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?

Solution(By Examveda Team)

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
$$ = {}^{10}\mathop C\nolimits_4 = \frac{{10!}}{{4! \times (10 - 4)!}}$$
$$ = \frac{{10 \times 9 \times 8 \times 7}}{{1 \times 2 \times 3 \times 4}}$$
= 210
Favorable outcomes = (selection of 2 green balls out of 2 balls) × (selection of 2 balls out of 3 blue balls)
$$\eqalign{ & = {}^2\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_2 \cr & = 1 \times 3 \cr & = 3 \cr} $$
∴ Required probability
  $$ = \frac{{{\text{Favorable outcomes}}}}{{{\text{Total possible outcomes}}}}$$
$$\eqalign{ & = \frac{3}{{210}} \cr & = \frac{1}{{70}} \cr} $$