Let's break down the problem step-by-step to understand what the data tells us so we can correctly fill in the blanks.
1. Finding [tex]\( P(\text{Boy}) \)[/tex]:
The probability of a randomly selected student being a boy is given by the ratio of the number of boys to the total number of students.
[tex]\[
P(\text{Boy}) = \frac{\text{Number of Boys}}{\text{Total Number of Students}} = \frac{160}{240} = \frac{2}{3} = 0.6666666666666666
\][/tex]
2. Finding [tex]\( P(\text{Boy} \mid \text{ Indoor Recess}) \)[/tex]:
The conditional probability [tex]\( P(\text{Boy} \mid \text{Indoor Recess}) \)[/tex] is the probability that a student is a boy given that the student prefers indoor recess. This is found by dividing the number of boys who prefer indoor recess by the total number of students who prefer indoor recess.
[tex]\[
P(\text{Boy} \mid \text{Indoor Recess}) = \frac{\text{Number of Boys Who Prefer Indoor Recess}}{\text{Total Number of Students Who Prefer Indoor Recess}} = \frac{64}{96} = \frac{2}{3} = 0.6666666666666666
\][/tex]
3. Determine if the events are independent:
Two events, A and B, are independent if [tex]\( P(A \cap B) = P(A) \times P(B) \)[/tex].
Here A is "being a boy" and B is "preferring indoor recess".
[tex]\[
P(\text{Boy and Indoor Recess}) = \frac{\text{Number of Boys Who Prefer Indoor Recess}}{\text{Total Number of Students}} = \frac{64}{240}
\][/tex]
[tex]\[
P(\text{Boy}) \times P(\text{Indoor Recess}) = P(\text{Boy}) \times \frac{\text{Total Students Who Prefer Indoor Recess}}{\text{Total Number of Students}} = \frac{2}{3} \times \frac{96}{240} = \frac{2}{3} \times \frac{2}{5}
\][/tex]
Simplify that:
[tex]\[
\frac{2}{3} \times \frac{2}{5} = \frac{4}{15} = \frac{64}{240}
\][/tex]
Since [tex]\( P(\text{Boy and Indoor Recess}) = P(\text{Boy}) \times P(\text{Indoor Recess}) \)[/tex], the events are independent.
Given all the calculations, we can draw the following conclusions:
1. [tex]\( P(\text{Boy}) = 0.6666666666666666 \)[/tex]
2. [tex]\( P(\text{Boy} \mid \text{Indoor Recess}) = 0.6666666666666666 \)[/tex]
3. The events of the student being a boy and the student preferring indoor recess are independent.
Placing these in the blanks:
[tex]\[
\begin{array}{l}
P(\text { Boy })= 0.6666666666666666 \\
P(\text { Boy } \mid \text { Indoor Recess })= 0.6666666666666666 \\
\end{array}
\][/tex]
The events of the student being a boy and the student preferring indoor recess are [tex]\( \text{independent} \)[/tex].
