A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If four marbles are picked at random, what is the probability that none is blue?

A. $$\frac{{17}}{{91}}$$

B. $$\frac{{33}}{{91}}$$

C. $$\frac{{51}}{{91}}$$

D. $$\frac{{65}}{{91}}$$

Answer: Option B

Solution(By Examveda Team)

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
When four marbles are picked at random, then the probability that none is blue is
$$\eqalign{ & = \frac{{{}^{12}{C_4}}}{{{}^{15}{C_4}}} \cr & = \frac{{12 \times 11 \times 10 \times 9}}{{15 \times 14 \times 13 \times 12}} \cr & = \frac{{11880}}{{32760}} \cr & = \frac{{33}}{{91}} \cr} $$