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200 balls are distributed into two urns based on their color, black or white.

We want to draw 85 balls in total, with a preference for black balls.

The first approach is to first draw 20 balls from the black urn and then mix the remaining balls together (black and white) and draw 65 more balls.

The second approach is to first mix all balls together (black and white), draw 65 balls, then put the remaining balls in their respective urns and draw 20 black balls.

What is the number of balls of each color knowing that the second approach increases the probability of drawing a black ball by 15%.

The probability of drawing a black ball following the first approach is equal to

$$P_1(B) = frac1{x_B} + frac{x_B-20}{180}$$

The probability of drawing a black ball following the second approach is equal to

$$P_2(B) = frac{x_B}{200}+frac1{x'_B}$$

$x_B$ being the number of black balls

$x_B'$ being the remaining number of balls after the first step

I am stuck trying to determine $x_B'$ so I can then solve $P_2(B)=P_1(B)+0.15$, any help would be greatly appreciated!