Chapter 2.8, Problem 95E

a.

To determine

Obtain the value of $P\left(A\cap B\right)$

Answer to Problem 95E

The value of $P\left(A\cap B\right)$ is 0.1.

Explanation of Solution

For any two events A and B, the addition rule of probability is $P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$.

Substitute $P\left(A\cup B\right)=0.4,P\left(A\right)=0.2,\text{ and }P\left(B\right)=0.3$

$\begin{array}{c}0.4=0.2+0.3-P\left(A\cap B\right)\\ P\left(A\cap B\right)=0.5-0.4\\ =0.1\end{array}$

Thus, the value of $P\left(A\cap B\right)$ is 0.1.

b.

To determine

Obtain the value of $P\left(\overline{A}\cup \overline{B}\right)$

Answer to Problem 95E

The value of $P\left(\overline{A}\cup \overline{B}\right)$ is 0.9

Explanation of Solution

DeMorgan’s laws:

For any two events A and B,

$\begin{array}{l}\left(\overline{A\cap B}\right)=\left(\overline{A}\right)\cup \left(\overline{B}\right)\\ \left(\overline{A\cup B}\right)=\left(\overline{A}\right)\cap \left(\overline{B}\right)\end{array}$

$\begin{array}{c}P\left(\overline{A}\cup \overline{B}\right)=P\left(\overline{A\cap B}\right)\\ =1-P\left(A\cap B\right)\\ =1-0.1\\ =0.9\end{array}$

Thus, the value of $P\left(\overline{A}\cup \overline{B}\right)$ is 0.9.

c.

To determine

Obtain the value of $P\left(\overline{A}\cap \overline{B}\right)$

Answer to Problem 95E

The value of $P\left(\overline{A}\cap \overline{B}\right)$ is 0.6

Explanation of Solution

Consider,

$\begin{array}{c}P\left(\overline{A}\cap \overline{B}\right)=P\left(\overline{A\cup B}\right)\\ =1-P\left(A\cup B\right)\\ =1-0.4\\ =0.6\end{array}$

Thus, the value of $P\left(\overline{A}\cap \overline{B}\right)$ is 0.6.

d.

To determine

Obtain the value of $P\left(\overline{A}|B\right)$

Answer to Problem 95E

The value of $P\left(\overline{A}|B\right)$ is 0.6667

Explanation of Solution

Consider,

$\begin{array}{c}P\left(\overline{A}|B\right)=\frac{P\left(\overline{A}\cap B\right)}{P\left(B\right)}\\ =\frac{P\left(B\right)-P\left(A\cap B\right)}{P\left(B\right)}\\ =\frac{0.3-0.1}{0.3}\\ =\frac{0.2}{0.3}\\ =0.6667\end{array}$

Thus, the value of $P\left(\overline{A}|B\right)$ is 0.6667