Agroup of 10 students participate in chess club, karate club, or neither. let event a = the student is in chess club. let event b = the student is in karate club. one of these students is randomly selected. what is p(a|b)? a. 2/6 ≈ 0.33 b. 4/6 ≈ 0.67 c. 6/10 ≈ 0.60 d. 2/10 ≈ 0.20 if you can, explain how you got your answer. i would like to know how you solved it.

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)

Number of student in karate club =6

P(B)

P(A/B)

Converting division into multiplication by reciprocating the term after division

P(A/B)

On solving we get ,

P(A/B)

P(A/B)

P(A/B)

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

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step-by-step explanation:

take it another