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

Answer: Option C

Total number of students = 20
Let S be the sample space
Then, n(S) = number of ways of three scored first mark
$$\begin{array}{rl}& \text{n}\left(\text{S}\right)={}^{20}{C}_3\\ & =\frac{20\times 19\times 18}{2\times 3}\\ & =20\times 19\times 3\end{array}$$
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
$$\begin{array}{rl}& \text{n}\left(\text{E}\right)={}^8{C}_1\times {}^{12}{C}_2\\ & =\frac{8\times 12\times 11}{1\times 2}\\ & =8\times 6\times 11\end{array}$$
Now, the required probability
$$\begin{array}{rl}& =\frac{\text{n}\left(\text{E}\right)}{\text{n}\left(\text{S}\right)}\\ & =\frac{8\times 6\times 11}{20\times 19\times 3}\\ & =\frac{44}{95}\end{array}$$