Chapter 5.2, Problem 53E

To determine

Probability of intersection of both events A and B.

Answer to Problem 53E

Probability of intersection,

  $P\left(\text{A}\cap \text{B}\right)=0.12$

Explanation of Solution

Given information:

  $P\left(\text{A}\right)=0.3$ ,

And

  $P\left(\text{B}\right)=0.4$ ,

And

  $P\left(\text{A}\cup \text{B}\right)=0.58$ .

Calculations:

For any two events, general addition rule:

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

Apply the above general addition rule:

  $P\left(\text{A}\cup \text{B}\right)=P\left(\text{A}\text{or}\text{B}\right)=P\left(\text{A}\right)+P\left(\text{B}\right)-P\left(\text{A}\text{and}\text{B}\right)=P\left(\text{A}\right)+P\left(\text{B}\right)-P\left(\text{A}\cap \text{B}\right)$

Thus,

The equation we obtain

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

Add $P\left(\text{C}\cap \text{D}\right)$ to both sides:

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

Subtract $P\left(\text{A}\cup \text{B}\right)$ from both sides:

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

Evaluate the above equation by substituting the values:

  $\begin{array}{l}P\left(\text{A}\cap \text{B}\right)=P\left(\text{A}\right)+P\left(\text{B}\right)-P\left(\text{A}\cup \text{B}\right)\\ =0.3+0.4-0.58\\ =0.12\end{array}$