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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

Feb 16, 2015

#2
+94558
+5

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

Feb 16, 2015

#1
+95360
+5

80-54-35=-9

so

80 children,    45 dance only,   26 sing only,    9 dance and sing

P(S|D)= 9/54 = 1/6

Feb 16, 2015
#2
+94558
+5

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

CPhill Feb 16, 2015