we can solve this problem by a direct method which is more time consuming.

i.e finding the no of ways that at least one women is selected.

However an easier approach is to find the no of ways in which only men are selected for both the position and reducing this probabilty by 1.

No. of ways in which only men can be selected for both the position is

\(3 * 2 = 6\) since there are \(3\) men and same person cannot hold both the position.

total no of possible outcome would be \(10 * 9 = 90\) This is because men + women = 10

therefore probability = \(\frac{6}{90} = \frac{2}{30} = \frac{1}{15}\)

Hence probability that at least a women is selected = \(\frac{15}{15} - \frac{1}{15} = \frac{14}{15}\)

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This is my response to the question and may be incorrect. Feel free to rectify any mistakes