To solve this problem, we need to find the probability that a student drives given that they are a senior. This is often written as [tex]\( P(\text{drives} \mid \text{senior}) \)[/tex].
We are given a table that shows the distribution of how the 100 students get to school according to their grade levels. Here is the table again for convenience:
[tex]\[
\begin{array}{|c|c|c|c|c|}
\hline
& \text{Drive} & \text{Bus} & \text{Walk} & \text{Total} \\
\hline
\text{Sophomore} & 2 & 25 & 3 & 30 \\
\hline
\text{Junior} & 13 & 20 & 2 & 35 \\
\hline
\text{Senior} & 25 & 5 & 5 & 35 \\
\hline
\text{Total} & 40 & 50 & 10 & 100 \\
\hline
\end{array}
\][/tex]
### Step-by-Step Solution:
1. Determine [tex]\( P(\text{senior}) \)[/tex]:
We need the probability that a student is a senior. We have 35 seniors out of a total of 100 students.
[tex]\[
P(\text{senior}) = \frac{\text{Number of seniors}}{\text{Total number of students}} = \frac{35}{100} = 0.35
\][/tex]
2. Determine [tex]\( P(\text{drives and senior}) \)[/tex]:
Next, we need the probability that a student both drives and is a senior. There are 25 such students out of a total of 100.
[tex]\[
P(\text{drives and senior}) = \frac{\text{Number of seniors who drive}}{\text{Total number of students}} = \frac{25}{100} = 0.25
\][/tex]
3. Calculate [tex]\( P(\text{drives} \mid \text{senior}) \)[/tex]:
We use the conditional probability formula:
[tex]\[
P(\text{drives} \mid \text{senior}) = \frac{P(\text{drives and senior})}{P(\text{senior})}
\][/tex]
Substituting in the values we have determined:
[tex]\[
P(\text{drives} \mid \text{senior}) = \frac{0.25}{0.35} \approx 0.7142857142857143
\][/tex]
4. Round to the nearest hundredth:
The last step is to round the result to the nearest hundredth:
[tex]\[
0.7142857142857143 \approx 0.71
\][/tex]
So, the probability that a student drives given that they are a senior is [tex]\( \boxed{0.71} \)[/tex].
