Question

Two cards are randomly chosen without replacement from an ordinary deck of$52$ cards. Let $B$ be the event that both cards are aces, let $A_s$be the event that the ace of spades is chosen, and let$A$ be the event that at least one ace is chosen. Find

(a)role="math" localid="1647789007426" $P\left(B\right|A_s)$

(b) $P\left(B\right|A)$

Short answer

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

(b) The probability of$\left(B\right|A)$is$0.0303$

Step-by-step solution

Step 1:Given Information(part a)

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 $P\left(B\cap A_s\right)$, there are three potential ways as an ace of spades and diamonds, aces of spades and hearts, or aces of spades and clubs.

Step 2:Explanation(part a)

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

$P\left(B\cap A_s\right)=\frac{3}{\left(\begin{array}{c}52\\ 2\end{array}\right)}$

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

Thus,

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

The computation as ,

$P\left(B\mid A_s\right)=\frac{P\left(B\cap A_s\right)}{P\left(A_s\right)}$

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

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

Step 3:Final Answer(part a)

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

Step 4:Given Information(part b)

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

$n(A\cap B)=n\left(B\right)$

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

Step 5:Explanation(part b)

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" $=6$

Step 6:Explanation of P(A)(part b)

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.

$P\left(A\right)=1-P\left(\text{Neither card is an ace}\right)$

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

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

$=1-\frac{1128}{1326}$

$=1-0.8507$

$=0.1493$

Step 7:Final Answer(part b)

Hence,

$P(B\mid A)=\frac{P\left(B\right)}{P\left(A\right)}$
$=\frac{6/\left(\begin{array}{l}52\\ 2\end{array}\right)}{0.1493}$

$=\frac{0.004525}{0.1493}$

$=0.0303$

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