A basket contains 6 blue, 2 red, 4 green, and 3 yellow balls. If three balls are picked up at random, what is the probability that none is yellow?

Answer: Option D

Total number of balls = (6 + 2 + 4 + 3) = 15
Let E be the event of drawing 3 non-yellow balls
Then, n(E) = $${}^{12}{C}_3$$   $$=\frac{12\times 11\times 10}{3\times 2\times 1}$$   = 220
Also, n(S) = $${}^{15}{C}_3$$   $$=\frac{15\times 14\times 13}{3\times 2\times 1}$$   = 455
$$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{220}{455}=\frac{44}{91}$$