I just came by this "easy" question but i am abit worried, ill tell you why.
We have a box with 2 yellow balls and 1 red ball. We have also a second box, with 2 red balls and 1 yellow. We randomly choose one box , and one ball from it. Then we pick another ball from it (we don't put the 1st ball back to it).
What is the chance that we choose two balls of different color?
1st and foremost i think that it doesn't matter which box we choose because they look symmetric to me. Now theoretically, if we (let's say) choose the 1st box, the possible picks are 2 {YY and YR} . Also we can see that we will always pick at least one Yellow ball. Therefore the chance is 50% to choose a pair of ball of different color.
Now i was quite bored at home, so decided to experiment a bit. I painted 3 paper balls (lol) just to see if this is true. i shacked them and in fact i only piked 1 of them, just to see what color is the pair left. Most of the times it was YR! I don't get this, seriously!!!!
You were on the right track noting that there is some symmetry. However, looking at the probability that you pick a yellow ball breaks that symmetry, because the proportion of yellows in each box is different.
The symmetry is that each box has two balls of one color and one ball of another. You're only asked about the probability that the two balls are different. So, in essence, which box you choose doesn't matter. Then just calculate the probability that the balls are different colors for one (either) of the boxes.
You have three possible combinations of balls $A, B, C$ that you can take out: $AB, BC, AC.$ If $A$ and $B$ are the ones that are the same color, then the probability of getting two different colors is $2/3$ (the two combinations that include $C$.)
For the explicit solution, first you choose the box, then you choose two balls from that box. Let's say that the first box contains $Y_1, Y_2, R$ and the second contains $R_1, R_2, Y.$
Your combinations are then, for the first box, $Y_1Y_2, Y_1R, Y_2R.$ For the second box, they're $R_1R_2, R_1Y, R_2Y.$ Four out of six have two different colors, which is also $2/3.$