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1. An urn contains 6 red, 4 blue, 2 green and 3 yellow marbles. If four marbles are picked up at random, what is the probability that 1 Is green, 2 are blue and 1 is red ?

A. $\frac{13}{35}$

B. $\frac{24}{455}$

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

D. $\frac{1}{13}$

Solution:

Total number of marbles = (6 + 4 + 2 + 3) = 15
Let E be the event of drawing 1 green, 2 blue and 1 red marble.
Then,
n (E) =

^{2} C_{1} \times^{4} C_{2} \times^{6} C_{1}

= 2 \times \frac{4 \times 3}{2 \times 1} \times 6

= 72
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{72}{1365}

= \frac{24}{455}

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