A three person comitee is chosen at random from a group of 8 women and 3 men. Find probability that the committee contains at least one man?

P(the committee contains no man)
= P(the committee contains 3 women)
= C(8,3) / C(11,3)
= (8!/5!3!) / (11!/8!3!)
= (8!/5!3!) × (8!3!/11!)
= (8×7×6) / (11×10×9)
= 56/165

P(the committee contains at least one man)
= 1 - P(the committee contains no man)
= 1 - (56/165)
= 109/165

For the first one picked, the probability of a man is 3/8
if it's not, the probability the second is a man is 3/7
if it's not, the probability the third is a man is 3/6

add the three together to get 0.375 + 0.4286 + 0.5 = 1.304