Question

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

Answer 1

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) $= \frac{2}{10}\div{\frac{6}{10}$

Converting division into multiplication by reciprocating the term after division

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

On solving we get ,

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

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

P(A/B) $=0.33$