Select the correct answer from each drop-down menu.

A survey asking about preference for recess location was randomly given to students in an elementary school. The results are recorded in the table below.

\begin{tabular}{|l|l|l|l|}
\hline
& Indoor Recess & Outdoor Recess & Total \\
\hline
Boys & 64 & 96 & 160 \\
\hline
Girls & 32 & 48 & 80 \\
\hline
Total & 96 & 144 & 240 \\
\hline
\end{tabular}

A student is randomly selected. Based on the data, what conclusions can be drawn?

[tex]\[
\begin{array}{l}
P(\text{Boy}) = \square \\
P(\text{Boy} \mid \text{Indoor Recess}) = \square
\end{array}
\][/tex]

[tex]$\square$[/tex] The events of the student being a boy and the student preferring indoor recess are [tex]$\square$[/tex]



Answer :

Let's analyze the given data from the table and draw conclusions step by step.

First, we need to address the probability of selecting a boy [tex]\(P(\text{Boy})\)[/tex]. We start by noting the total number of students and the number of boys.

- Total students = 240
- Total boys = 160

The probability [tex]\(P(\text{Boy})\)[/tex] is calculated by dividing the number of boys by the total number of students:
[tex]\[ P(\text{Boy}) = \frac{\text{Number of Boys}}{\text{Total Number of Students}} = \frac{160}{240} = \frac{2}{3} \approx 0.6667 \][/tex]

Next, we look at the conditional probability of a student being a boy given that they prefer indoor recess [tex]\(P(\text{Boy} \mid \text{Indoor Recess})\)[/tex].

- Total students who prefer indoor recess = 96
- Boys who prefer indoor recess = 64

The conditional probability [tex]\(P(\text{Boy} \mid \text{Indoor Recess})\)[/tex] is calculated by dividing the number of boys who prefer indoor recess by the total number of indoor recess students:
[tex]\[ P(\text{Boy} \mid \text{Indoor Recess}) = \frac{\text{Number of Boys who prefer Indoor Recess}}{\text{Total Students who prefer Indoor Recess}} = \frac{64}{96} = \frac{2}{3} \approx 0.6667 \][/tex]

Finally, we need to determine whether the events of being a boy and preferring indoor recess are independent. Events A and B are independent if:
[tex]\[ P(A \cap B) = P(A) \times P(B) \][/tex]

In our case:
- [tex]\( P(\text{Boy} \cap \text{Indoor Recess}) \)[/tex] is the probability of being a boy and preferring indoor recess, which is:
[tex]\[ P(\text{Boy} \cap \text{Indoor Recess}) = \frac{\text{Number of Boys who prefer Indoor Recess}}{\text{Total Number of Students}} = \frac{64}{240} \][/tex]
- [tex]\( P(\text{Indoor Recess}) \)[/tex] is the probability of preferring indoor recess:
[tex]\[ P(\text{Indoor Recess}) = \frac{96}{240} = \frac{2}{5} \][/tex]
- Given that [tex]\( P(\text{Boy}) \)[/tex] is:
[tex]\[ P(\text{Boy}) = \frac{160}{240} = \frac{2}{3} \][/tex]

We compare [tex]\( P(\text{Boy} \cap \text{Indoor Recess}) \)[/tex] and [tex]\( P(\text{Boy}) \times P(\text{Indoor Recess}) \)[/tex]:
[tex]\[ P(\text{Boy} \cap \text{Indoor Recess}) = \frac{64}{240} = 0.2667 \][/tex]
[tex]\[ P(\text{Boy}) \times P(\text{Indoor Recess}) = \frac{2}{3} \times \frac{2}{5} = \frac{4}{15} = 0.2667 \][/tex]

Since [tex]\( P(\text{Boy} \cap \text{Indoor Recess}) = P(\text{Boy}) \times P(\text{Indoor Recess}) \)[/tex], the events are indeed independent.

Thus, the correct answers are:
[tex]\[ P(\text{Boy}) = 0.6667 \][/tex]
[tex]\[ P(\text{Boy} \mid \text{Indoor Recess}) = 0.6667 \][/tex]
The events of the student being a boy and the student preferring indoor recess are independent.