Probability

Let A,B and C be the event that a student is from section A,B and C.
So, P(A)=0.2,P(B)=0.3,P(C)=0.5
Now, section A : 20 girls, 30 boys, total 50 students
Section B : 40 girls, 20 boys, total 60 students
Section C : 10 girls, 30 boys, total 40 students
P(G∕A)‌=‌

20

50

=0.4,P(G∕B)=‌

40

60

=‌

2

3

P(G∕C)‌=‌

10

40

=0.25
P(G)‌=P(G∕A)⋅P(A)+P(G∕B)⋅P(B)+P(G∕C)⋅P(C)
‌=0.4×0.2+‌

2

3

×0.3+0.25×0.5
‌=0.08+0.2+0.125
‌=0.405
Now, P(G∩A)=P(G∕A)⋅P(A)
=0.4×0.2=0.08
Using Bayes' theorem,
‌P(A∕G)=‌

P(A∩G)

P(G)

‌=‌

0.08

0.405

=‌

80

405

=‌

16

81

So, the probability that a randomly selected girl belongs to section A is ‌

16

81

.