Hello our question is,
If a group of 80 schoolgirls consists of 54 dancers and 35 singers, each member of the group is either a dancer or singer or both. The probability that a randomly selected student is a singer given that she is a dancer is?
Thanks 
+128408 Let A be the the probability that she is a dancer = 54/80 = 27/40
Let B be the probability that she is a singer = 35/80 = 7/16
And the probability of A and B = (A ∩ B) = 9/80
So the probability of B, given A, is just
P(B l A ) = P ( A ∩ B) / P(A) = (9/80) /(27/40) = (9/80)(40/27) =(1/2)(1/3) = 1/6
This makes sense....there are 54 dancers and 9 of them sing, as well. So...9/54 = 1/6
+118608 80-54-35=-9
so
80 children, 45 dance only, 26 sing only, 9 dance and sing
P(S|D)= 9/54 = 1/6
+128408 Let A be the the probability that she is a dancer = 54/80 = 27/40
Let B be the probability that she is a singer = 35/80 = 7/16
And the probability of A and B = (A ∩ B) = 9/80
So the probability of B, given A, is just
P(B l A ) = P ( A ∩ B) / P(A) = (9/80) /(27/40) = (9/80)(40/27) =(1/2)(1/3) = 1/6
This makes sense....there are 54 dancers and 9 of them sing, as well. So...9/54 = 1/6
