Chapter 4, Problem 4.1.6RE

Rolling Two Dice When two dice are rolled, find the probability of getting A sum of 5 or 6

a. A sum greater than 9

b. A sum less than 4 or greater than 9

c. A sum that is divisible by 4

d. A sum of 14

e. A sum less than 13

To determine

The probability of getting a sum of 5 or 6.

Answer to Problem 4.1.6RE

The probability of gettinga sum of 5 or 6 is $\frac{1}{4}$.

Explanation of Solution

Given info:

Two dices are rolled.

Calculation:

When two dices are rolled the sample space 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 5.

Hence, the possible outcomes for getting a sum 5 are,

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

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

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

Let event B 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 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 5 for ‘Number of outcomes in B’ and 36 for ‘Total number of outcomes in the sample space’,

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

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 $\frac{4}{\text{36}}$ for $P\left(A\right)$ and $\frac{5}{\text{36}}$ for $P\left(B\right)$

$\begin{array}{c}P\left(A\text{ or }B\right)=\frac{4}{\text{36}}+\frac{5}{36}\\ =\frac{4+5}{36}\\ =\frac{9}{36}\\ =\frac{1}{4}\end{array}$

Thus, the probability that the outcome is sum of 5 or 6 is $\frac{1}{4}$.

To determine

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

Answer to Problem 4.1.6RE

The probability of getting a sum greater than 9 is $\underset{\_}{\frac{1}{6}}$.

Explanation of Solution

Calculation:

Let event C denote getting a sum greater than 9.

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

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

That is, there are 6 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 6 for ‘Number of outcomes in C’ and 36 for ‘Total number of outcomes in the sample space’,

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

Thus, the probability of getting a sum greater than 9is $\underset{\_}{\frac{1}{6}}$.

To determine

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

Answer to Problem 4.1.6RE

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

Explanation of Solution

Calculation:

Let event D denote getting a less than 4.

Hence, the possible outcomes for getting a sum less than 4are,

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

That is, there are 3 outcomes 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 3 for ‘Number of outcomes in D’ and 36 for ‘Total number of outcomes in the sample space’,

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

Let event E denote getting a sum greater than 9.

Hence, the possible outcomes for getting a sum greater than 9are,

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

That is, there are 6 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 6 for ‘Number of outcomes in E’ and 36 for ‘Total number of outcomes in the sample space’,

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

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 $\frac{3}{\text{36}}$ for $P\left(D\right)$ and $\frac{6}{\text{36}}$ for $P\left(E\right)$

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

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

To determine

To obtain: The probability of getting a sum divisible 4.

Answer to Problem 4.1.6RE

The probability of getting a sum divisible 4is $\frac{1}{4}$.

Explanation of Solution

Calculation:

Let event E denote getting a sum divisible by4.

Hence, the possible outcomes for getting a sum divisible 4are,

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

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

$\begin{array}{c}P\left(E\right)=\frac{9}{\text{36}}\\ =\frac{1}{4}\end{array}$

Thus, the probability of getting a sum divisible 4is. $\frac{1}{4}$

To determine

To obtain: The probability of getting a sum of 14.

Answer to Problem 4.1.6RE

The probability of getting a sum of 14is0.

Explanation of Solution

Calculation:

Let event F denote getting a sum of 14.

Here, the number of possible outcomes for getting a sum of 14 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’,

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

Thus, the probability of getting a sum of 14is0.

To determine

To obtain: The probability of getting a sum less than 13.

Answer to Problem 4.1.6RE

The probability of getting a sum less than 13is1.

Explanation of Solution

Calculation:

Let event G denote getting a sum less than 13.

Hence, the possible outcomes for getting a sum less than 13are,

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

That is, there are 36 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 36 for ‘Number of outcomes in F’ and 36 for ‘Total number of outcomes in the sample space’,

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

Thus, the probability of getting a sum less than 13is1.