0 like 0 dislike
A group of 10 student participate in chess club, karate club, or neither.

Let even A = The student is in karate club
Let event B = The student is in chess club

One of these students is randomly selected. What is P(A|B)?

0 like 0 dislike
Total number of students = 10

As we have to find

P(A/B) = Probability( A when B has happend)

P(A/B)= P(A intersection B)/P(B)

According to given figure only yolanda and Rob are in both club

Therefore,P(A intersection B) $$=\frac{2}{10}$$

Number of student in karate club =6

P(B)$$=\frac{6}{10}$$

P(A/B) $$=\frac{2}{10}$$ ÷ $$\frac{6}{10}$$

Converting division into multiplication by reciprocating the term after division

P(A/B) $$=\frac{2}{10}$$ × $$\frac{10}{6}$$

On solving we get ,

P(A/B) $$=\frac{2}{6}$$

P(A/B)$$=\frac{1}{3}$$

P(A/B) $$=0.33$$

- hope this helps! -
by