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Sagot :
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.
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.
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