Doubt in finding probability of an event from the conditional probability

Question

I am studying about conditional probability and see a problem given as

In the solution$P(B_1)=P(B_2)=P(B_3)=1/3$ is taken.

From set point of view, event $B_1$ can be seen as ${\{R,R,R,R,G,G,G,B,B,B,B,B}\}$ and $R\cap B_1={\{R,R,R,R}\}$.
So $P(R\cap B_1)=|R\cap B_1|/|B_1|=P(R\cap B_1)/P(B_1)$.

Here our sample space is like ${\{R,...,R,G,...,G,B,...,B}\}$.

If we have to calculate P(R), so shouldn't we have to use $P(B_1)=|B_1|/|S|= 12/30=4/10$ instead of 1/3.
Because the formula used in the solution, holds only when we don't change our sample space from ${\{R,...,R,G,...,G,B,...,B}\}$ to ${\{B_1,B_2,B_3}\}$.

Also in the solution probability of choosing red ball is incorrect not equal to number of red balls divided by this total number of balls. If we use $P(B)=|\text{Number of balls in box B}|/|\text{Total number of balls}|$, then we get $P(R)= 1/3$.

Is the solution incorrect, or my way of thinking is incorrect. Please clarify the doubt.

Answer 1

Your way of thinking is incorrect. You are assuming that since the total number of balls in the 3 boxes is 30, each of the balls has a $(1/30)$ chance of being picked.

This is wrong. In fact, because box 2 contains only 8 balls and box 1 contains 12 balls, each ball in box 2 is 1.5 times more likely to be chosen than a ball in box 1.

To take an extreme case, suppose box 2 had 1 ball (of any color), and boxes 1 and 3 each had 100 balls (of any color). Then the ball in box 2 would be 100 times more likely to be chosen than any of the individual balls in either box 1 or box 3.

Addendum
Per OP's request.
Responding to his comment/question to Math Lover:

You questioned the formula

$$p(R) ~=~ \sum_{i=1}^3 p(B_i)p(R|B_i).\tag1 $$

Note that I have rearranged the factors in each term, with the $p(B_i)$ factor shown 1st to make the formula easier to intuit.

If I understand your question, you are questioning what would happen if the number of red balls in one of the boxes were to change.

Let's take an extreme case.

Suppose that Box 1 has 100 balls, all blue.
Suppose that Box 2 has 1 ball, which is red.
Suppose that Box 3 has 100 balls, all green.

According to the summation formula, the chance that a red ball would be selected is

$$\left(\frac{1}{3} \times \frac{0}{100}\right) ~+~ \left(\frac{1}{3} \times \frac{1}{1}\right) ~+~ \left(\frac{1}{3} \times \frac{0}{100}\right) = \frac{1}{3}.$$

In fact, in this extreme case, the probability that the ball chosen is red is in fact $\frac{1}{3},$ so the formula continues to be accurate.