Chapter 8.6, Problem 60E

(a).

To determine

To find:

The probability that the numbers are both even.

Answer to Problem 60E

The required probability is $\frac{1}{4}$ .

Explanation of Solution

Given:

Two integers from $1$ through $40$ are chosen by a random number generator.

Concept Used:

Apply the formula of the probability:

  $\begin{array}{c}P\left(A\right)=\frac{NumberofFavourableOutcome}{TotalNumberofFavourableOutcome}\\ =\frac{n\left(A\right)}{n\left(S\right)}\end{array}$

Calculation:

Consider the total number is $40$ .

Assume the event $A$ be even numbers.

In which $20$ number is even.

Thus, $n\left(S\right)=40$ and $n\left(A\right)=20$ .

  $\begin{array}{c}P\left(botheven\right)=\frac{20}{40}\cdot \frac{20}{40}\\ =\frac{1}{4}\end{array}$

Conclusion:

Hence, the required probability is $\frac{1}{4}$ .

(b).

To determine

To find:

The probability that the one number is even and one number is odd.

Answer to Problem 60E

The required probability is $\frac{1}{2}$ .

Explanation of Solution

Given:

Two integers from $1$ through $40$ are chosen by a random number generator.

Concept Used:

Apply the formula of the probability:

  $\begin{array}{c}P\left(A\right)=\frac{NumberofFavourableOutcome}{TotalNumberofFavourableOutcome}\\ =\frac{n\left(A\right)}{n\left(S\right)}\end{array}$

Calculation:

Consider the total number is $40$ .

In which $20$ number is even.

In which $20$ number is odd.

  $\begin{array}{c}P\left(even\right)=\frac{20}{40}\\ =\frac{1}{2}\end{array}$

  $\begin{array}{c}P\left(odd\right)=\frac{20}{40}\\ =\frac{1}{2}\end{array}$

  $\begin{array}{c}P\left(oneoddandone\text{ even}\right)=P\left(1stodd\cdot 2nd\text{ even}\right)+P\left(1st\text{ even}\cdot 2ndodd\right)\\ =\frac{1}{2}\cdot \frac{1}{2}+\frac{1}{2}\cdot \frac{1}{2}\\ =\frac{1}{4}+\frac{1}{4}\\ =\frac{1}{2}\end{array}$

Conclusion:

Hence, the required probability is $\frac{1}{2}$ .

(c).

To determine

To find:

The probability that the both numbers are less than $30$ .

Answer to Problem 60E

The required probability is $0.5256$ .

Explanation of Solution

Given:

Two integers from $1$ through $40$ are chosen by a random number generator.

Concept Used:

Apply the formula of the probability:

  $\begin{array}{c}P\left(A\right)=\frac{NumberofFavourableOutcome}{TotalNumberofFavourableOutcome}\\ =\frac{n\left(A\right)}{n\left(S\right)}\end{array}$

Calculation:

Consider the total number is $40$ .

Total number less than $30$ is $29$ .

Assume that the event $A$ is number less than $30$ .

Thus, $n\left(S\right)=40$ and $n\left(A\right)=29$ .

  $\begin{array}{c}P\left(A\right)=\frac{29}{40}\cdot \frac{29}{40}\\ =\frac{841}{1600}\\ =0.5256\end{array}$

Conclusion:

Hence, the required probability is $0.5256$ .

(d).

To determine

To find:

The probability that the same number is chosen twice.

Answer to Problem 60E

The required probability is $0.513$ .

Explanation of Solution

Given:

Two integers from $1$ through $40$ are chosen by a random number generator.

Concept Used:

Apply the formula of the probability:

  $\begin{array}{c}P\left(A\right)=\frac{NumberofFavourableOutcome}{TotalNumberofFavourableOutcome}\\ =\frac{n\left(A\right)}{n\left(S\right)}\end{array}$

Calculation:

There are $C{}_2=780$ different possible combinations of $2$ randomly generated numbers from $1$ to $40$ .

There are $40$ different ways to select the same number twice.

Therefore, the probability of both numbers being same is:

  $\begin{array}{c}P\left(bothsame\right)=\frac{40}{780}\\ =0.513\end{array}$

Conclusion:

Hence, the required probability is $0.513$ .