Chapter 5.2, Problem 46E

(a)

To determine

Two − way table describing sample space in terms of events R and N.

Explanation of Solution

Given information:

R: get a red disk

N: get a disk with the number 9

For two − way table,

We are required to make two rows and two columns.

Let us name the rows and columns according to the events R and N.

Rows: red, not red.

Columns: nine, not nine.

Since there are 36 disks in total, we have to mention 36 in the bottom left corner of the table.

We know that

There are 9 disks of red, blue, green and yellow each.

This implies

There are 9 disks which are red in color.

And

Remaining 27 disks of blue, green and yellow color.

We also know that

Each color disks are numbered 1 − 9.

This implies

There is only 1 disk which is colored red and numbered nine.

And

Remaining 3 disks are numbered 9 and of different color.

This also implies

With only 1 disk of number 9 and colored red, remaining 8 disks of red color are numbered 1 − 8.

Since 3 disks of different color are numbered 9, remaining 24 disks of different color are numbered 1 − 8.

Two − way table is shown as:

  

(b)

To determine

Probabilities of both events R and N.

Answer to Problem 46E

Probabilities:

For event R,

  $P\left(\text{R}\right)=0.25$

For event N,

  $P\left(\text{N}\right)\approx 0.1111$

Explanation of Solution

Given information:

R: get a red disk

N: get a disk with the number 9

We know that

9 out of 36 disks are red in color.

When number of favorable outcomes are divided by the number of possible outcomes, we get the probability.

  $P\left(\text{R}\right)=\frac{\text{Number}\text{of}\text{favourable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{9}{36}=\frac{1}{4}=0.25$

We also know that

Each color disks are numbered 1 − 9.

With 4 colors, 4 disks are numbered 9 among 36 disks.

  $P\left(\text{N}\right)=\frac{\text{Number}\text{of}\text{favourable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}=\frac{4}{36}=\frac{1}{9}\approx 0.1111$

(c)

To determine

Discuss the event “R and N” and also find its probability.

Answer to Problem 46E

Probability for event,

  $P\left(\text{R}\text{and}\text{N}\right)\approx 0.0278$

Explanation of Solution

Given information:

R: get a red disk

N: get a disk with the number 9

“R and N” represents the event that the disk is numbered 9 and red in color.

We know that

Out of total 36 disks, there are 9 disks which are red in color.

Since each color disks are numbered 1 − 9.

This implies

There is only one disk which is numbered 9 and red in color among all 36 disks and possesses the event “R and N”.

Now,

For the probability,

Number of favorable outcomes are divided by the number of possible outcomes.

  $P\left(\text{R}\text{and}\text{N}\right)=\frac{\text{Number}\text{of}\text{favourable}\text{outcomes}}{\text{Number}\text{of}\text{possible}\text{outcomes}}\approx 0.0278$

(d)

To determine

Discuss the probability of event “R or N” is not equal to the sum of individual probabilities of both events. Furthermore, general addition rule to be used for the probability of event “R or N”.

Answer to Problem 46E

Probability for event,

  $P\left(\text{R}\text{or}\text{N}\right)\approx 0.3333$

Explanation of Solution

Given information:

R: get a red disk

N: get a disk with the number 9

For any two events,

General addition rule:

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

According to the statement,

  $P\left(\text{R}\text{or}\text{N}\right)\ne P\left(\text{R}\right)+P\left(\text{N}\right)$

This is due to the events R and N are not mutually exclusive events.

That means

The occurrence of both the events at the same time is possible.

Such that

  $P\left(\text{R}\text{and}\text{N}\right)\ne 0$

From Part (b) and (c),

We have

Probabilities,

For event R:

  $P\left(\text{R}\right)=\frac{9}{36}$

For event N:

  $P\left(\text{N}\right)=\frac{4}{36}$

For event “R and N”

  $P\left(\text{R}\text{and}\text{N}\right)=\frac{1}{36}$

Apply general addition rule for any two events:

  $\begin{array}{l}P\left(\text{R}\text{or}\text{N}\right)=P\left(\text{R}\right)+P\left(\text{N}\right)-P\left(\text{R}\text{and}\text{N}\right)\\ =\frac{9}{36}+\frac{4}{36}-\frac{1}{36}\\ =\frac{12}{36}\\ =\frac{1}{3}\\ \approx 0.3333\end{array}$