A basket contains 6 blue, 2 red,4 green and 3 yellow balls. If four balls are picked up at random, what is the probability that 2 are red 2 are green?

Answer: Option D

Total number of balls = (6 + 2 + 4 + 3) = 15
Let E be the event of drawing 4 balls such that 2 are red and 2 are green.
Then, n(E) = $$\left({}^2{C}_2\times {}^4{C}_2\right)$$   $$=\left(1\times \frac{4\times 3}{2\times 1}\right)$$   = 6
And, n(S) = $${}^{15}{C}_4=$$   $$\frac{15\times 14\times 13\times 12}{4\times 3\times 2\times 1}$$     = 1365
$$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{6}{1365}=\frac{2}{455}$$