Chapter 18, Problem 28E

(a)

To determine

Section 1:

To find: The number of outcomes with the sum of the spots on the up faces equal to 6.

Answer to Problem 28E

Solution: There are 5 outcomes for which the sum of the spots on the up faces is equal to 6.

Explanation of Solution

Calculation:

There are 6 spots on the six faces of a die numbered from 1 to 6. So, when 2 dice are rolled the possible number of outcomes or the sample space is as follows:

$S=\left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\\ \left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right),\\ \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\\ \left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right),\\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right),\\ \left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\end{array}\right\}$

The possible number of outcomes such that the sum on the up faces of the die is equal to 6 is as follows:

$\left\{\left(1,5\right),\left(5,1\right),\left(2,4\right),\left(4,2\right),\left(3,3\right)\right\}$

Hence, there are 5 such outcomes whose sum on the up faces will be equal to 6.

Interpretation: In the throw of two dice, there will be 5 such outcomes out of the total of 36 possibilities whose sum on the up faces will be 6.

Section 2:

To find: The probability such that the sum on the up faces of two dice is 6.

Solution: The probability of the sum of the points on the up faces of the two dice is equal to 6 is 0.1389.

Explanation:

Calculation:

From section 1 of part (a), the possible number of outcomes whose sum of the points on the up faces is 6 is as follows:

$\left\{\left(1,5\right),\left(5,1\right),\left(2,4\right),\left(4,2\right),\left(3,3\right)\right\}$

Since it is provided that each outcome is equally likely and also the occurrence of the outcomes is independent, the required probability is calculated as follows:

$\begin{array}{c}P\left(\text{Sum on up faces}=6\right)=\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\\ =\frac{5}{36}\\ =0.1389\end{array}$

Interpretation: Hence, there are 13.89% chances that the sum of the points on the up faces of the two dice is equal to 6.

(b)

Section 1:

To determine

To find: The number of outcomes with at least one of the up faces displaying a single spot.

Answer to Problem 28E

Solution: The required number of outcomes is 11.

Explanation of Solution

Calculation:

There are 6 spots on the six faces of a die numbered from 1 to 6. So, when 2 dice are rolled the possible number of outcomes or the sample space is as follows:

$S=\left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\\ \left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right)\\ \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right)\\ \left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right)\\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right)\\ \left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\end{array}\right\}$

The outcomes with at least one of the up face displaying a single spot are as follows:

$\left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\\ \left(2,1\right),\left(3,1\right),\left(4,1\right),\left(5,1\right),\left(6,1\right)\end{array}\right\}$

Hence, there are 11 such outcomes with at least one of the up face displaying a single spot.

Interpretation: In the throw of two dice, there will be 11 such outcomes out of the total of 36 possibilities when at least one of the dice will display 1 on the up face.

Section 2:

To find: The probability that at least one of the up face in a throw of 2 dice will display a single spot.

Solution: The probability of at least one of the dice will display a single point on the up face is 0.3056.

Explanation:

Calculation:

From section 1 of part (b), the possible number of outcomes with at least one die displaying a single spot on the up face is as follows:

$\left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\\ \left(2,1\right),\left(3,1\right),\left(4,1\right),\left(5,1\right),\left(6,1\right)\end{array}\right\}$

Since it is provided that each outcome is equally likely and also the occurrence of the outcomes is independent, the required probability is calculated as follows:

$\begin{array}{c}P\left(\text{At least one die displaying a single spot}\right)=\left(\begin{array}{l}\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\\ \frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\end{array}\right)\\ =\frac{11}{36}\\ =0.3056\end{array}$

Interpretation: Hence, there are approximately 30.56% chances that at least one of the dice will display a single point on the up face.

(c)

To determine

Section 1:

To find: The number of outcomes with the sum of the spots on the up faces as 6 and at least one of the up face with a single spot..

Answer to Problem 28E

Solution: The required number of outcomes is 2.

Explanation of Solution

Calculation:

There are 6 spots on the six faces of a die numbered from 1 to 6. So, when 2 die are rolled the possible number of outcomes or the sample space are as follows:

$S=\left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\\ \left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right)\\ \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right)\\ \left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right)\\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right)\\ \left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\end{array}\right\}$

The possible number of outcomes such that the sum on the up faces of the die is equal to 6 and at least one of the up face displaying a single spot are as follows:

$\left\{\left(1,5\right),\left(5,1\right)\right\}$

Hence, there are 2 such outcomes such that the sum on the up faces of the dice is equal to 6 and at least one of the up face displays a single spot.

Interpretation: In the throw of two dice there will be 2 such outcomes out of the total of 36 possibilities such that the sum on the up faces of the die is equal to 6 and at least one of the up faces displaying a single spot.

