Chapter 13.5, Problem 37E

To determine

To calculate: The probability that sum of two cards drawn is greater than 18 if at least one card is queen of diamonds.

Answer to Problem 37E

The probability that sum of two cards drawn is greater than 18 if at least one card is queen of diamonds is $\frac{19}{51}$ .

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 greater than 18 the following ways if at least one of them is queen of diamonds.

The other card needs to have value either nine or ten points.

In a deck of cards there are 4 cards that have 9 as the value.

There are 16 cards that have 10 has the value.

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

One queen of diamonds is already drawn which is a part of cards that have value as 10.

There are $4+16-1=19$ options for selecting the second card.

Recall that 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 $P\left(Q\right)\ne 0$ .

Let A be the event that sum is greater than 18 and B be the event of selecting a queen of diamonds.

Then probability of selecting a queen of diamonds is,

$P\left(B\right)=\frac{1}{52}\cdot \frac{51}{51}$

Then probability of selecting a queen of diamonds and sum is greater than 18 is,

$P\left(A\text{ and }B\right)=\frac{1}{52}\cdot \frac{19}{51}$

The probability that sum of two cards drawn is greater than 18 if at least one card is queen of diamonds.

So,

  $\begin{array}{c}P\left(A|B\right)=\frac{\frac{1}{52}\cdot \frac{19}{51}}{\frac{1}{52}\cdot \frac{51}{51}}\\ =\frac{19}{51}\end{array}$

Thus, the probability that sum of two cards drawn is greater than 18 if at least one card is queen of diamonds is $\frac{19}{51}$ .