A selection committee consisting of 10 members is to be formed from a group of employees at a firm. 20 of this employees, 10 are female.
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We need to eliminate the probabilities of the whole committee consisting of women, and of there being only one man on the committee. There are equal numbers of men and women in the employees.

Probability of all women: (10/20)(9/19)(8/18)...(3/13)(2/12)(1/11)=(10!/20!)×10!=1/184756.

Probability of one man: (10/20)(10/19)(9/18)(8/17)...(2/11)=(1/2)(10!/19!)×10!=5/92378.

We need to multiply this by 10, because there are 10 permutations of the man within the committee: 25/46189.

When we subtract the sum of these from 1 we are left with the probability of at most 8 women on the committee.

1-101/184756=184655/184756=0.99945333 approx. That’s 99.95%.

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