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

Answer: Option B

Total number of marbles = (6 + 4 + 2 + 3) = 15
Let E be the event of drawing 4 marbles such that none is blue.
Then, n(E) = number of ways of drawing 4 marbles out of 11 non-blue.
$${}^{11}{C}_4=\frac{11\times 10\times 9\times 8}{4\times 3\times 2\times 1}$$     = 330
And n(S) = $${}^{15}{C}_4=$$   $$\frac{15\times 14\times 13\times 12}{4\times 3\times 2\times 1}$$     = 1365
$$\therefore P(E)=\frac{n(E)}{n(S)}$$    $$=\frac{330}{1365}$$  $$=\frac{22}{91}$$
∴ Required probality =$$\left(1-\frac{22}{91}\right)$$   $$=\frac{69}{91}$$