91Ó°ÊÓ

Two cards are drawn from a deck of $52$cards, Let $B$be the event that the two cards are aces. Allow$A_s$to be the event that the trump card is picked, Let $A$be the event that something like one ace is picked.

For registering the $P\left(Bâˆ\copyright A_s\right)$, there are three potential ways as an ace of spades and diamonds, aces of spades and hearts, or aces of spades and clubs.

The deck of cards ,there are $\left(\begin{array}{c}52\\ 2\end{array}\right)$ways of selecting two cards. Hence,

The event $A_s$,there $51$ways to draw a card.

Thus,

The computation as ,

$P\left(Bâˆ\pounds A_s\right)=\frac{P\left(Bâˆ\copyright A_s\right)}{P\left(A_s\right)}$

$=\frac{\frac{3}{\left(\begin{array}{c}52\\ 2\end{array}\right)}}{\frac{51}{\left(\begin{array}{c}52\\ 2\end{array}\right)}}$

$=\frac{3}{51}$
$=\frac{1}{17}$

The probability of$\left(B|A_s\right)$is$\frac{1}{17}$

Let $Aâˆ\copyright B$is the occasion that something like one ace is picked the two cards are aces which is excess. That is, $Aâˆ\copyright B=B$

$n(Aâˆ\copyright B)=n\left(B\right)$

The given data $B$means the occasion that the two cards are aces.

Out of$4$ aces, two cards are drawn, the conceivable number of cases is as displayed beneath:
$=\left(\begin{array}{l}4\\ 2\end{array}\right)$

$=\frac{4!}{2!(4−2)!}$

$=\frac{24}{4}$

role="math" localid="1647791327857" $=6$

Presently, find neither one of the cards is an ace out of the $48$ non-aces, we need to pick $2$ of them and of the $4$ aces.

Let $P\left(A\right)$ is the probability that no less than one of the cards is an ace.

$=1−\frac{\left(\begin{array}{c}48\\ 2\end{array}\right)}{\left(\begin{array}{l}52\\ 2\end{array}\right)}$

$=1−\frac{\frac{48!}{2!(48−2)!}}{\frac{52!}{2!(52−2)!}}$

$=1−\frac{1128}{1326}$

$=1-0.8507$

$=0.1493$

Hence,

$=\frac{0.004525}{0.1493}$

$=0.0303$

Hence ,the value of$P\left(B\right|A)$is$0.0303$