JEE Advance - Mathematics (2004 - No. 6)
A box contains $$12$$ red and $$6$$ white balls. Balls are drawn from the box one at a time without replacement. If in $$6$$ draws there are at least $$4$$ white balls, find the probability that exactly one white is drawn in the next two draws. (binomial coefficients can be left as such)
($$\binom{6}{4}\binom{12}{2} + \binom{6}{5}\binom{12}{1} + \binom{6}{6}\binom{12}{0})/\binom{18}{6}$$
$$\frac{\binom{6}{4}\binom{12}{2}}{\binom{18}{6}}$$
$$\frac{(\binom{6}{4}\binom{12}{2} + \binom{6}{5}\binom{12}{1} + \binom{6}{6}\binom{12}{0})}{\binom{18}{6}} \times \frac{\binom{2}{1}\binom{2}{1}}{\binom{12}{2}}$$
$$(\frac{\binom{6}{4}\binom{12}{2} + \binom{6}{5}\binom{12}{1} + \binom{6}{6}\binom{12}{0}}{\binom{18}{6}}) \times (\frac{\binom{2}{1} \binom{10}{1}}{\binom{12}{2}} + \frac{\binom{12}{1} \binom{2}{1}}{\binom{12}{2}})$$
$$(\frac{\binom{6}{4}\binom{12}{2} + \binom{6}{5}\binom{12}{1} + \binom{6}{6}\binom{12}{0}}{\binom{18}{6}}) \times (\frac{\binom{2}{1} \binom{10}{1}}{\binom{12}{2}})$$
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