There are 3 red and 7 black balls in first pot. There are 41 red and 59 black balls in second pot. And there are 481 red and 519 black balls in third pot. One of the three pots is chosen according to the random principle and after that one ball is blindly chosen. How big is the probability to get a red ball?

Solution: $\frac{397}{1000}$

How big is the probability of taking a red ball from a fourth pot, when previously the content of the first three pots is placed in empty fourth pot?

Solution: $\frac{35}{74}$

My idea:

I was thinking to use formula: $P(A|B)=\frac{P(B|A)P(A)}{P(B)} $ But I am not sure how to connect this formula with the things that are wanted in this example.

I calculated only: $P(Red)=\frac{3}{10}$ and $P(Black)=\frac{7}{10}$
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$$\log\_2 x +\log\_4 x = 2$$
$$\frac{1}{\log\_ x2} +\frac{1}{\log\_ x4} = 2$$
$$\frac{1}{\log\_ x2} +\frac{1}{2\log\_ x2} = 2$$
$$\frac{3}{2\log\_ x2} = 2$$
$$\frac{1}{\log\_ x2} =\frac{4}{3}$$
$$\log\_ 2x =\frac{4}{3}$$
$$x=2^{4/3}$$
we write $$\frac{\ln(x)}{\ln(2)}+\frac{\ln(x)}{2\ln(2)}=2$$ multiplying by $$2\ln(2)$$ we obtain
$$2\ln(x)+\ln(x)=4\ln(2)$$thus we get
$$\ln(x)=\frac{4}{3}\ln(2)$$ can you finish this?