Chapter 4, Problem 137E

(a)

To determine

To find: The probabilities of the events $A\text{ and }B$.

Answer to Problem 137E

Solution: The probabilities are as follows:

$\begin{array}{c}P\left(A\right)=\frac{1}{36}\\ P\left(B\right)=\frac{15}{36}\end{array}$

Explanation of Solution

Given: $A=\left\{2\text{ on the first roll}\right\},B=\left\{8\text{ or more on the first roll}\right\}.$

Explanation:

Calculation: The sample space can be shown as follows:

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

So, using the sample space, a table can be formed as follows:

Sum	Favorable Outcomes
2	1
3	2
4	3
5	4
6	5
7	6
8	5
9	4
10	3
11	2
12	1

Using the table, for event $A$, there is only one way of getting a sum of two on the first roll and there are total 36 outcomes possible. Therefore:

$\begin{array}{c}P\left(A\right)=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}\\ =\frac{1}{36}\end{array}$

For event $B$, there are 15 ways of getting a sum of 8 or more on the first roll and there are total 36 outcomes possible. Therefore:

$\begin{array}{c}P\left(B\right)=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}\\ =\frac{5+4+3+2+1}{36}\\ =\frac{15}{36}\end{array}$

Hence, the probability $P\left(A\right)$ is $\frac{1}{36}$ and the probability $P\left(B\right)$ is $\frac{15}{36}$.

(b)

To determine

To find: The probabilities of events $A\text{ and }B$.

Answer to Problem 137E

Solution: The probabilities are as follows:

$\begin{array}{c}P\left(A\right)=\frac{1}{36}\\ P\left(B\right)=\frac{15}{36}\end{array}$

Explanation of Solution

Given: $A=\left\{2\text{ on the first roll}\right\},B=\left\{8\text{ or more on the second roll}\right\}.$

Explanation:

Calculation: By using the table in part (a), for event $A$, there is only one way of getting a sum of two on the first roll and there are total 36 outcomes possible. Therefore:

$\begin{array}{c}P\left(A\right)=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}\\ =\frac{1}{36}\end{array}$

By using the table in part (a), for event $B$, there are 15 ways of getting a sum of 8 or more on the second roll and there are total 36 outcomes possible. Therefore:

$\begin{array}{c}P\left(B\right)=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}\\ =\frac{5+4+3+2+1}{36}\\ =\frac{15}{36}\end{array}$

Hence, the probability $P\left(A\right)$ is $\frac{1}{36}$ and the probability $P\left(B\right)$ is $\frac{15}{36}$.

(c)

To determine

To find: The probabilities of events $A\text{ and }B$.

Answer to Problem 137E

Solution: The probabilities are as follows:

$\begin{array}{c}P\left(A\right)=\frac{10}{36}\\ P\left(B\right)=\frac{6}{36}\end{array}$

Explanation of Solution

Given: $A=\left\{5\text{ or less on second roll}\right\},B=\left\{4\text{ or less on the first roll}\right\}.$

Explanation:

Calculation: By using the table in part (a), for event $A$, there are 10 ways of getting a sum of 5 or less on the second roll and there are total 36 outcomes. Therefore:

$\begin{array}{c}P\left(A\right)=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}\\ =\frac{4+3+2+1}{36}\\ =\frac{10}{36}\end{array}$

By using the table in part (a), for event $B$, there are 6 ways of getting a sum of 4 or less on the first roll and there are total 36 outcomes possible. Therefore:

$\begin{array}{c}P\left(B\right)=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}\\ =\frac{3+2+1}{36}\\ =\frac{6}{36}\end{array}$

Hence, the probability $P\left(A\right)$ is $\frac{10}{36}$ and the probability $P\left(B\right)$ is $\frac{6}{36}$.

(d)

To determine

To find: The probabilities of events $A\text{ and }B$.

Answer to Problem 137E

Solution: The probabilities are as follows:

$\begin{array}{c}P\left(A\right)=\frac{10}{36}\\ P\left(B\right)=\frac{6}{36}\end{array}$

Explanation of Solution

Given: $A=\left\{5\text{ or less on second roll}\right\},B=\left\{4\text{ or less on the second roll}\right\}.$

Explanation:

Calculation: By using the table in part (a), for event $A$, there are 10 ways of getting a sum of 5 or less on the second roll and there are total 36 outcomes possible. Therefore:

$\begin{array}{c}P\left(A\right)=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}\\ =\frac{4+3+2+1}{36}\\ =\frac{10}{36}\end{array}$

By using the table in part (a), for event $B$, there are 6 ways of getting a sum of 4 or less on the second roll and there are total 36 outcomes possible. Therefore:

$\begin{array}{c}P\left(B\right)=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}\\ =\frac{3+2+1}{36}\\ =\frac{6}{36}\end{array}$

Hence, the probability $P\left(A\right)$ is $\frac{10}{36}$ and the probability $P\left(B\right)$ is $\frac{6}{36}$.