Chapter 4.2, Problem 23E

Selecting a Card If one card is drawn from an ordinary deck of cards, find the probability of getting each event:

a. A 7 or an 8 or a 9

b. A spade or a queen or a king

c. A club or a face card

d. An ace or a diamond or a heart

e. A 9 or a 10 or a spade or a club

To determine

To obtain: The probability of getting a 7 or an 8 or a 9.

Answer to Problem 23E

The probability of getting a 7 or an 8 or a 9 is $\frac{3}{13}$.

Explanation of Solution

Calculation:

In ordinary deck of cards there are 4 suits. They are hearts, clubs, diamonds and spades. In each suite there are 13 cards. In 13 cards, nine cards are numbers from 2 to 10 and remaining four cards are king, queen, ace and jack cards.

Here, the number ‘7, 8, 9’ areoutcomes.

The total number of outcomes is 52.

Let event A denote that the outcome is 7.

The number of 7 number cards is 4.

The formula for probability of event A is,

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Substitute 4 for ‘Number of outcomes inA’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(A\right)=\frac{4}{\text{52}}$

Let event B denote that the outcome is 8.

The number of 8 number cards is 4.

The formula for probability of event A is,

$P\left(B\right)=\frac{\text{Number of outcomes in }B}{\text{Total number of outcomes in the sample space}}$

Substitute 4 for ‘Number of outcomes inC’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(C\right)=\frac{4}{\text{52}}$

Let event C denote that the outcome is 9.

The number of 9 number cards is 4.

The formula for probability of event C is,

$P\left(C\right)=\frac{\text{Number of outcomes in C}}{\text{Total number of outcomes in the sample space}}$

Substitute 4 for ‘Number of outcomes inC’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(C\right)=\frac{4}{\text{52}}$

The formula for probability of getting A or B or C is,

$P\left(A\text{ or }B\text{ or }C\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)$

Substitute $\frac{4}{\text{52}}$ for $P\left(A\right)$, $\frac{4}{\text{52}}$ for $P\left(B\right)$ and $\frac{4}{\text{52}}$ for $P\left(C\right)$

$\begin{array}{c}P\left(A\text{ or }B\text{ or }C\right)=\frac{4}{52}+\frac{4}{52}+\frac{4}{52}\\ =\frac{12}{52}\\ =\frac{3}{13}\end{array}$

Thus, the probability of getting a 7 or an 8 or a 9 is $\frac{3}{13}$.

To determine

To obtain: The probability of getting a spade or a queen or a king.

Answer to Problem 23E

The probability of getting a spade or a queen or a king is $\frac{19}{52}$.

Explanation of Solution

Calculation:

Let event A denote that the outcome is spade, event B denote that the outcome is queen and event C denote that the outcome is king.

The formula for probability of getting A or B or C is,

$P\left(A\text{ or }B\text{ or }C\right)=\left[\begin{array}{l}P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\text{ and B}\right)-P\left(B\text{ and }C\right)\\ -P\left(A\text{ and }C\right)+P\left(A\text{ and }B\text{ and }C\right)\end{array}\right]$

The number of spade cards is 13.

The total number of outcomes is 52.

The formula for probability of event A is,

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Substitute 13 for ‘Number of outcomes inA’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(A\right)=\frac{13}{\text{52}}$

The number of queen cards is 4.

The total number of outcomes is 52.

The formula for probability of event B is,

$P\left(B\right)=\frac{\text{Number of outcomes in }B}{\text{Total number of outcomes in the sample space}}$

Substitute 4 for ‘Number of outcomes inB’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(B\right)=\frac{4}{\text{52}}$

The number of king cards is 4.

The total number of outcomes is 52.

