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1. 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?

A. $\frac{4}{15}$

B. $\frac{69}{91}$

C. $\frac{11}{15}$

D. $\frac{22}{91}$

Solution:

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

∴ P (E) = \frac{n (E)}{n (S)}

= \frac{330}{1365}

= \frac{22}{91}

∴ Required probality =

(1 - \frac{22}{91})

= \frac{69}{91}

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