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)?

1 Answer

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

Number of student in karate club =6


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]=0.33[/tex]

- hope this helps! -
Welcome to AskTheTask.com, where understudies, educators and math devotees can ask and respond to any number related inquiry. Find support and replies to any numerical statement including variable based math, geometry, calculation, analytics, geometry, divisions, settling articulation, improving on articulations from there, the sky is the limit. Find solutions to numerical problems. Help is consistently 100 percent free!


No related questions found