Chapter 4.2, Problem 23E

a.

To determine

To obtain: The probability of getting a King or a queen or a jack.

Answer to Problem 23E

The probability of getting a King or a queen or a jack is $\underset{\_}{\frac{3}{13}}$.

Explanation of Solution

Given info:

One card is drawn from an ordinary deck of cards.

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 cards are a King or a queen or a jack.

The total number of outcomes is 52.

Let event A denote that getting a King.

The number of elements in A is 4.

The formula for probability of event A is,

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

Substitute 4 for ‘Number of outcomes in A’ and 52 for ‘Total number of outcomes’,

$\begin{array}{c}P\left(A\right)=\frac{4}{52}\\ =\frac{1}{13}\end{array}$

Let event B denote that getting a queen.

The number of elements in B is 4.

The formula for probability of event B is,

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

Substitute 4 for ‘Number of outcomes in B’ and 52 for ‘Total number of outcomes’,

$\begin{array}{c}P\left(A\right)=\frac{4}{52}\\ =\frac{1}{13}\end{array}$

Let event C denote that getting jack.

The number elements in  C is 4.

The formula for probability of event C is,

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

Substitute 4 for ‘Number of outcomes in C’ and 52 for ‘Total number of outcomes’,

$\begin{array}{c}P\left(A\right)=\frac{4}{52}\\ =\frac{1}{13}\end{array}$

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

$P(\text{getting a king or a queen or ajack)=P}\left(A\right)+P\left(B\right)+P\left(C\right)$

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

$\begin{array}{c}P(\text{getting a king or a queen or ajack)=}\frac{1}{13}+\frac{1}{13}+\frac{1}{13}\\ =\frac{3}{13}\end{array}$

Thus, the probability of getting a King or a queen or a jack is $\frac{3}{13}$.

b.

To determine

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

Answer to Problem 23E

The probability of getting a club or a heart or a spade is $\underset{\_}{\frac{3}{4}}$.

Explanation of Solution

Calculation:

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

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

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

The number of club cards is 13.

The total number of outcomes is 52.

The formula for probability of event A is,

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

Substitute 13 for ‘Number of outcomes in A’ and 52 for ‘Total number of outcomes’,

$\begin{array}{c}P\left(A\right)=\frac{13}{52}\\ =\frac{1}{4}\end{array}$

The number of heart cards is 13.

The total number of outcomes is 52.

The formula for probability of event B is,

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

Substitute 13   for ‘Number of outcomes in B’ and 52 for ‘Total number of outcomes’,

$\begin{array}{c}P\left(B\right)=\frac{13}{52}\\ =\frac{1}{4}\end{array}$

The number of spade cards is 13.

The total number of outcomes is 52.

The formula for probability of event C is,

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

Substitute 13 for ‘Number of outcomes in C’ and 52 for ‘Total number of’,

$\begin{array}{c}P\left(C\right)=\frac{13}{52}\\ =\frac{1}{4}\end{array}$

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

$\begin{array}{c}P(A\text{or }B\text{ or }C\text{)=}\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\\ =\frac{3}{4}\end{array}$

Thus, the probability of getting a club or a heart or a spade is $\underset{\_}{\frac{3}{4}}$.

c.

To determine

To obtain: The probability of getting a King or a queen or a diamond.

Answer to Problem 23E

The probability of getting a King or a queen or a diamond is $\underset{\_}{\frac{19}{52}}$.

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 cards are a King or a queen or a diamond.

The total number of outcomes is 52.

Let event A denote that getting a King.

The number of elements in A is 4.

The formula for probability of event A is,

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

Substitute 4 for ‘Number of outcomes in A’ and 52 for ‘Total number of outcomes’,

$\begin{array}{c}P\left(A\right)=\frac{4}{52}\\ =\frac{1}{13}\end{array}$

Let event B denote that getting a queen.

The number of elements in B is 4.

The formula for probability of event B is,

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

Substitute 4 for ‘Number of outcomes in B’ and 52 for ‘Total number of outcomes’,

$\begin{array}{c}P\left(B\right)=\frac{4}{52}\\ =\frac{1}{13}\end{array}$

Let event C denote that getting diamond.

The number of diamond cards is 13.

The total number of outcomes is 52.