Section 2:

To find: The probability such that the sum of the spots on the up faces in the throw of two dice is 6 and at least one up face has only a single point.

Solution: The probability of the sum of the points on the up faces of two coins is 6 and at least one die displays only 1 at its up face is 0.0556.

Explanation:

Calculation:

From section 1 of part (c), the possible number of outcomes such that the sum on the up faces of the die is equal to 6 and at least one of the up face displaying a single spot is as follows:

$\left\{\left(1,5\right),\left(5,1\right)\right\}$

Since it is provided that each outcome is equally likely and also the occurrence of the outcomes is independent, the required probability is calculated as follows:

$\begin{array}{c}P\left(\begin{array}{l}\text{Sum on up faces is }6\text{ and at }\\ \text{least one upface displays only 1}\end{array}\right)=\frac{1}{36}+\frac{1}{36}\\ =\frac{2}{36}\\ =0.0556\end{array}$

Interpretation: Hence, there are approximately 5.56% chances that the sum of the points on the up faces of two coins is 6 and at least one die displays only 1 at its up face.

(d)

Section 1:

To determine

To find: The number of outcomes such that the sum of the spots on the up faces in the throw of two dice is 6 or at least one of the up faces displaying only a single point.

Answer to Problem 28E

Solution: The required number of outcomes is 14.

Explanation of Solution

Calculation:

There are 6 spots on the six faces of a die numbered from 1 to 6. So, when 2 dice are rolled the possible number of outcomes or the sample space is as follows:

$S=\left\{\begin{array}{l}\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\\ \left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right)\\ \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right)\\ \left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right)\\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right)\\ \left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\end{array}\right\}$

The possible number of outcomes such that the sum of the spots on the up faces in the throw of two dice is 6 or at least one of the up faces displaying only a single point is as follows:

$\left\{\begin{array}{l}\left(1,5\right),\left(5,1\right),\left(2,4\right),\left(4,2\right),\left(3,3\right),\\ \left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,6\right),\\ \left(2,1\right),\left(3,1\right),\left(4,1\right),\left(6,1\right)\end{array}\right\}$

Hence, there are 14 such outcomes whose sum of the spots on the up faces in the throw of two dice is 6 or at least one of the up faces displaying only a single point.

Interpretation: In the throw of two dice, there will be 14 such outcomes out of the total of 36 possibilities whose sum of the spots on the up faces in the throw of two dice is 6 or at least one of the up faces displaying only a single point.

Section 2:

To find: The probability that sum of the points on the up faces is either 6 or at least one up face displays a single spot.

Solution: The probability of the sum of the points on the up faces is either 6 or at least one up face displays a single spot is 0.3889.

Explanation:

Calculation:

From the results of section 1 of part (d), the possible number of outcomes whose sum of the points on the up faces is either 6 or at least one up face displays a single spot is as follows:

$\left\{\begin{array}{l}\left(1,5\right),\left(5,1\right),\left(2,4\right),\left(4,2\right),\left(3,3\right),\\ \left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,6\right),\\ \left(2,1\right),\left(3,1\right),\left(4,1\right),\left(6,1\right)\end{array}\right\}$

Since it is provided that each outcome is equally likely and also the occurrence of the outcomes is independent, the required probability is calculated as follows:

$\begin{array}{c}P\left(\begin{array}{l}\text{Sum on up faces is }6\text{ or atleast }\\ \text{one up face displays a single spot}\end{array}\right)=\left(\begin{array}{l}\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\\ \frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}+\frac{1}{36}\end{array}\right)\\ =\frac{14}{36}\\ =0.3889\end{array}$

Interpretation: Hence, there are approximately 38.89% chances that the sum of the points on the up faces is either 6 or at least one up face displays a single spot.

(e)

To determine

To test: The addition rule of probability using the results of part (a), (b), (c), and (d).

Answer to Problem 28E

Solution: The sum of the probabilities of parts (a) and (b) minus the probability obtained in part (c) is equal to the probability obtained in part (d). This verifies the addition rule of probability.

Explanation of Solution

Calculation:

The addition rule of probability states that the probability of occurrence of either of the two events is the sum of their individual probability minus the probability of their simultaneous occurrence. To verify the addition rule of probability, add the probabilities obtained in parts (a) and (b) and subtract the result of part (c) as follows:

$\begin{array}{c}P\left(\text{Part a}\right)+P\left(\text{Part b}\right)-P\left(\text{Part c}\right)=0.1389+0.3056-0.0556\\ =0.4445-0.0556\\ =0.3889\end{array}$

The above obtained quantity is equal to the probability obtained in part (d). Hence, this verifies the addition rule of probability.

Conclusion: From the results of parts (a), (b), (c), and (d), the addition rule of probability is verified.