Math probability with job applicants

Question

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.)

Answer 1

90/300 = 9/30= 3/10 = 30% chance of a woman selected once. For all women in 3 times the odds are .3x.3x.3 = 0.027 = 2.7% chance of 3 women

odds of 2 women, exactly, is .3x.3x..7 + .3x7x.3x + .7x.3x3x = 3 times .7(.3)(.3) = 3(.21)(.3) = .9(.21) = 0.189 = 18.9%

odds of 1 women are .3x.7x.7 + .7x.3x.7 + .7x.7x.3 = 3 times .49(.3) = .9(.49) = .441 = 44.1%

odds of no women are .7x.7x.7 = .49(.7) = .343= 34.3%