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

Answer: Option C

Total number of marbles = (6 + 4 + 2 + 3) = 15
Let E be the event of drawing 2 blue and 1 yellow marble.
Then, n(E) = $$\left({}^4{C}_2\times {}^3{C}_1\right)$$   $$=\left(\frac{4\times 3}{2\times 1}\times 3\right)$$   = 18
Also, n(S) = $${}^{15}{C}_3=$$   $$\frac{15\times 14\times 13}{3\times 2\times 1}$$   = 455
$$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{18}{455}$$