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(B) The total number of persons is $2 + 2 = 4$. Out of these $4$ persons,$2$ can be selected in $^{4}C_{2}$ ways. $^{4}C_{2} = \frac{4 	imes 3}{2 	i

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$2/3$

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(B) The total number of persons is $2 + 2 = 4$. Out of these $4$ persons,$2$ can be selected in $^{4}C_{2}$ ways. $^{4}C_{2} = \frac{4 	imes 3}{2 	imes 1} = 6$. For the committee to have exactly one man,it must also have exactly one woman. One man out of $2$ can be selected in $^{2}C_{1} = 2$ ways. One woman out of $2$ can be selected in $^{2}C_{1} = 2$ ways. Total favorable ways $= ^{2}C_{1} 	imes ^{2}C_{1} = 2 	imes 2 = 4$. Therefore,$P(	ext{one man}) = \frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{4}{6} = \frac{2}{3}$.

$A$ committee of two persons is selected from two men and two women. What is the probability that the committee will have exactly one man?

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