Chapter 4.1, Problem 17E

a.

To determine

To obtain: The probability of getting a Queen.

Answer to Problem 17E

The probability of getting a Queen  is $\frac{1}{13}$.

Explanation of Solution

Given info:

One card is drawn from the deck.

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. There is a total of 4 queen in a deck.

The total number of outcomes is 52.

Let event A denote getting a queen.

The possible number of outcomes for getting a queen is 4.

That is, there are 4 outcomes for getting an event A.

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 in A’ and 52 for ‘Total number of outcomes in the sample space’,

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

Thus, the probability of getting a queen is $\frac{1}{13}$.

b.

To determine

To obtain: The probability of getting a club.

Answer to Problem 17E

The probability of getting a club is $\frac{1}{4}$.

Explanation of Solution

Calculation:

Let event B denote getting a club.

The possible number of outcomes for getting a club is 13.

That is, there are 13 outcomes for getting an event B.

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 in B’ and 52 for ‘Total number of outcomes in the sample space’,

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

Thus, the probability of getting a club is $\frac{1}{4}$.

c.

To determine

To obtain: The probability of getting a queen of cubs.

Answer to Problem 17E

The probability of getting a queen of cubs is $\frac{1}{52}$.

Explanation of Solution

Calculation:

Let event C denote getting a queen of cubs.

The possible number of outcomes for getting a queen of cubs is 1.

That is, there is 1 outcome for getting an event C.

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

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

Thus, the probability of getting a queen of cubs is $\frac{1}{52}$.

d.

To determine

To obtain: The probability of getting a 3 or an 8.

Answer to Problem 17E

The probability of getting a 3 or an 8 is $\frac{2}{13}$.

Explanation of Solution

Calculation:

Let event D denote getting a 3.

The possible number of outcomes for getting a 3 is 4.

That is, there are 4 outcomes for getting an event D.

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

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

Let event E denote getting an 8.

The possible number of outcomes for getting an 8 is 4.

That is, there is 4 outcomes for getting an event E.

The formula for probability of event E is,

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

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

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

Addition Rule:

The formula for probability of getting event D or event E is,

$P\left(D\text{ or }E\right)=P\left(D\right)+P\left(E\right)$

Substitute $\frac{1}{13}$ for ‘ $P\left(D\right)$’, $\frac{1}{13}$ for ‘ $P\left(E\right)$’

$\begin{array}{c}P\left(D\text{ or }E\right)=\frac{1}{13}+\frac{1}{13}\\ =\frac{2}{13}\end{array}$

Thus, the probability of getting a 3 or an 8 is $\frac{2}{13}$

e.

To determine

To obtain: The probability that getting a 6 or a spade.

Answer to Problem 17E

The probability that getting a 6 or a spade is $\frac{4}{13}$.

Explanation of Solution

Calculation:

Let event F denote getting a 6.

The possible number of outcomes for getting a 6 is 4.

That is, there are 4 outcomes for getting an event F.

The formula for probability of event F is,

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

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

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

Let event G denote getting a spade.

The possible number of outcomes for getting a spade is 13.

That is, there is 13 outcomes for getting an event G.

The formula for probability of event G is,

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

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

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

The formula for probability of event F and G is,

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

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

$P\left(F\text{ and }G\right)=\frac{1}{52}$

Addition Rule:

The formula for probability of getting event F or event G is,

$P\left(F\text{ or }G\right)=P\left(F\right)+P\left(G\right)-P\left(F\cap G\right)$

Substitute $\frac{4}{52}$ for ‘ $P\left(F\right)$’, $\frac{13}{52}$ for ‘ $P\left(F\right)$’, and $\frac{4}{52}$ for $P\left(F\text{ and }G\right)$

$\begin{array}{c}P\left(F\text{ or }G\right)=\frac{4}{52}+\frac{13}{52}-\frac{1}{52}\\ =\frac{16}{52}\\ =\frac{4}{13}\end{array}$

Thus, the probability of getting a 6 or a spade is $\frac{4}{13}$