$$P(A|B) = \frac{P(A\cap B)}{P(B)}$$
$$= \frac{P(2 \,red \,balls\, AND \,even\, number\, of\, red\, balls\,)}{P(0\, red\, balls\, OR\, 2\, red\, balls)}$$
$$= \frac{P(2\,red\,balls)}{P(0\, red\, balls\, OR\, 2\, red\, balls)}$$
$$=\frac{\frac{\binom{7}{2}\binom{9}{1}}{\binom{16}{3}}}{\frac{\binom{9}{3}+\binom{7}{2}\binom{9}{1}}{\binom{16}{3}}}$$
$$=\frac{\binom{7}{2}\binom{9}{1}}{\binom{9}{3}+\binom{7}{2}\binom{9}{1}}$$
The blindfolded is just supposed to be the same as a random selection. What is the chance the first two picks were red? Without the information that there was at least one red ball in the first two picks, it is $\frac 12 \cdot \frac {n-1}{2n-1}$. The chance of one of each is $\frac n{2n-1}$, again without knowing that there was at least one red in the first two. We need to scale these up based on the new information, so the chance of two red is $\frac 12 \cdot \frac {n-1}{2n-1}\cdot \frac{4n-2}{3n-1}$. Similarly the chance the first two picks are opposite is $\frac n{2n-1}\cdot \frac{4n-2}{3n-1}.$ After the first two picks, either there are $n-1$ of each color and the chance the third is red is $\frac 12$, or there are $n-2$ red ones and $n$ green ones and the chance of the third is red is $\frac {n-2}{2n}$
Best Answer
The total number of (equally likely) outcomes is $14 \choose 4$ For a outcome to be favorable, you need to get 2 red balls out of 5, 1 blue out of 4, 1 green out of 3 and 0 yellow out of 2, so the total number of favorable outcomes is: ${5\choose 2}{4\choose 1}{3\choose 1}{2\choose 0}$ So the probability you are looking for is:
$$\frac{{5\choose 2}{4\choose 1}{3\choose 1}{2\choose 0}}{14 \choose 4}$$