10 પુરુષ અને 5 સ્ત્રીમાંથી ચાર સભ્યોની એક સમિ | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{1}{11}$

Vedclass · Answer

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.

$\mathrm{P}(1 \mathrm{M}, 3 \mathrm{W})+\mathrm{P}(0 \mathrm{M}, 4 \mathrm{W})$

$\mathrm{P}(3 \mathrm{M}, 1 \mathrm{W})+\mathrm{P}(2 \mathrm{M}, 2 \mathrm{W})+\mathrm{P}(1 \mathrm{M}, 3 \mathrm{W})+\mathrm{P}(\mathrm{OM}, 4 \mathrm{W})$

$=\frac{\frac{105}{1365}}{\frac{1155}{1365}}=\frac{1}{11}$

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