Chapter 4, Problem 26CQ

a.

To determine

To obtain: The probability of getting a sum 6 or 7.

Answer to Problem 26CQ

The probability of getting a sum 6 or 7 is $\frac{11}{36}$.

Explanation of Solution

Given info:

A pair of six-sided dice is rolled.

Calculation:

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 getting a sum 6.

Hence, the possible outcomes for getting a sum 6 are,

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

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’,

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

Let event B denote getting a sum 7.

Hence, the possible outcomes for getting a sum 7 are,

$\left\{\left(1,6\right),\left(2,5\right),\left(3,4\right),\left(4,3\right),\left(5,2\right),\left(6,1\right)\right\}$.

That is, there are 6 outcomes for eventB.

The formula for 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 6 for ‘Number of outcomes in B’ and 36 for ‘Total number of outcomes in the sample space’,

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

Addition Rule:

The formula for probability of getting event A or event B is,

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

Substitute $\frac{5}{\text{36}}$ for $P\left(A\right)$ and $\frac{6}{\text{36}}$ for $P\left(B\right)$

$\begin{array}{c}P\left(A\text{ or }B\right)=\frac{5}{36}+\frac{6}{36}\\ =\frac{5+6}{36}\\ =\frac{11}{36}\end{array}$

Thus, the probability that the outcome is sum less than 9 is $\frac{13}{18}$.

b.

To determine

To obtain: The probability of getting a sum greater than 8.

Answer to Problem 26CQ

The probability of getting a sum greater than 8 is $\frac{5}{8}$.

Explanation of Solution

Calculation:

Let event C denote getting a sum greater than 8.

Hence, the possible outcomes for getting a sum greater than 8 are,

$\left\{\left(3,6\right),\left(4,5\right),\left(4,6\right),\left(5,4\right),\left(5,5\right),\left(5,6\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\right\}$.

That is, there are 10 outcomes for eventC.

The formula for 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 10 for ‘Number of outcomes in C’ and 36 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(C\right)=\frac{10}{\text{36}}\\ =\frac{5}{18}\end{array}$

Thus, the probability of getting a sum greater than 8 is $\frac{5}{8}$.

c.

To determine

To obtain: The probability of getting a sum less than 3 or greater than 8.

Answer to Problem 26CQ

The probability of getting a sum less than 3 or greater than 8 is $\frac{11}{36}$.

Explanation of Solution

Calculation:

Let event D denote getting a less than 3.

Hence, the possible outcomes for getting a sum less than 3 are,

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

That is, there are 1 outcome for eventD.

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

$P\left(D\right)=\frac{1}{\text{36}}$

Let event E denote getting a sum greater than 8.

Hence, the possible outcomes for getting a sum greater than 8 are,

$\left\{\left(3,6\right),\left(4,5\right),\left(4,6\right),\left(5,4\right),\left(5,5\right),\left(5,6\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\right\}$.

That is, there are 10 outcomes for eventE.

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 10 for ‘Number of outcomes in E’ and 36 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{10}{\text{36}}$

Addition Rule:

The formula for probability of getting event A or event B is,

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

Substitute $\frac{1}{\text{36}}$ for $P\left(D\right)$ and $\frac{10}{\text{36}}$ for $P\left(E\right)$

$\begin{array}{c}P\left(C\text{ or }D\right)=\frac{1}{36}+\frac{10}{36}\\ =\frac{1+10}{36}\\ =\frac{11}{36}\end{array}$

Thus, the probability of getting a sum less than 3 or greater than 8 is $\frac{11}{36}$.

d.

To determine

To obtain: The probability of getting a sum divisible 3.

Answer to Problem 26CQ

The probability of getting a sum divisible 3 is $\frac{1}{3}$.

Explanation of Solution

Calculation:

Let event E denote getting a sum greater than 8.

Hence, the possible outcomes for getting a sum divisible 3 are,

$\left\{\left(1,2\right),\left(1,5\right),\left(2,1\right),\left(2,4\right),\left(3,3\right),\left(3,6\right),\left(4,2\right),\left(4,5\right),\left(5,1\right),\left(5,4\right),\left(6,3\right),\left(6,6\right)\right\}$.

That is, there are 12 outcomes for eventE.

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 12 for ‘Number of outcomes in E’ and 36 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(E\right)=\frac{12}{\text{36}}\\ =\frac{1}{3}\end{array}$

Thus, the probability of getting a sum divisible 3 is $\frac{1}{3}$.

e.

To determine

To obtain: The probability of getting a sum of 16.

Answer to Problem 26CQ

The probability of getting a sum of 16 is0.

Explanation of Solution

Calculation:

Let event F denote getting a sum of 16.

Here, the number of possible outcomes for getting a sum of 16 is 0.

That is, there are 0 outcomes for eventF.

The formula for probability of event F is,

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

Substitute 0 for ‘Number of outcomes in F’ and 36 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(F\right)=\frac{0}{\text{36}}\\ =0\end{array}$

Thus, the probability of getting a sum of 16 is0.

f.

To determine

To obtain: The probability of getting a sum less than 11.

Answer to Problem 26CQ

The probability of getting a sum less than 11 is $\frac{11}{12}$.

Explanation of Solution

Calculation:

Let event G denote getting a sum less than 11.

Hence, the possible outcomes for getting a sum divisible 3 are,

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

That is, there are 33 outcomes for eventG.

The formula for probability of event G is,

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

Substitute 33 for ‘Number of outcomes in F’ and 36 for ‘Total number of outcomes in the sample space’,

$\begin{array}{c}P\left(F\right)=\frac{33}{\text{36}}\\ =\frac{11}{12}\end{array}$

Thus, the probability of getting a sum less than 11 is $\frac{11}{12}$.