An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked up at random, what is the probability that 1 Is green, 2 are blue and 1 is red ?

Answer: Option B

Total number of marbles = (6 + 4 + 2 + 3) = 15
Let E be the event of drawing 1 green, 2 blue and 1 red marble.
Then,
n (E) = $${}^2{C}_1\times {}^4{C}_2\times {}^6{C}_1$$   $$=2\times \frac{4\times 3}{2\times 1}\times 6$$     = 72
And, n (S) = $${}^{15}{C}_4=$$   $$\frac{15\times 14\times 13\times 12}{4\times 3\times 2\times 1}$$     = 1365
$$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{72}{1365}$$      $$=\frac{24}{455}$$