Chapter CSB, Problem 4.18E

To determine

To find: the probability $P\left(\text{both either black or face}\right)$ .

Answer to Problem 4.18E

The probability $P\left(\text{both either black or face}\right)$ is $\frac{65}{221}$ .

Explanation of Solution

Given information:

As mentioned in problem two cards are drawn out of deck of cards.

Formula:

Total number of black cards is $26$ and face is $12$ .

  $\text{Number of probability}=\frac{\text{No}.\text{ of outcome}}{\text{Total no}.\text{ of outcome}}.$

Total probability $P\left(\text{both either black or face}\right)$ = $p\left(A\cup B\right)=p\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$ .

Calculation:

Consider the statement that two cards are drawn out of deck of cards.

Since total number of cards in the deck is $52$ .

Two cards are drawn from the deck of cards.

Therefore $n\left(s\right)$ =event of two cards drawn ${=}^{52}{C}_2\Rightarrow \frac{52\times 51}{2}=1326.$

Recall that there are $26$ black card and $12$ face card.

Event of both black card are $n\left(A\right){=}^{26}{C}_2\Rightarrow \frac{26\times 25}{2}=325.$

Event of ace card are $n\left(B\right){=}^{12}{C}_2=\frac{12\times 11\times 10!}{10!\times 2!}=\frac{12\times 11}{2}=66.$

Event of both black and face card is $n\left(A\cap B\right)$${=}^2{C}_2=1$ .

Recall $\text{number of probability of black cards}=\frac{\text{No}.\text{ of outcome}}{\text{Total no}.\text{ of outcome}}=p\left(A\right)=\frac{n\left(A\right)}{n\left(s\right)}=\frac{325}{1326}$ .

Recall $\text{number of probability of face cards}=\frac{\text{No}.\text{ of outcome}}{\text{Total no}.\text{ of outcome}}=p\left(B\right)=\frac{n\left(B\right)}{n\left(s\right)}=\frac{66}{1326}.$

Recall that probability of both black and ace card $p\left(A\cap B\right)=\frac{n\left(A\cap B\right)}{n\left(s\right)}=\frac{1}{1326}.$

Therefore the probability $P\left(\text{both either black or ace}\right)$ is $p\left(A\cup B\right)=p\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$ .

Therefore $p\left(A\cup B\right)=\frac{325}{1326}+\frac{66}{1326}-\frac{1}{1326}=\frac{390}{1326}=\frac{65}{221}.$

Therefore probability $P\left(\text{both either black or ace}\right)$ is $\frac{65}{221}$ .