Chapter 9.7, Problem 49E

a.

To determine

To find: the probability that both numbers are even.

Answer to Problem 49E

  $\frac{1}{4}$

Explanation of Solution

Concept Used:

Let A and B be two independent events, then

  $P\left(A\text{ and }B\right)=P\left(A\right)\times P\left(B\right)$

Let E be the event that both numbers are even.

Total number of even numbers from 1 to 40 = 20

Let A be the event of choosing an even number out of 1 to 40.

Then,

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

Here, both the numbers are chosen independently. So, the probability that both numbers are even is:

  $\begin{array}{c}P\left(\text{both numbers even}\right)=P\left(\text{first number is even}\right)\times P\left(\text{second number is even}\right)\\ =\frac{1}{2}\times \frac{1}{2}\\ =\frac{1}{4}\end{array}$

b.

To determine

To find: the probability that both one number is even and one number is odd.

Answer to Problem 49E

  $\frac{1}{2}$

Explanation of Solution

Concept Used:

Let A and B be two independent events, then

  $P\left(A\text{ and }B\right)=P\left(A\right)\times P\left(B\right)$

Let E be the event that one number is even and one number is odd.

Total number of even numbers from 1 to 40 = 20

Total number of odd numbers from 1 to 40 = 20

Let A be the event of choosing an even number out of 1 to 40.

Then,

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

Let B be the event of choosing an odd number out of 1 to 40.

Then,

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

So, the probability that both one number is even and one number is odd is:

  $\begin{array}{c}P\left(\text{E}\right)=P\left(\text{first number even and second odd}\right)+P\left(\text{first number odd and second even}\right)\\ =P\left(\text{first even}\right)\times P\left(\text{second odd}\right)+P\left(\text{first odd}\right)\times P\left(\text{second even}\right)\\ =\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{2}\end{array}$

  $\begin{array}{c}=\frac{1}{4}+\frac{1}{4}\\ =\frac{1}{2}\end{array}$

c.

To determine

To find: the probability that both numbers are less than 30.

Answer to Problem 49E

  $\frac{841}{1600}$

Explanation of Solution

Concept Used:

Let A and B be two independent events, then

  $P\left(A\text{ and }B\right)=P\left(A\right)\times P\left(B\right)$

Let E be the event that both numbers are less than 30.

Total number from1 to 40 which are less than 30 = 29

Let A be the event of choosing a number less than 30 out of 1 to 40.

Then,

  $P\left(A\right)=\frac{29}{40}$

Here, both the numbers are chosen independently. So, the probability that both numbers are less than 30 is:

  $\begin{array}{c}P\left(\text{both numbers less than 30}\right)=P\left(\text{first number less than 30}\right)\times P\left(\text{second number less than 30}\right)\\ =\frac{29}{40}\times \frac{29}{40}\\ =\frac{841}{1600}\end{array}$

b.

To determine

To find: the probability that the same number is selected twice.

Answer to Problem 49E

  $\frac{1}{40}$

Explanation of Solution

Concept Used:

Let A and B be two independent events, then

  $P\left(A\text{ and }B\right)=P\left(A\right)\times P\left(B\right)$

Let E be the event that the same number is selected twice.

Now let 1^st number be randomly chosen from 1 to 40. So, there any number out of 40 can be chosen.

So,

  $\begin{array}{c}P\left(\text{first number}\right)=\frac{40}{40}\\ =1\end{array}$

Now, 2^nd number needs to be same as the first number. So, there is only 1 choice out of 40 choices.

Then,

  $P\left(\text{second number same as first}\right)=\frac{1}{40}$

So, the probability that both numbers are same is:

  $\begin{array}{c}P\left(\text{E}\right)=P\left(\text{first number}\right)\times P\left(\text{second number same as first}\right)\\ =1\times \frac{1}{40}\\ =\frac{1}{40}\end{array}$