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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