The formula for probability of event C is,

$P\left(C\right)=\frac{\text{Number of outcomes in C}}{\text{Total number of outcomes in the sample space}}$

Substitute 4 for ‘Number of outcomes inC’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(C\right)=\frac{4}{\text{52}}$

The formula for probability of event A and B is,

$P\left(A\text{ and }B\right)=\frac{\text{Number of outcomes in }A\text{ and }B}{\text{Total number of outcomes in the sample space}}$

Substitute 1 for ‘Number of outcomes inA and B’ and 52 for ‘Total number of outcomes in the sample space’,

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

The formula for probability of event A and C is,

$P\left(A\text{ and }C\right)=\frac{\text{Number of outcomes in }A\text{ and }C}{\text{Total number of outcomes in the sample space}}$

Substitute 1 for ‘Number of outcomes inA and C’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(A\text{ and }C\right)=\frac{1}{\text{52}}$

There is no intersection part for events B and C. Hence $P\left(B\text{ and }C\right)=0$ and $P\left(A\text{ and }B\text{ and }C\right)=0$

Substitute $\frac{13}{\text{52}}$ for $P\left(A\right)$, $\frac{4}{\text{52}}$ for $P\left(B\right)$, $\frac{4}{\text{52}}$ for $P\left(C\right)$, $\frac{1}{\text{52}}$ for $P\left(A\text{ and }B\right)$ , $\frac{1}{\text{52}}$ for $P\left(A\text{ and }C\right)$, 0 for $P\left(B\text{ and }C\right)$, 0 for $P\left(A\text{ and }B\text{ and }C\right)$.

$\begin{array}{c}P\left(A\text{ or }B\text{ or }C\right)=\frac{13}{52}+\frac{4}{52}+\frac{4}{52}-\frac{1}{52}-\frac{1}{52}-0+0\\ =\frac{19}{52}\end{array}$

Thus, the probability of getting a spade or a queen or a king is $\frac{19}{52}$.

To determine

To obtain: The probability of getting a club or a face card.

Answer to Problem 23E

The probability of getting a club or a face card is $\frac{11}{26}$.

Explanation of Solution

Calculation:

Let event A denote that the outcome is club card, event B denote that the outcome is face card.

The formula for probability of getting A or B is,

$P\left(A\text{ or }B\right)=P\left(A\right)+P\left(B\right)-P\left(A\text{ and }B\right)$

The number of spade club cards is 13.

The total number of outcomes is 52.

The formula for probability of event A is,

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Substitute 13 for ‘Number of outcomes inA’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(A\right)=\frac{13}{\text{52}}$

The cards king, queen and jack are face cards.

The number of face cards is in deck of cards is 12.

The total number of outcomes is 52.

The formula for probability of event B is,

$P\left(B\right)=\frac{\text{Number of outcomes in }B}{\text{Total number of outcomes in the sample space}}$

Substitute 12 for ‘Number of outcomes inB’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(B\right)=\frac{12}{\text{52}}$

The number of king cards is 4.

The total number of outcomes is 52.

The formula for probability of event A and B is,

$P\left(A\text{ and }B\right)=\frac{\text{Number of outcomes in }A\text{ and }B}{\text{Total number of outcomes in the sample space}}$

Substitute 3 for ‘Number of outcomes inA and B’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(A\text{ and }B\right)=\frac{3}{\text{52}}$

Substitute $\frac{13}{\text{52}}$ for $P\left(A\right)$, $\frac{12}{\text{52}}$ for $P\left(B\right)$ and $\frac{3}{\text{52}}$ for $P\left(A\text{ and }B\right)$

$\begin{array}{c}P\left(A\text{ or }B\right)=\frac{13}{52}+\frac{12}{52}-\frac{3}{52}\\ =\frac{22}{52}\\ =\frac{11}{26}\end{array}$

Thus, the probability of getting a club or a face card is $\frac{11}{26}$.

To determine

To obtain: The probability of getting an ace or a diamond or a heart.

Answer to Problem 23E

The probability of getting an ace or a diamond or a heart is $\frac{7}{13}$.

Explanation of Solution

Calculation:

Let event A denote that the outcome is ace, event B denote that the outcome is diamond and event C denote that the outcome is club.

