Chapter 12, Problem 32SGR

To determine

To find: the probability that the selected house has fewer than $\text{4}$ televisions.

Answer to Problem 32SGR

The probability that selected house has fewer than $\text{4}$ televisions is $0.78$ .

Explanation of Solution

Given information:

X=number of televisions	probability
$1$	$0.18$
$2$	$0.36$
$3$	$0.34$
$4$	$0.08$
$5+$	$0.04$

Calculation: to find the probability add the probability which are fewer than $\text{4}$ televisions.,

  $\begin{array}{l}P\left(<4\right)=0.18+0.36+0.34\\ P\left(<4\right)=0.78\text{ [add]}\end{array}$

Thus, the probability is

  $0.78$ .