Sol:

There are four possibilities

 First Ball Second Ball Event White White E1 White Black E2 Black White E3 Black Black E4
$$P(E_{1}) = \frac{2}{4} * \frac{1}{3} = \frac{1}{6}$$ $$P(E_{2}) = \frac{2}{4} * \frac{2}{3} = \frac{1}{3}$$ $$P(E_{3}) = \frac{2}{4} * \frac{2}{5} = \frac{1}{5}$$ $$P(E_{4}) = \frac{2}{4} * \frac{3}{5} = \frac{1}{10}$$

Let E denotes the event of drawing a black ball in the third attempt

$$P(\frac{E}{E_{1}}) = 1$$. Because there is no white ball left to be selected.

$$P(\frac{E}{E_{2}}) = \frac{3}{4}$$.Because there are 3 black ball and 1 white ball left.

$$P(\frac{E}{E_{3}}) = \frac{3}{4}$$. Because again there are 3 black and 1 white ball left.

$$P(\frac{E}{E_{4}}) = \frac{4}{6} = \frac{2}{3}$$. Because there are 4 black balls and 2 white balls.

$$P(E) = P(E_{1}) P (\frac{E}{E_{1}}) + P(E_{2}) P (\frac{E}{E_{2}}) + P (E_{3}) P (\frac{E}{E_{3}}) + P (E_{4}) P (\frac{E}{E_{4}})$$ $$\Rightarrow \frac{1}{6} * 1 + \frac{1}{3} * \frac{3}{4} + \frac{1}{5} * \frac{3}{4} + \frac{3}{10} + \frac{2}{3}$$ $$\Rightarrow \frac{1}{6} + \frac{1}{4} + \frac{3}{20} + \frac{1}{5}$$ $$\Rightarrow \frac{23}{30}$$

Therefore, the value of K is $$\frac{23}{30}$$.

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