Chapter 4.2, Problem 24E

Rolling Die Two dice are rolled. Find the probability of getting

a. A sum of 8, 9, or 10

b. Doubles or a sum of 7

c. A sum greater than 9 or less than 4

d. Based on the answers to a, b, and c, which is least likely to occur?

To determine

To obtain: The probability of getting a sum of 8, 9, or 10.

Answer to Problem 24E

The probability of getting a sum of 8, 9, or 10 is $\frac{1}{3}$.

Explanation of Solution

Calculation:

Mutually exclusive events:

The two events A and B are mutually exclusive events, then probability $P\left(A\text{ and }B\right)=0$

The possibilities for rolling a pair of six-sided dice is,

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

Thus, the total number of outcomes is 36.

Let event A denote that the outcome as sum of 8. Hence, the possible outcomes to get the sum 8 are ‘$\left(2,6\right),\left(3,5\right),\left(4,4\right),\left(5,3\right),\left(6,2\right)$’. That is, there are 5 outcomes for event A.

The formula for probability of event A is,

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Substitute 5 for ‘Number of outcomes in A’ and 36 for‘Total number of outcomes in the sample space’,

$P\left(A\right)=\frac{5}{\text{36}}$

Here, the event B is defined getting sum of 9.

Hence, the possible outcomes are ‘$\left(3,6\right),\left(4,5\right),\left(5,4\right),\left(6,2\right)$’. That is, there are 4 outcomes for event B.

The probability of event B is,

$P\left(B\right)=\frac{\text{Number of outcomes in }B}{\text{Total number of outcomes in the sample space}}$

Substitute 4 for ‘Number of outcomes in B’ and 36 for ‘Total number of outcomes in the sample space’,

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

Here, the eventC is defined as getting a sum of 10. Hence, the possible outcomes for sum 10 are ‘$\left(4,6\right),\left(5,5\right),\left(6,4\right)$’. That is, there are 5 outcomes for event C.

The probability of event C is,

$P\left(C\right)=\frac{\text{Number of outcomes in }C}{\text{Total number of outcomes in the sample space}}$

Substitute 3 for ‘Number of outcomes in C’ and 36 for ‘Total number of outcomes in the sample space’,

$P\left(C\right)=\frac{3}{36}$

Addition Rule:

Here, the events are mutually exclusive. Thus, the formula for finding the probability of getting event A or event B or event C is,

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

Substitute $\frac{5}{36}$ for $P\left(A\right)$, $\frac{4}{36}$ for $P\left(B\right)$ and $\frac{3}{36}$ for $P\left(C\right)$

$\begin{array}{c}P\left(A\text{ or }B\text{ or }C\right)=\frac{5}{36}+\frac{4}{36}+\frac{3}{36}\\ =\frac{12}{36}\\ =\frac{1}{3}\end{array}$

Thus, the probability of getting a sum of 8, 9, or 10 is $\frac{1}{3}$.

To determine

To obtain: The probability of getting doubles or a sum of 7.

Answer to Problem 24E

The probability of getting doubles or a sum of 7 is $\frac{1}{18}$.

Explanation of Solution

Calculation:

Let event D denote that the outcomes sum of 7. Hence, the possible outcomes to get the sum 7 are ‘$\left(1,6\right),\left(2,5\right),\left(3,4\right),\left(4,3\right),\left(5,2\right),\left(6,1\right)$’. That is, there are 6 outcomes for event A.

The formula for probability of event D is,

$P\left(D\right)=\frac{\text{Number of outcomes in }D}{\text{Total number of outcomes in the sample space}}$

Substitute 6 for ‘Number of outcomes in A’ and 36 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(D\right)=\frac{6}{\text{36}}\\ =\frac{1}{6}\end{array}$

Let the event E denote getting doubles.

Hence, the possible outcomes are ‘$\left\{\left(1,1\right),\left(2,2\right),\left(3,3\right),\left(4,4\right),\left(5,5\right),\left(6,6\right)\right\}$’. That is, there are 6 outcomes for event E.

The formula for probability of event E is,

$P\left(E\right)=\frac{\text{Number of outcomes in }E}{\text{Total number of outcomes in the sample space}}$

Substitute 6 for ‘Number of outcomes in D’ and 36 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(D\right)=\frac{6}{\text{36}}\\ =\frac{1}{6}\end{array}$

Addition Rule:

Here, the events are mutually exclusive. Thus, the formula for finding the probability of getting event E or event D,

$P\left(D\text{ or }E\right)=P\left(D\right)+P\left(E\right)$

Substitute $\frac{1}{36}$ for $P\left(A\right)$, $\frac{1}{36}$ for $P\left(B\right)$

$\begin{array}{c}P\left(D\text{ or }E\right)=\frac{1}{36}+\frac{1}{36}\\ =\frac{2}{36}\\ =\frac{1}{18}\end{array}$

Thus, the probability of getting doubles or a sum of 7 is $\frac{1}{18}$.

To determine

To obtain: The probability of getting sum greater than 9 or a less than 4.

Answer to Problem 24E

The probability of getting sum greater than 9 or a less than 4 is $\frac{1}{4}$.

Explanation of Solution

Calculation:

Let event A denote that the outcome as sum greater than 9. Hence, the possible outcomes to get the sum greater than 9 are ‘$\left(4,6\right),\left(6,4\right),\left(5,5\right),\left(6,5\right),\left(5,6\right),\left(6,6\right)$’. That is, there are 6 outcomes for event A.

The formula for probability of event A is,

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Substitute 6 for ‘Number of outcomes in A’ and 36 for ‘Total number of outcomes in the sample space’,

$P\left(A\right)=\frac{6}{\text{36}}$

Here, the event B is defined getting sum is less than 4.

Hence, the possible outcomes are ‘$\left(1,2\right),\left(2,1\right),\left(1,1\right)$’. That is, there are 3 outcomes for event B.

The probability of event B is,

$P\left(B\right)=\frac{\text{Number of outcomes in }B}{\text{Total number of outcomes in the sample space}}$

Substitute 3 for ‘Number of outcomes in B’ and36 for ‘Total number of outcomes in the sample space’,

$P\left(B\right)=\frac{3}{36}$

Addition Rule:

Here, the events are mutually exclusive. Thus, the formula for finding the probability of getting event A or event B,

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

Substitute $\frac{6}{36}$ for $P\left(A\right)$ and $\frac{3}{36}$ for $P\left(B\right)$

$\begin{array}{c}P\left(A\text{ or }B\right)=\frac{6}{36}+\frac{3}{36}\\ =\frac{9}{36}\\ =\frac{1}{4}\end{array}$

Thus, the probability of getting sum greater than 9 or a less than 4 is $\frac{1}{4}$.

To determine

To identify: The event which is least likely to occur.

Answer to Problem 24E

The event b is least likely to occur.

Explanation of Solution

From the probabilities of events a, b and c, it can be observed that the probability of event b is less the other probabilities of a and c. That is, $0.055<0.250<0.333$. Thus, the event b is least likely to occur