ABC26GN5379 · Probability

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