To determine the conditional probability [tex]\( P(A \mid B) \)[/tex], which is the probability that a student is in the chess club given that they are in the karate club, we apply the formula for conditional probability:
[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \][/tex]
Where:
- [tex]\(P(A \cap B)\)[/tex] is the probability that a student is in both the chess club and the karate club.
- [tex]\(P(B)\)[/tex] is the probability that a student is in the karate club.
According to the problem statement, there are 10 students in total.
We know the following details:
1. There are 2 students in both the chess club and the karate club.
2. There are 4 students who are only in the chess club.
3. There are 2 students who are only in the karate club.
4. The remaining 2 students are in neither club.
This means:
- The number of students in both clubs ([tex]\(A \cap B\)[/tex]) is 2.
- The total number of students in karate club ([tex]\(B\)[/tex]) is the sum of those in both clubs and those only in karate:
[tex]\[ \text{Students in karate club} = 2 \ (\text{both clubs}) + 2 \ (\text{only karate}) = 4 \ (\text{total in karate}) \][/tex]
Substituting these values into the conditional probability formula:
[tex]\[ P(A \mid B) = \frac{\text{number of students in both clubs}}{\text{number of students in karate club}} = \frac{2}{6} \approx 0.33 \][/tex]
Thus, the correct answer is:
C. [tex]\(\frac{2}{6} \approx 0.33\)[/tex]