The formula for probability of event C is,

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

Substitute 13 for ‘Number of outcomes in C’ and 52 for ‘Total number of’,

$\begin{array}{c}P\left(C\right)=\frac{13}{52}\\ =\frac{1}{4}\end{array}$

Events A and B cannot be occur together. Therefore $P\left(A\cap B\right)=0$

The formula for probability of event A and C is,

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

Substitute 1 for ‘Number of outcomes in A and C’ and 52 for ‘Total number of outcomes’,

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

The formula for probability of event B and C is,

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

Substitute 1 for ‘Number of outcomes in B and C’ and 52 for ‘Total number of outcomes’,

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

There is no element is common to the events A, B, and C. Therefore $P\left(A\cap B\cap C\right)=0$

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

$P(\text{getting a king or a queen or diamond)=}\left[\begin{array}{l}\text{P}\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\cap B\right)-\\ P\left(A\cap C\right)-P\left(B\cap C\right)+P\left(A\cap B\cap C\right)\end{array}\right]$

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

$\begin{array}{c}P(\text{getting a king or a queen or diamond)=}\frac{1}{13}+\frac{1}{13}+\frac{1}{4}-0-\frac{1}{52}-\frac{1}{52}+0\\ =\frac{19}{52}\end{array}$

Thus, the probability of getting a King or a queen or a diamond is $\underset{\_}{\frac{19}{52}}$.

d.

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 $\underset{\_}{\frac{7}{13}}$.

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,

$P(\text{getting an ace or a diamond or heart)=}\left[\begin{array}{l}\text{P}\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\cap B\right)-\\ P\left(A\cap C\right)-P\left(B\cap C\right)+P\left(A\cap B\cap C\right)\end{array}\right]$

The number of ace cards is 4.

The total number of outcomes is 52.

The formula for probability of event A is,

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

Substitute 4 for ‘Number of outcomes in A’ and 52 for ‘Total number of outcomes’,

$\begin{array}{c}P\left(A\right)=\frac{4}{52}\\ =\frac{1}{13}\end{array}$

The number of diamond cards is 13.

The total number of outcomes is 52.

The formula for probability of event B is,

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

Substitute 13 for ‘Number of outcomes in B’ and 52 for ‘Total number of outcomes’,

$\begin{array}{c}P\left(B\right)=\frac{13}{52}\\ =\frac{1}{4}\end{array}$

The number of heart cards is 13.

The total number of outcomes is 52.

The formula for probability of event C is,

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

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

$\begin{array}{c}P\left(C\right)=\frac{13}{52}\\ =\frac{1}{4}\end{array}$

The formula for probability of event A and B is,

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

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

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

The formula for probability of event A and C is,

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

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

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

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

Substitute $\frac{4}{52}$ for $P\left(A\right)$, $\frac{13}{52}$ for $P\left(B\right)$, $\frac{13}{42}$ for $P\left(C\right)$, $\frac{1}{52}$ for $P\left(A\text{and}C\right),\text{and }P\left(A\text{and}B\right)$, 0 for $P\left(B\text{and}C\right),\text{and }P\left(A\text{and}B\text{and}C\right)$.

$\begin{array}{c}P(\text{getting a king or a queen or diamond)=}\frac{4}{52}+\frac{13}{52}+\frac{13}{52}-\frac{1}{52}-\frac{1}{52}-0+0\\ =\frac{28}{52}\\ =\frac{7}{13}\end{array}$

Thus, the probability of getting an ace or a diamond or a heart is $\underset{\_}{\frac{7}{13}}$.

e.

To determine

To obtain: The probability of getting a 9 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 $\underset{\_}{\frac{15}{26}}$.

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,

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

The number of 9 cards is 4.

The total number of outcomes is 52.

The formula for probability of event A is,

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

Substitute 4 for ‘Number of outcomes in A’ and 52 for ‘Total number of outcomes’,

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

The number of 10 cards is 4.

The total number of outcomes is 52.

The formula for probability of event B is,

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

Substitute 4 for ‘Number of outcomes in B’ and 52 for ‘Total number of outcomes’,

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

The number of spade cards is 13.

The total number of outcomes is 52.

The formula for probability of event C is,

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

Substitute 13 for ‘Number of outcomes in C’ and 52 for ‘Total number of outcomes’,

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

The number of club cards is 13.

The total number of outcomes is 52.

The formula for probability of event D is,

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

Substitute 13 for ‘Number of outcomes in D’ and 52 for ‘Total number of outcomes’,

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

The formula for probability of event A and C is,

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

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

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

The formula for probability of event A and D is,

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

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

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

The formula for probability of event B and C is,

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

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

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

The formula for probability of event B and D is,

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

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

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

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

Substitute the probabilities of the events in the above equation.

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

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