Question

Choose an American household at random and record the number of vehicles they own. Here is the probability model if we ignore the few households that own more than $5$ cars: A housing company builds houses with two-car garages. What percent of households have more cars than the

garage can hold? $\left(a\right)7\%\left(b\right)13\%\left(c\right)20\%\left(d\right)45\%\left(e\right)55\%$

Short answer

The percentage is 20%.

Step-by-step solution

Step 1. Given

The table is:

Step 2. Concept 

The probability of an event=$\frac{numberoffavourableoutcomes}{totalnumberofoutcomes}$

Step 3. Calculation

The percentage of houses with cards that could be held by the garage is calculated as follows:

$$\begin{gathered} P(X>2)=P(X=3)+P(X=4)+P(X=5) \\ =0.13+0.05+0.02 \\ =0.20 \\ =20\% \end{gathered}$$

As a result, the needed percentage is $20\%$

As a result, the best solution is (c).