A committee of 3 members is to be selected out of 3 men and 2 women. What is the probability that the committee has at least 1 woman ?

Answer: Option D

Total number of persons = (3 + 2) = 5
$$\therefore n(S)={}^5{C}_3={}^5{C}_2$$     $$=\frac{5\times 4}{2\times 1}$$   = 10
Let E be the event of selecting 3 members having at least 1 women
Then, n(E) = n [(1 women and 2 men ) or (2 women and 1 man)]
= n (1 woman and 2 men) + n (2 women and 1 man)
$$\begin{array}{rl}& =\left({}^2{C}_1\times {}^3{C}_2\right)+\left({}^2{C}_2\times {}^3{C}_1\right)\\ & =\left({}^2{C}_1\times {}^3{C}_1\right)+\left(1\times {}^3{C}_1\right)\\ & =\left(2\times 3\right)+\left(1\times 3\right)\\ & =\left(6+3\right)\\ & =9\end{array}$$
$$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{9}{10}$$