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A group of 10 students participates in the chess club, karate club, or neither.

Let event [tex]$A$[/tex] be the student is in the chess club.
Let event [tex]$B$[/tex] be the student is in the karate club.

One of these students is randomly selected. What is [tex]$P(A \mid B)$[/tex]?

A. [tex]\frac{2}{10}=0.20[/tex]

Sagot :

To determine the conditional probability [tex]\(P(A \mid B)\)[/tex], which is the probability that a randomly selected student is in the chess club given that they are in the karate club, we can use the formula for conditional probability:

[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \][/tex]

Here, [tex]\(P(A \cap B)\)[/tex] is the probability that a student is in both the chess club and the karate club, and [tex]\(P(B)\)[/tex] is the probability that a student is in the karate club.

Based on the information provided:
- There are 10 students in total.
- 4 students are in the chess club.
- 3 students are in the karate club.
- 2 students are in both clubs.

First, we need to calculate [tex]\(P(A \cap B)\)[/tex]:

[tex]\[ P(A \cap B) = \frac{\text{Number of students in both clubs}}{\text{Total number of students}} = \frac{2}{10} = 0.2 \][/tex]

Next, we calculate [tex]\(P(B)\)[/tex]:

[tex]\[ P(B) = \frac{\text{Number of students in the karate club}}{\text{Total number of students}} = \frac{3}{10} = 0.3 \][/tex]

Now, we can calculate [tex]\(P(A \mid B)\)[/tex]:

[tex]\[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.2}{0.3} \][/tex]

[tex]\[ P(A \mid B) = \frac{2}{3} \][/tex]

Converting [tex]\(\frac{2}{3}\)[/tex] to a decimal:

[tex]\[ \frac{2}{3} \approx 0.6667 \][/tex]

Thus, the conditional probability [tex]\(P(A \mid B)\)[/tex] is approximately [tex]\(0.6667\)[/tex]. The given answer of [tex]\(\frac{2}{10} = 0.20\)[/tex] is incorrect.