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?

A. $$\frac{{4}}{{15}}$$

B. $$\frac{{5}}{{27}}$$

C. $$\frac{{1}}{{3}}$$

D. $$\frac{{2}}{{445}}$$

E. None of these

Answer: Option D

Solution(By Examveda Team)

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) = $$\left( {{}^2\mathop C\nolimits_2 \times {}^4\mathop C\nolimits_2 } \right)$$   $$ = \left( {1 \times \frac{{4 \times 3}}{{2 \times 1}}} \right)$$   = 6
And, n(S) = $${}^{15}\mathop C\nolimits_4 = $$   $$\frac{{15 \times 14 \times 13 \times 12}}{{4 \times 3 \times 2 \times 1}}$$     = 1365
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{6}{{1365}} = \frac{2}{{455}}$$