Chapter 13.5, Problem 36E

To determine

To calculate: The probability that sum of two cards drawn is less than or equal to 7 if at least one card is ace.

Answer to Problem 36E

The probability that sum of two cards drawn is less than or equal to 7 if at least one card is ace is $\frac{23}{51}$ .

Explanation of Solution

Given information:

In a deck of 52 cards 10 points are for each face card, 1 point for each ace and all number cards have value equal to their number.

Formula used:

Probability of an event E is $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

Given an event Q, the conditional probability of event A is $P\left(A|Q\right)=\frac{P\left(A\text{ and Q}\right)}{P\left(Q\right)}$ where $P\left(Q\right)\ne 0$ .

Calculation:

Consider the provided information that in a deck of 52 cards 10 points are for each face card, 1 point for each ace and all number cards have value equal to their number.

The sum of total of two cards is equal to less than 7 in the following ways if at least one of them is ace.

Either the second card can be ace, there are 3 more aces in the deck.

Either the second card can be 2, there are 4 twos in the deck.

Either the second card can be 3, there are 4 threes in the deck.

Either the second card can be 4, there are 4 fours in the deck.

Either the second card can be 5, there are 4 fives in the deck.

Either the second card can be 6, there are 4 six in the deck.

Since one card is already drawn from the deck of 52 cards so remaining cards are 51.

There are $3+4+4+4+4+4=23$ options for selecting the second card.

Recall that Probability of an event E is $P\left(\text{E}\right)=\frac{\text{number of outcome}}{\text{total number of outcome}}$ .

Let A be the event sum of total of two cards is equal to less than 7 in the following ways if at least one of them is ace.

So,

  $P\left(A\right)=\frac{23}{51}$

Thus, the probability that sum of two cards drawn is less than or equal to 7 if at least one card is ace is $\frac{23}{51}$ .