The formula for probability of getting A or B or C is,

$P\left(A\text{ or }B\text{ or }C\right)=\left[\begin{array}{l}P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\text{ and B}\right)-P\left(B\text{ and }C\right)\\ -P\left(A\text{ and }C\right)+P\left(A\text{ and }B\text{ and }C\right)\end{array}\right]$

The number of ace cards is 4.

The total number of outcomes is 52.

The formula for probability of event A is,

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Substitute 4 for ‘Number of outcomes inA’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(A\right)=\frac{4}{\text{52}}$

The number of diamond cards is 13.

The total number of outcomes is 52.

The formula for probability of event B is,

$P\left(B\right)=\frac{\text{Number of outcomes in }B}{\text{Total number of outcomes in the sample space}}$

Substitute 13 for ‘Number of outcomes inB’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(B\right)=\frac{13}{\text{52}}$

The number of heart cards is 13.

The total number of outcomes is 52.

The formula for probability of event C is,

$P\left(C\right)=\frac{\text{Number of outcomes in C}}{\text{Total number of outcomes in the sample space}}$

Substitute 13 for ‘Number of outcomes inC’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(C\right)=\frac{13}{\text{52}}$

The formula for probability of event A and B is,

$P\left(A\text{ and }B\right)=\frac{\text{Number of outcomes in }A\text{ and }B}{\text{Total number of outcomes in the sample space}}$

Substitute 1 for ‘Number of outcomes inA and B’ and 52 for ‘Total number of outcomes in the sample space’,

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

The formula for probability of event A and C is,

$P\left(A\text{ and }C\right)=\frac{\text{Number of outcomes in }A\text{ and }C}{\text{Total number of outcomes in the sample space}}$

Substitute 1 for ‘Number of outcomes inA and C’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(A\text{ and }C\right)=\frac{1}{\text{52}}$

There is no intersection part for events B and C. Hence $P\left(B\text{ and }C\right)=0$ and $P\left(A\text{ and }B\text{ and }C\right)=0$

Substitute $\frac{4}{\text{52}}$ for $P\left(A\right)$, $\frac{13}{\text{52}}$ for $P\left(B\right)$, $\frac{13}{\text{52}}$ for $P\left(C\right)$, $\frac{1}{\text{52}}$ for $P\left(A\text{ and }B\right)$ , $\frac{1}{\text{52}}$ for $P\left(A\text{ and }C\right)$, 0 for $P\left(B\text{ and }C\right)$, 0 for $P\left(A\text{ and }B\text{ and }C\right)$.

$\begin{array}{c}P\left(A\text{ or }B\text{ or }C\right)=\frac{4}{52}+\frac{13}{52}+\frac{13}{52}-\frac{1}{52}-\frac{1}{52}-0+0\\ =\frac{28}{52}\\ =\frac{7}{13}\end{array}$

Thus, the probability of getting an ace or a diamond or a heart is $\frac{19}{52}$.

To determine

To obtain: The probability of getting a9 or a 10 or a spade or a club.

Answer to Problem 23E

The probability of getting a 9 or a 10 or a spade or a club is $\frac{15}{26}$.

Explanation of Solution

Calculation:

Let event A denote that the outcome is a 9, event B denote that the outcome is a 10,event C denote that the outcome is a spade and D denote that the outcome is a club.

The formula for probability of getting A or B or C or D is,

$P\left(A\text{ or }B\text{ or }C\right)=\left[\begin{array}{l}P\left(A\right)+P\left(B\right)+P\left(C\right)+P\left(D\right)-P\left(A\text{ and B}\right)-P\left(A\text{ and }C\right)\\ -P\left(A\text{ and }D\right)-P\left(B\text{ and }C\right)-P\left(B\text{ and }D\right)-P\left(C\text{ and }D\right)+\\ P\left(A\text{ and }B\text{ and }C\right)+P\left(A\text{ and }B\text{ and }D\right)+P\left(B\text{ and }C\text{ and }D\right)\\ +P\left(C\text{ and }D\text{ and }A\right)-P\left(A\text{ and }B\text{ and }C\text{ and D}\right)\end{array}\right]$

The number of 9 cards is 4.

