To solve this problem, we need to find the conditional probability that a student is in the 9th grade given that they enjoy watching sports the most.
The formula for conditional probability [tex]\( P(A|B) \)[/tex] is given by
[tex]\[
P(A|B) = \frac{P(A \cap B)}{P(B)}
\][/tex]
where:
- [tex]\( A \)[/tex] is the event that the student is in the 9th grade.
- [tex]\( B \)[/tex] is the event that the student enjoys watching sports the most.
- [tex]\( P(A \cap B) \)[/tex] is the probability that both events A and B occur.
- [tex]\( P(B) \)[/tex] is the probability of event B occurring.
First, we need to determine the values for [tex]\( P(A \cap B) \)[/tex] and [tex]\( P(B) \)[/tex].
1. From the table, the total number of students who enjoy watching sports the most is 35.
2. The number of 9th-grade students who enjoy watching sports the most is 13.
So, the probability [tex]\( P(B) \)[/tex], the probability that a student enjoys watching sports the most, is given by:
[tex]\[
P(B) = \frac{\text{Total number of students who enjoy watching sports the most}}{\text{Total number of students}} = \frac{35}{100} = 0.35
\][/tex]
The probability [tex]\( P(A \cap B) \)[/tex], the probability that a student is in the 9th grade and enjoys watching sports the most, is given by:
[tex]\[
P(A \cap B) = \frac{\text{Number of 9th-grade students who enjoy watching sports the most}}{\text{Total number of students who enjoy watching sports the most}} = \frac{13}{35}
\][/tex]
Now we can calculate the conditional probability:
[tex]\[
P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{13/35}{35/100}
\][/tex]
Simplify the fraction:
[tex]\[
P(A|B) = \frac{13}{35} \times \frac{100}{35} = 0.37142857142857144
\][/tex]
Thus, the probability that a randomly selected student is in the 9th grade, given that they enjoy watching sports the most, is approximately [tex]\( 0.37 \)[/tex].
Therefore, the correct answer is:
C. 0.37
