Chapter 12.5, Problem 4CYP

To determine

To find the probability that a randomly selected household has a dog or a cat.

Answer to Problem 4CYP

The probability that a randomly selected household has a dog or a cat is $\frac{2612}{5200}\text{ or 50}\text{.2\%}$ .

Explanation of Solution

Given information:

The given statement is:

“Out of $5200$ households surveyed, $2107$ had a dog, $807$ had a cat, and $303$ had both a dog and a cat. What is the probability that a randomly selected household has a dog or a cat?”

Formula used:

The following formula is used:

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

Calculation :

Since, some households has both a dog and a cat, the events are not mutually exclusive.

  $\begin{array}{l}\text{P}\left(\text{dog}\right)=\frac{2107}{5200}\\ \text{P}\left(\text{cat}\right)=\frac{807}{5200}\\ \text{P}\left(\text{dog and cat}\right)=\frac{302}{5200}\\ \text{P}\left(\text{dog or cat}\right)=\frac{2107}{5200}+\frac{807}{5200}-\frac{302}{5200}\\ \text{P}\left(\text{dog or cat}\right)=\frac{2107+807-302}{5200}\\ \text{P}\left(\text{dog or cat}\right)=\frac{2612}{5200}\text{ or 50}\text{.2\%}\end{array}$

Hence, the probability that a randomly selected household has a dog or a cat is $\frac{2612}{5200}\text{ or 50}\text{.2\%}$ .