Chapter 2.4, Problem 4E

(a)

To determine

(i)

Find the probability that the sum of the two die rolls is at least 5.

(ii)

Find the probability that the sum of the two die rolls is no more than 8.

(iii)

Find the probability of $E_3$.

Find the probability of $E_4$.

Find the probability of $E_5$.

Answer to Problem 4E

(i)

The probability that the sum of the two die rolls is at least 5 is $\frac{5}{6}$.

(ii)

The probability that the sum of the two die rolls is no more than 8 is $\frac{13}{18}$.

(iii)

The probability of $E_3$ is 1.

The probability of $E_4$ is $\frac{5}{18}$.

The probability of $E_5$ is 0.

Explanation of Solution

Calculation:

(i)

The probability of their union equals the sum of their probabilities is,

$P\left({\displaystyle \underset{i=1}{\overset{\infty }{\cup }}{E}_i}\right)={\displaystyle \sum _{i=1}^{\infty }P\left(E_i\right)}$

In the formula $E_1,E_2,\dots$ are sequence of disjoint events.

The event $E_1$ denotes the sum of the two die rolls is at least 5. That is, $E_1=\left\{5,6,7,8,9,10,11,12\right\}$.

The probability that the sum of the two die rolls is at least 5 is,

$\begin{array}{c}P\left(E_1\right)=p\left(5\right)+p\left(6\right)+p\left(7\right)+p\left(8\right)+p\left(9\right)+p\left(10\right)+p\left(11\right)+p\left(12\right)\\ =\frac{4}{36}+\frac{5}{36}+\frac{6}{36}+\frac{5}{36}+\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36}\\ =\frac{30}{36}\\ =\frac{5}{6}\end{array}$

Hence, the probability that the sum of the two die rolls is at least 5 is $\frac{5}{6}$.

(ii)

The event $E_2$ denotes the sum of the two die rolls is no more than 8. That is, $E_2=\left\{2,3,4,5,6,7,8\right\}$.

The probability that the sum of the two die rolls is no more than 8 is,

$\begin{array}{c}P\left(E_2\right)=p\left(2\right)+p\left(3\right)+p\left(4\right)+p\left(5\right)+p\left(6\right)+p\left(7\right)+p\left(8\right)\\ =\frac{1}{36}+\frac{2}{36}+\frac{3}{36}+\frac{4}{36}+\frac{5}{36}+\frac{6}{36}+\frac{5}{36}\\ =\frac{26}{36}\\ =\frac{13}{18}\end{array}$

Hence, the probability that the sum of the two die rolls is no more than 8 is $\frac{13}{18}$.

(iii)

The event $E_3=E_1\cup E_2$. That is, $E_3=\left\{2,3,4,5,6,7,8,9,10,11,12\right\}$.

The probability of $E_3$ is,

$\begin{array}{c}P\left(E_3\right)=p\left(2\right)+p\left(3\right)+p\left(4\right)+p\left(5\right)+p\left(6\right)+p\left(7\right)+p\left(8\right)+p\left(9\right)+p\left(10\right)+p\left(11\right)+p\left(12\right)\\ =\frac{1}{36}+\frac{2}{36}+\frac{3}{36}+\frac{4}{36}+\frac{5}{36}+\frac{6}{36}+\frac{5}{36}+\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36}\\ =\frac{36}{36}\\ =1\end{array}$

Hence, the probability of $E_3$ is 1.

The event $E_4=E_1-E_2$. That is, $E_4=\left\{9,10,11,12\right\}$.

The probability of $E_4$ is,

$\begin{array}{c}P\left(E_4\right)=p\left(9\right)+p\left(10\right)+p\left(11\right)+p\left(12\right)\\ =\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36}\\ =\frac{10}{36}\\ =\frac{5}{18}\end{array}$

Hence, the probability of $E_4$ is $\frac{5}{18}$.

The event $E_5=E_1^c\cap E_2^c$. That is, $E_4=\varnothing$.

The probability of $E_5$ is,

$P\left(E_5\right)=0$

Hence, the probability of $E_5$ is 0.

(b)

To determine

Find the probability of $E_3$.

Answer to Problem 4E

The probability of $E_3$ is 1.

Explanation of Solution

Calculation:

The formula for union of two events is,

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

The event $E_3=E_1\cup E_2$. The samples for event $E_1\cap E_2$ is $E_1\cap E_2=\left\{5,6,7,8\right\}$.

The probability of $E_3$ is,

$\begin{array}{c}P\left(E_3\right)=P\left(E_1\cup E_2\right)\\ =P\left(E_1\right)+P\left(E_2\right)-P\left(E_1\cap E_2\right)\\ =\frac{30}{36}+\frac{26}{36}-\left(\frac{4}{36}+\frac{5}{36}+\frac{6}{36}+\frac{5}{36}\right)\\ =\frac{56}{36}-\frac{20}{36}\end{array}$

            $\begin{array}{c}=\frac{36}{36}\\ =1\end{array}$

Hence, the probability of $E_3$ is 1.

(c)

To determine

Find the probability of $E_5$.

Answer to Problem 4E

The probability of $E_5$ is 0.

Explanation of Solution

Calculation:

The formula for complement of event A is,

$P\left(A^c\right)=1-P\left(A\right)$

The event $E_5=E_1^c\cap E_2^c$.

The probability of $E_5$ is,

$\begin{array}{c}P\left(E_5\right)=P\left(E_1^c\cap E_2^c\right)\\ =P\left[{\left(E_1\cup E_2\right)}^c\right]\\ =P\left[{\left(E_3\right)}^c\right]\\ =1-P\left(E_3\right)\end{array}$

            $\begin{array}{c}=1-1\\ =0\end{array}$

Hence, the probability of $E_5$ is 0.