A group 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, please explain how you got your answer. I would like to know how you solved it.

A group 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 o class=

Respuesta :

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}\div{\frac{6}{10}[/tex]


Converting division into multiplication by reciprocating the term after division



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

Answer:

There is another version of this question on ap3x and the answer is 2/4=0.50

Step-by-step explanation:

just took the ap3x quiz