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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)?
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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) [tex]=\frac{2}{10}[/tex]

Number of student in karate club =6

P(B)[tex]=\frac{6}{10}[/tex]

P(A/B) [tex]=\frac{2}{10}[/tex] ÷ [tex]\frac{6}{10}[/tex]

Converting division into multiplication by reciprocating the term after division

P(A/B) [tex]=\frac{2}{10}[/tex] × [tex]\frac{10}{6}[/tex]

On solving we get ,

P(A/B) [tex]=\frac{2}{6}[/tex]

P(A/B)[tex]=\frac{1}{3}[/tex]

P(A/B) [tex]=0.33[/tex]

- hope this helps! -
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