The total number of outcomes is 52.

The formula for probability of event A is,

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Substitute 4 for ‘Number of outcomes inA’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(A\right)=\frac{4}{\text{52}}$

The number of 10 cards is 4.

The total number of outcomes is 52.

The formula for probability of event B is,

$P\left(B\right)=\frac{\text{Number of outcomes in }B}{\text{Total number of outcomes in the sample space}}$

Substitute 4 for ‘Number of outcomes inB’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(B\right)=\frac{4}{\text{52}}$

The number of spade cards is 13.

The total number of outcomes is 52.

The formula for probability of event C is,

$P\left(C\right)=\frac{\text{Number of outcomes in C}}{\text{Total number of outcomes in the sample space}}$

Substitute 13 for ‘Number of outcomes inC’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(C\right)=\frac{13}{\text{52}}$

The number of club cards is 13.

The total number of outcomes is 52.

The formula for probability of event D is,

$P\left(D\right)=\frac{\text{Number of outcomes in D}}{\text{Total number of outcomes in the sample space}}$

Substitute 13 for ‘Number of outcomes inD’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(D\right)=\frac{13}{\text{52}}$

The formula for probability of event A and C is,

$P\left(A\text{ and }C\right)=\frac{\text{Number of outcomes in }A\text{ and }C}{\text{Total number of outcomes in the sample space}}$

Substitute 1 for ‘Number of outcomes inA and B’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(A\text{ and }C\right)=\frac{1}{\text{52}}$

The formula for probability of event A and D is,

$P\left(A\text{ and }D\right)=\frac{\text{Number of outcomes in }A\text{ and }D}{\text{Total number of outcomes in the sample space}}$

Substitute 1 for ‘Number of outcomes inA and D’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(A\text{ and }D\right)=\frac{1}{\text{52}}$

The formula for probability of event B and C is,

$P\left(B\text{ and }C\right)=\frac{\text{Number of outcomes in }B\text{ and }C}{\text{Total number of outcomes in the sample space}}$

Substitute 1 for ‘Number of outcomes inA and B’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(B\text{ and }C\right)=\frac{1}{\text{52}}$

The formula for probability of event A and D is,

$P\left(B\text{ and }D\right)=\frac{\text{Number of outcomes in }B\text{ and }D}{\text{Total number of outcomes in the sample space}}$

Substitute 1 for ‘Number of outcomes inA and D’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(B\text{ and }D\right)=\frac{1}{\text{52}}$

The formula for probability of event A and B and C is,

$P\left(A\text{ and }B\text{ and }C\right)=\frac{\text{Number of outcomes in }A\text{ and }B\text{ and }C}{\text{Total number of outcomes in the sample space}}$

There is no intersection part for events A and B, C and D. Hence $P\left(A\text{ and }B\right)=0$, $P\left(C\text{ and }D\right)=0$, $P\left(B\text{ and }C\text{ and D}\right)=0$, $P\left(C\text{ and }D\text{ and A}\right)=0$, $P\left(A\text{ and }B\text{ and }C\right)=0$, $P\left(A\text{ and }B\text{ and }D\right)=0$ and $P\left(A\text{ and }B\text{ and }C\text{ and D}\right)=0$.

Substitute the probabilities of the events in $P\left(A\text{ or }B\text{ or }C\text{ or D}\right)$.

$\begin{array}{c}P\left(A\text{ or }B\text{ or }C\text{ or D}\right)=\left(\begin{array}{l}\frac{4}{52}+\frac{4}{52}+\frac{13}{52}+\frac{13}{52}-0-\frac{1}{52}-\frac{1}{52}-\frac{1}{52}\\ -\frac{1}{52}-0+0+0+0+0-0\end{array}\right)\\ =\frac{30}{52}\\ =\frac{15}{26}\end{array}$

Thus, the probability of getting a 9 or a 10 or a spade or a club is $\frac{15}{26}$.