There are 12 boys and 8 girls in a tuition centre. If three of them scored first marks, then what is the probability that one of the three is a girl and the other two are boys?

Total number of students = 20.
Let S be the sample space.
Then, n(S) = number of ways of three scored first mark
n(S) = ²⁰C₃
= 20 x 19 x 18 / 2 x 3
= 20 x 19 x 3

Let,
E be the event of 1 girl and 2 boys.
Therefore, n(E) = number of possible of 1 girl out of 8 and 2 boys out of 12.
n(E) = ⁸C₁ x ¹²C₂
= 8 x 12 x 11 / 1 x 2
= 8 x 6 x 11.

Now, the required probability = n(E)/n(S)
= (8 x 6 x 11)/(20 x 19 x 3)
= 44/95.