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A committee of two persons is selected from two men and two women. The probability that the committee will have exactly one woman is
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$\frac{2}{3}$
We know, $\mathrm{P}(\mathrm{E})=\frac{\text { No. of favourable outcomes }}{\text { Total no. of outcomes }}$
Number of possible outcomes $={ }^{4} \mathrm{C}_{2}=\frac{4 \times 3}{2 \times 1}=6$
(selecting 2 people from 4 people $)$ Number of favourable outcomes
$={ }^{2} \mathrm{C}_{1} \times{ }^{2} \mathrm{C}_{1}=2 \times 2=4$
(selecting 1 from 2 men, 1 from 2 women)
$\mathrm{P}(\mathrm{E})=\frac{4}{6}=\frac{2}{3}$
Number of possible outcomes $={ }^{4} \mathrm{C}_{2}=\frac{4 \times 3}{2 \times 1}=6$
(selecting 2 people from 4 people $)$ Number of favourable outcomes
$={ }^{2} \mathrm{C}_{1} \times{ }^{2} \mathrm{C}_{1}=2 \times 2=4$
(selecting 1 from 2 men, 1 from 2 women)
$\mathrm{P}(\mathrm{E})=\frac{4}{6}=\frac{2}{3}$
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