Answer: Option C
Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that both marbles are blue
$$\eqalign{
& = \frac{{{}^3{C_2}}}{{{}^{15}{C_2}}} \cr
& = \frac{{3 \times 2}}{{15 \times 14}} \cr
& = \frac{1}{{35}} \cr} $$
Probability that both are yellow
$$\eqalign{
& = \frac{{{}^2{C_2}}}{{{}^{15}{C_2}}} \cr
& = \frac{{2 \times 1}}{{15 \times 14}} \cr
& = \frac{1}{{105}} \cr} $$
Probability that one blue and other is yellow
$$\eqalign{
& = \frac{{{}^3{C_1} \times {}^2{C_1}}}{{{}^{15}{C_2}}} \cr
& = \frac{{2 \times 3 \times 2}}{{15 \times 14}} \cr
& = \frac{2}{{35}} \cr} $$
∴ Required probability
$$\eqalign{
& \frac{1}{{35}} + \frac{1}{{105}} + \frac{2}{{35}} \cr
& = \frac{{3 + 1 + 6}}{{105}} \cr
& = \frac{{10}}{{105}} \cr
& = \frac{2}{{21}} \cr} $$