$S$ - one card is the ace of spades.

Every ordered pair of different cards is equally likely to be the drawn pair: $52·51$possibilities.

If one card is an ace of spades, the other can be any of the $51$other cards, taking into consideration the order -

$2·51$equally likely possibilities in the reduced outcome space

If additionally both cards are aces, there are $3$possibilities for the other card and $2$for the order $-6$possibilities.

Therefore:

The required probability is$P\left(B\right|S)=\frac{1}{17}$.

$B-$ both cards are aces.

$A_1-$ first card in the ace.

The number of events in the reduced outcome space:

The first card can be any of the $4$ aces, and the second one any of the remaining $51-4·51$

If both cars are aces:

First can be any of the $4$ aces, and the second any of the remaining $3-4·3$

The required probability is$P\left(B\right|A_1)=\frac{1}{17}$.

$B-$both cards are aces.

$A_2-$ second card is an ace.

This is identical as b), because two cards are drawn randomly

The required probability is$P\left(B\right|A_2)=\frac{1}{17}$.

$B-$both cards are aces.

$A=A_1\cup A_2$ - one of the cards is an ace.

According to the formula of inclusion and exclusion for the number of elements in set:

From sections b) and c) the number of events in $A_1-n\left(A_1\right)$ and $A_{2}-$A_2-n\left(A_2\right)$ are $4·51$. And there are $4·3$ draws in which both cards are aces $A_1{A}_2$.

And from those events, there are mentioned $4·3$ possibilities that both cards are aces - this makes:

The required probability is $P\left(B\right|A)=\frac{1}{33}$.