<s>[INST] Why is this intuitive method valid? Problem. There are $2$ white and $3$ black balls in the urn. A person randomly picked $2$ balls and put $1$ white ball. What is the probability of the event that the next randomly-picked ball would be white.
To solve this problem formally you have to consider conditional probabilities and so on. But in the childhood I did not know such words so I was solving problems of this type in the following way.

Intuitive solution. Let's grind balls into powder and mix it. We will have $5$ (whatever you want, for instance) kilograms of powder. $\frac{2}{5}$ of each kilogram is white and $\frac{3}{5}$ is black. A person randomly took $2$ balls in my model means that he took $2$ kilogram of powder. Putting $1$ white ball means that he put $1$ kilogram of white powder.
After all actions there are $\left(5 \cdot\frac35-2\cdot \frac35\right)=\frac95$ kilograms of black powder and $\left(5 \cdot\frac25-2\cdot \frac25\right) + 1=\frac95=\frac{11}5$ kilograms of white powder. So the final probability of picking white ball is $\frac{11}{5}/\left(\frac{11}{5}+\frac{9}{5}\right) = \frac{11}{20}$.

You can check that this is the right answer, but the question is why is this intuitive method valid?
 [/INST] Grinding balls into powder is mathematically the same as calculating expected value. 
In fact, the reason why mathematicians often speak of expected value calculations as the sum of the possible values, weighted by their probabilities, is exactly why your intuition of weights of grinded balls is relevant. 

The expected number of black balls taken out, $b$, is 
$$E(b) = P(b=0)\cdot 0 + P(b=1) \cdot 1 + P(b=2) \cdot 2$$
The $P(b=k)$ for $k=0,1,2$ are
$$P(b=0) = \frac{1}{\binom{5}{2}} = \frac{1}{10}$$
$$P(b=1) = \frac{2 \cdot 3}{\binom{5}{2}} = \frac{3}{5}$$
$$P(b=2) = \frac{\binom{3}{2}}{\binom{5}{2}} = \frac{3}{10}$$
This gives us
$$E(b) = 0 + \frac{3}{5} + 2 \cdot \frac{3}{10} = \frac{6}{5}$$
This means that the expected number of black balls left after two are taken out is 
$$3-E(b) = \frac{9}{5}$$
Similarly, the expected number of white balls taken out, $w$, is
$$E(w) = \frac{4}{5}$$
And the expected number of white balls remaining:
$$2-E(w) = \frac{6}{5}$$
You can calculate this directly, or notice that $$(3-E(b)) + (2-E(w)) = 5-E(b+w) = 5-2 = 3$$

After adding one white ball, the new expected number of white balls is
$$1+(2-E(w)) = 1 + \frac{6}{5} = \frac{11}{5}$$
There is now a total of $4$ balls, so the probability of choosing a white ball is
$$\frac{\frac{11}{5}}{4} = \frac{11}{20}$$
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