A bag contains 12 red balls and 6 white balls, six balls are drawn one by one without replacement of which atleast 4 balls are white. Find the probability that in the next two draws exactly one white ball is drawn

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$\frac{3}{5}$

$\frac{1}{5}$

$\frac{2}{119}$

$\frac{4}{5}$

answer is C.

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Detailed Solution

Let E be the event that atleast 4 White balls are drawn in the first six draws,
with replacement
Let $E_{4}$ be the event that exactly 4 white balls are drawn

Let $E_{5}$ be the event that exactly 5 white balls are drawn

And $E_{6}$ be the event that exactly 6 white balls are drawn

If A is the event that out of the next two draws exactly one white ball is drawn, then

$P (A) = P (A \cap E_{4}) + P (A \cap E_{5}) + P (A \cap E_{6})$

$= P (E_{4}) P (A / E_{4}) + P (E_{5}) P (A / E_{5}) + P (E_{6}) P (A / E_{6})$

$= \frac{6_{C 4} 12 C_{2}}{18 C_{6}} . \frac{2 C_{1} . 10 C_{1}}{12 C_{2}} + \frac{6 C_{5} 12 C_{1}}{18 C_{6}} \frac{1 C_{1} 11 C_{1}}{12 C_{2}} + \frac{6 C_{6}}{18 C_{6}} .0 = \frac{2}{119}$