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 ?

A. $$\frac{{13}}{{35}}$$

B. $$\frac{{1}}{{5}}$$

C. $$\frac{{18}}{{455}}$$

D. $$\frac{{7}}{{15}}$$

E. None of these

Answer: Option C

Solution(By Examveda Team)

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) = $$\left( {{}^4\mathop C\nolimits_2 \times {}^3\mathop C\nolimits_1 } \right)$$   $$ = \left( {\frac{{4 \times 3}}{{2 \times 1}} \times 3} \right)$$   = 18
Also, n(S) = $${}^{15}\mathop C\nolimits_3 = $$   $$\frac{{15 \times 14 \times 13}}{{3 \times 2 \times 1}}$$   = 455
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{18}}{{455}}$$