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A bag contains, $7$ different Black balls .and $10$ different Red balls, if one by one ball are randomely drawn untill all black balls are not drawn, then probability that this process is completed in $12 ^{th}$ draw, is equal to
$\frac{{^7{C_6}{\,^{10}}{C_6}}}{{^{17}{C_{12}}}} - \frac{{^1{C_1}}}{{^5{C_1}}}$
$\frac{{^7{C_6}{\,^{10}}{C_5}}}{{^{17}{C_{11}}}} - \frac{{^1{C_1}}}{{^6{C_1}}}$
$\frac{{^7{C_6}{\,^{10}}{C_10}}}{{^{17}{C_{11}}}} - \frac{{^1{C_1}}}{{^6{C_1}}}$
None
Solution
$\boxed{\frac{{{\,^{11}}{C_6} \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 10 \times 9 \times 8 \times 7 \times 6}}{{17 \times 16 \times 15 \times 14 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7}}} \times \boxed{\frac{1}{6}}$
$ \Downarrow $ $ \Downarrow $
(Prob. of $6$ black and $5$ red in frist $11$ draw) Prob. of $7$ black ball in ${12^{th}}$ draw
$ = \frac{{{\,^7}{C_6} \times {\,^{10}}{C_5}}}{{{\,^{17}}{C_{11}}}} \times \frac{{{\,^1}{C_1}}}{{{\,^6}{C_1}}}$
