hundred people apply for three jobs. 110 of the applicants are women.
(a) If three persons are selected at random, what is the probability that two are women? (Round the answer to six decimal places.)
3C2(110/400)^2(290/400)=0.164484 wrong answer
To find the probability that two out of three selected applicants are women, we need to consider both the number of ways to choose two women from the pool of 110 women and the number of ways to choose one applicant from the remaining 290 applicants who are not women.
To calculate this, we can use the combination formula, also known as "nCr," where n represents the total number of applicants and r represents the number of applicants we want to choose. In this case, n = 400 (total number of applicants) and r = 3 (number of applicants to be selected).
The number of ways to choose two women from 110 women (n1Cr1) is: 110C2 = (110! / (2! * (110-2)!))
Similarly, the number of ways to choose one applicant from the remaining 290 applicants who are not women (n2Cr2) is: 290C1 = (290! / (1! * (290-1)!))
Now, we multiply these two combinations together and divide by the total number of ways to choose three applicants from the entire pool of 400 applicants (nCr): 400C3 = (400! / (3! * (400-3)!))
So, the probability of selecting two women out of three randomly chosen applicants is:
P(two women) = (110C2 * 290C1) / 400C3
Now, let's calculate the value:
110C2 = (110! / (2! * (110-2)!)) = (110! / (2! * 108!)) = ((110 * 109) / (2 * 1)) = 5,945
290C1 = (290! / (1! * (290-1)!)) = (290! / (1! * 289!)) = 290
400C3 = (400! / (3! * (400-3)!)) = (400! / (3! * 397!)) = 91,200
Probability = (5,945 * 290) / 91,200 = 0.189750
Rounded to six decimal places, the probability that two out of three randomly chosen applicants are women is 0.189750.