Bunuel wrote:

What is the probability that a 4-person committee chosen at random from a group consisting of 6 men, 7 women, and 5 children contains at least 1 woman?

A. 11/102

B. 77/204

C. 77/102

D. 91/102

E. 31/34

The probability of at least one woman on the four-person team is the same as (1 - probability of no women on the team).

Thus, the total probability is \(1 - (\frac{11}{18})(\frac{10}{17})(\frac{9}{16})(\frac{8}{15}) = 1-\frac{11}{102} = \frac{91}{102}\)

The logic behind the calculation is simple:

The probability of selecting a man or child for the first position is \(\frac{(5+6)}{(5+6+7)}=\frac{11}{18}\).

Now, the key is that each time we draw another person, we reduce both the pool we are drawing from--the number of men and children--

and the total population by 1, as there is no replacement. Thus,

The probability of selecting a man or child for the second position is \(\frac{(11-1)}{(18-1)} = \frac{10}{17}\)

The third is \(\frac{(10-1)}{(17-1)} = \frac{9}{16}\), and so on