Let

*b* = number of black marbles

*w* = number of white marbles

Note:

The probability of removing *n* marbles (1 marble at a time) in succession without replacement is equal to the probability of removing *n* marbles at once.

**2 Marbles are Removed**

*P*_{First 2 are white} = 1/3

**3 Marbles are Removed**

*P*_{First 2 are white} *P*_{Third is white} = 1/6

(1/3) *P*_{Third is white} = 1/6

*P*_{Third is white} = 1/2

Therefore, the number of black and white marbles are equal after removing 2 white marbles.

*b* = *w* - 2

total number of marbles = *b* + *w*

total number of marbles = 2*w* - *2*

**First Draw White, Second Draw White**

$\dfrac{w}{2w - 2} \cdot \dfrac{w - 1}{2w - 3} = \dfrac{1}{3}$

$3w(w - 1) = (2w - 2)(2w - 3)$

$3w^2 - 3w = 4w^2 - 10w + 6$

$w^2 - 7w + 6 = 0$

$(w - 1)(w - 6) = 0$

$w = 1 ~ \text{ and } ~ 6$

Use *w* = 6 marbles

$b = 6 - 2$

$b = 4 ~ \text{ black marbles}$ ← *answer*