Why probability of retrieving a specific colored ball is not the same as retrieving it from distributed boxes? The probability of retrieving one red ball without looking out of 10 blue and 2 red balls is 2/12.
If we put the blue balls in 10 boxes individually and put the 2 red balls in any 2 random boxes the probability of retrieving one red ball will be (selecting the box with red ball) * (retrieving red ball) = 2/10 * 1/2 = 1/10.
As we are retrieving one red ball in both cases out of 12 total balls, why the probability is not the same. I need help to understand the reasoning behind it?
 A: Look at a simpler example, with $2$ blue balls and $1$ red ball. Clearly if you put all $3$ balls into a box and draw one ball at random the probability of drawing a red ball is $\frac13$.
Suppose, however, that you perform a different experiment. You put the red ball and one blue ball in one box and the second blue ball in another box. You then choose one of the boxes at random and draw a ball from it — at random if it’s the box with two balls. What has to happen in order for you to get the red ball? First you must pick the box that contains the red ball, but that’s not enough: then you must choose the red ball from it instead of the blue one. The probability that you will choose the right box is $\frac12$, and if you’ve done so, the probability that you will draw the red ball is $\frac12$. The probability that both of these things happen is the product of their probabilities, or $\frac12\cdot\frac12=\frac14$.
If you think in frequentist terms, you can think of it like this: on average you will pick the right box half the time, and on average you will then pick the red ball half of those times, so you’ll end up with the red ball only a quarter of the time on average.
In both experiments there are three possible outcomes, but the probabilities of those outcomes aren’t the same in the two experiments. When you put all $3$ balls in one box and draw one at random, the balls are equally likely to be drawn, so each is drawn with probability $\frac13$. In the other experiment, however, the probability of getting the blue ball that is in its own box is $\frac12$, so the other two balls have to ‘share’ the other $\frac12$ of the total probability of $1$. It’s pretty clear that they are equally likely to be chosen, so they must split it equally: each of them has a probability of $\frac14$ of being drawn.
The situation in your problem is entirely similar; the numbers are just bigger. The idea remains the same: in the second experiment a ball that shares a box with another ball is less likely to be drawn than a ball that has a box all to itself, whereas in the first experiment the balls are all equally likely to be drawn.
A: It is because the method of drawing the balls introduces a bias.

In the second case the probabilities for selecting individual balls are not equal.
The ten boxes are each equally likely to be selected.  But there is a bias towards selecting a ball that is the only ball in its box.
Eight of the blue balls are the only ball in their boxes; if you select any among those boxes you then certainly pull out a blue ball.   The remaining two blue balls are in boxes with a red ball; if you select any among those boxes you then have probability of 1/2 for pulling out a blue ball.
A: When you pick balls out of another container, you sum each container's probability of drawing a target ball color. If the balls from all of the containers were combined into 1 container, then it would be a different situation.
A: The equal number of Balls is not a necessary requirement for equivalence of experiments. You can always have one red ball and the probability of drawing one red ball is one and have N boxes and distribute the ball and the probability of picking it is now 1/N.
