In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that the selected students are 2 boys and 1 girls, is:

Answer: Option A

Let S be the sample space and let E be the event of selecting 2 boys and 1 girl.
Then, n(S) = number of ways of selecting 3 students out of 25
= $${}^{25}{C}_3=$$   $$\frac{25\times 24\times 23}{3\times 2\times 1}$$   = 2300
And, n(E) = $$\left({}^{15}{C}_2\times {}^{10}{C}_1\right)$$   $$=\left(\frac{15\times 14}{2\times 1}\times 10\right)$$    = 1050
$$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{1050}{2300}=\frac{21}{46}$$