A bag contains 3 blue, 2 green and 5 red balls. If four balls are picked at random, what is the probability that two are green and two are blue?

Answer: Option B

Number of blue balls = 3
Number of green balls = 2
Numbers of red balls = 5
Total balls in the bag = 3 + 2 + 5 = 10
Total possible outcomes = Selection of 4 balls out of 10 balls
$$={}^{10}{C}_4=\frac{10!}{4!\times (10-4)!}$$
$$=\frac{10\times 9\times 8\times 7}{1\times 2\times 3\times 4}$$
= 210
Favorable outcomes = (selection of 2 green balls out of 2 balls) × (selection of 2 balls out of 3 blue balls)
$$\begin{array}{rl}& ={}^2{C}_2\times {}^3{C}_2\\ & =1\times 3\\ & =3\end{array}$$
∴ Required probability
  $$=\frac{\text{Favorable outcomes}}{\text{Total possible outcomes}}$$
$$\begin{array}{rl}& =\frac{3}{210}\\ & =\frac{1}{70}\end{array}$$