math
posted by CARMEN .
i have no clue how to do this problem. any help would be appreciated!! :
Three hundred people apply for three jobs. 90 of the applicants are women.
(a) If three persons are selected at random, what is the probability that all are women? (Round the answer to six decimal places.)
(b) If three persons are selected at random, what is the probability that two are women? (Round the answer to six decimal places.)
(c) If three persons are selected at random, what is the probability that one is a woman? (Round the answer to six decimal places.)
(d) If three persons are selected at random, what is the probability that none is a woman? (Round the answer to six decimal places.)
(e) If you were an applicant, and the three selected people were not of your gender, should the above probabilities have an impact on your situation? Why?
Yes, the probabilities indicate the presence of gender discrimination.
No, because in the hiring process all outcomes are not equally likely.

so we have 210 men and 90 women,
Must choose 3
a) (90/300)(89/299)(88/298) = .026370
or
C(90,3)/C(300,3) = .026370
b) so you want permutations of WWM
= 3(90/300)(89/299)(210/298) = .188784
or
C(90,2)C(210,1)/C(300,3) = .188784
c) short way:
C(90,1)C(210,2)/C(300,3) = .....
d) none is woman > all 3 men
C(210,3)/C(300,3) = ...
e) your call.