91Ó°ÊÓ

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.

$A=A_1∪A_2$ - 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 $P\left(B\right|A)=\frac{1}{33}$.