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

E. None of these

Answer: Option D

Solution(By Examveda Team)

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) = $$\left( {{}^2\mathop C\nolimits_1 + {}^3\mathop C\nolimits_2 } \right)$$   $$ = \left( {1 + {}^3\mathop C\nolimits_1 } \right)$$   = (1 + 3) = 4
And, n (S) = $${}^{15}\mathop C\nolimits_2 = $$   $$\frac{{15 \times 14}}{{2 \times 1}}$$   = 105
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{4}{{105}}$$