Probability Math Question...? Bag A contains 3 white and 2 red balls. Bag B contains 6 white and 3 red balls. One of the two bags will be chosen at random, and then two balls will be drawn from that bag at random without replacement. What is the probability that the two balls drawn will the be the same color? Express your answer as a common fraction.
 A: Assuming each bag has equal chances of being chosen, probability = $$\frac{1}{2}\frac{\left({3\choose 2}+{2\choose 2}\right)}{{5\choose 2}}+\frac{1}{2}\frac{\left({6\choose 2}+{3\choose 2}\right)}{{9\choose 2}}$$
First, probability of choosing first bag is $1/2$, then after selecting first bag , we need to select two balls of same color which has prob.= $\frac{\left({3\choose 2}+{2\choose 2}\right)}{{5\choose 2}}$ as we can choose both white balls in ${3\choose 2}$ ways or both red balls in ${2\choose 2}$ ways while total number of ways of selecting 2 balls out of $5$ balls in first bag is ${5\choose 2}$. You can follow same argument for second bag. 
A: More directly:
$$P\left(AWW\right)+P\left(ARR\right)+P\left(BWW\right)+P\left(BRR\right)=\frac{1}{2}\frac{3}{5}\frac{2}{4}+\frac{1}{2}\frac{2}{5}\frac{1}{4}+\frac{1}{2}\frac{6}{9}\frac{5}{8}+\frac{1}{2}\frac{3}{9}\frac{2}{8}$$
Here e.g.  $AWW$ stands for the event that bag $A$ is chosen, a white
ball is taken out at the first draw and a white ball is taken out
at the second draw.
