Answer :
To determine the value of [tex]\( P(A \mid B) \)[/tex], which is the probability that a student is in the karate club given that the student is in the chess club, we need to analyze the provided options and their interpretations.
Let's first understand the concept of conditional probability. The formula for conditional probability is:
[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \][/tex]
Here, [tex]\( P(A \mid B) \)[/tex] represents the probability that a student is in the karate club ([tex]\( A \)[/tex]) given that they are already in the chess club ([tex]\( B \)[/tex]).
Now let's examine the options given:
Option A: [tex]\( P(A \mid B) = \frac{3}{10} = 0.30 \)[/tex]
Option B: [tex]\( P(A \mid B) = \frac{3}{2} = 1.50 \)[/tex]
Probabilities are values between 0 and 1 (inclusive). Therefore, a probability value cannot exceed 1. Option B suggests that the probability [tex]\( P(A \mid B) = 1.50 \)[/tex], which is not valid as it exceeds the maximum possible probability value of 1. Thus, Option B cannot be correct.
On the other hand, Option A suggests that [tex]\( P(A \mid B) = 0.30 \)[/tex], which is within the valid range for probabilities (between 0 and 1). This value indicates a 30% chance of a student being in the karate club given that they are already in the chess club. This is a reasonable and valid probability.
Therefore, the correct probability [tex]\( P(A \mid B) \)[/tex] is:
[tex]\[ \boxed{0.30} \][/tex]
Let's first understand the concept of conditional probability. The formula for conditional probability is:
[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \][/tex]
Here, [tex]\( P(A \mid B) \)[/tex] represents the probability that a student is in the karate club ([tex]\( A \)[/tex]) given that they are already in the chess club ([tex]\( B \)[/tex]).
Now let's examine the options given:
Option A: [tex]\( P(A \mid B) = \frac{3}{10} = 0.30 \)[/tex]
Option B: [tex]\( P(A \mid B) = \frac{3}{2} = 1.50 \)[/tex]
Probabilities are values between 0 and 1 (inclusive). Therefore, a probability value cannot exceed 1. Option B suggests that the probability [tex]\( P(A \mid B) = 1.50 \)[/tex], which is not valid as it exceeds the maximum possible probability value of 1. Thus, Option B cannot be correct.
On the other hand, Option A suggests that [tex]\( P(A \mid B) = 0.30 \)[/tex], which is within the valid range for probabilities (between 0 and 1). This value indicates a 30% chance of a student being in the karate club given that they are already in the chess club. This is a reasonable and valid probability.
Therefore, the correct probability [tex]\( P(A \mid B) \)[/tex] is:
[tex]\[ \boxed{0.30} \][/tex]
