A basket contains 6 blue, 2 red, 4 green and 3 yellow balls. If 5 balls are picked up at random, what is the probability that at least one is blue?

Answer: Option A

Total number of balls = (6 + 2 + 4 + 3) = 15
Let E be the event of drawing 5 balls out of 9 non-blue balls.
$$={}^9{C}_5={}^9{C}_{\left(9-5\right)}={}^9{C}_4$$    $$=\frac{9\times 8\times 7\times 6}{4\times 3\times 2\times 1}$$    = 126
And,
n(S) = $${}^{15}{C}_5=$$   $$\frac{15\times 14\times 13\times 12\times 11}{5\times 4\times 3\times 2\times 1}$$     = 3003
$$\therefore P(E)=\frac{n(E)}{n(S)}=$$    $$\frac{126}{3003}=\frac{6}{143}$$
∴ Required Probability
$$\begin{array}{rl}& =\left(1-\frac{6}{143}\right)\\ & =\frac{137}{143}\end{array}$$