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

A. $\frac{5}{91}$

B. $\frac{1}{35}$

C. $\frac{1}{3}$

D. $\frac{4}{105}$

Solution:

Total number of marbles = (6 + 4 + 2 + 3) = 15
Let E be the event of drawing 2 marbles such that either both are green or both are yellow.
Then,
n (E) =

(^{2} C_{1} +^{3} C_{2})

= (1 +^{3} C_{1})

= (1 + 3) = 4
And,n (S) =

^{15} C_{2} =

\frac{15 \times 14}{2 \times 1}

= 105

∴ P (E) = \frac{n (E)}{n (S)} = \frac{4}{105}

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