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From a group of $10$ men and $5$ women, four member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men, is
$\frac{21}{220}$
$\frac{3}{11}$
$\frac{1}{11}$
$\frac{2}{23}$
Solution
Probability of $4$ member committee which obntain atleast one woman.
$\Rightarrow \mathrm{P}(3 \mathrm{M}, 1 \mathrm{W})+\mathrm{P}(2 \mathrm{M}, 2 \mathrm{W})+$
$\mathrm{P}(1 \mathrm{M}, 3 \mathrm{W})+\mathrm{P}(0 \mathrm{M}, 4 \mathrm{W})$
$ \Rightarrow \frac{{{\,^{10}}{C_3}{\,^5}{C_1}}}{{^{15}{C_4}}} + \frac{{{\,^{10}}{C_2}{\,^5}{C_2}}}{{^{15}{C_4}}} + \frac{{{\,^{10}}{C_1}{\,^5}{C_3}}}{{^{15}{C_4}}} + \frac{{{\,^{10}}{C_0}{\,^5}{C_4}}}{{^{15}{C_4}}}$
$\Rightarrow \frac{600}{1365}+\frac{450}{1365}+\frac{100}{1365}+\frac{5}{1365}$
$\Rightarrow \quad \frac{1155}{1365}$
Probability of committees to have more women than men.
$\mathrm{P}(1 \mathrm{M}, 3 \mathrm{W})+\mathrm{P}(0 \mathrm{M}, 4 \mathrm{W})$
$\mathrm{P}(3 \mathrm{M}, 1 \mathrm{W})+\mathrm{P}(2 \mathrm{M}, 2 \mathrm{W})+\mathrm{P}(1 \mathrm{M}, 3 \mathrm{W})+\mathrm{P}(\mathrm{OM}, 4 \mathrm{W})$
$=\frac{\frac{105}{1365}}{\frac{1155}{1365}}=\frac{1}{11}$