The two-way table shows the number of sport utility vehicles with certain features for sale at the car lot.

\begin{tabular}{|c|c|c|c|}
\hline & \begin{tabular}{c}
4-Wheel \\
Drive
\end{tabular} & \begin{tabular}{c}
No 4-Wheel \\
Drive
\end{tabular} & Total \\
\hline Third-Row Seats & 18 & 12 & 30 \\
\hline No Third-Row Seats & 7 & 28 & 35 \\
\hline Total & 25 & 40 & 65 \\
\hline
\end{tabular}

What is the probability that a randomly selected car with no 4-wheel drive has third-row seats?

A. 0.3
B. 0.4
C. 0.7
D. 0.8



Answer :

To solve the given problem, we need to find the probability that a randomly selected car with no 4-wheel drive has third-row seats. We follow these steps:

1. Identify the relevant data from the table:
- The total number of cars with no 4-wheel drive: 40
- The number of cars with no 4-wheel drive that have third-row seats: 12

2. Set up the probability formula:
The probability of an event is given by the ratio of the number of favorable outcomes to the total number of possible outcomes.

Here, the event is selecting a car with no 4-wheel drive that has third-row seats. So the formula is:
[tex]\[ \text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \][/tex]

3. Substitute the values into the formula:
[tex]\[ \text{Probability} = \frac{12}{40} \][/tex]

4. Simplify the fraction:
[tex]\[ \frac{12}{40} = \frac{3}{10} = 0.3 \][/tex]

So, the probability that a randomly selected car with no 4-wheel drive has third-row seats is [tex]\( 0.3 \)[/tex].

Thus, the correct answer is [tex]\( 0.3 \)[/tex].