# Probability Math Question…? [closed]

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.

-

## closed as off-topic by ᴡᴏʀᴅs, Claude Leibovici, hardmath, drhab, martiniJul 2 '14 at 8:49

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – ᴡᴏʀᴅs, Claude Leibovici, hardmath, drhab, martini
If this question can be reworded to fit the rules in the help center, please edit the question.

This looks like homework--is it? That is, should we be telling you the answer, or giving you hints to help you learn and solve it yourself? – Jonathan Christensen Jan 25 '13 at 2:46
I hate fractions. They are common, vulgar, or improper. – Dilip Sarwate Jan 25 '13 at 2:49
@DilipSarwate Is there an example of an uncommon, proper, civilized number that isn't a fraction? – Joe Z. Jan 25 '13 at 3:21

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.

-
It isn't graded homework, we are trying to learn how to solve the problems. I have an idea of how to solve this one but I'm not entirely sure... so hints would be appreciated. – Abs Jan 25 '13 at 2:52

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.

-