Among the N = 18 advisory members of an education program, A = 12 are university
ID: 3063981 • Letter: A
Question
Among the N = 18 advisory members of an education program, A = 12 are university faculty members and the other N - A = 6 are business managers. A random sample of n = 6 advisory members is going to be selected. Use X to denote the number of advisory members who are university faculty members. Keep at least 4 decimal digits if the result has more decimal digits.
17. The probability that exactly 5 advisory members are university faculty members is closest to
(A) 0.0498 (B) 0.2370 (C) 0.2560 (D) 0.3058 (E) 0.6942
18. The probability that all 6 advisory members are university faculty members is closest to
(A) 0.0498 (B) 0.2370 (C) 0.2560 (D) 0.3058 (E) 0.6942
19. The probability that at least 5 advisory members are university faculty members is closest to
(A) 0.0498 (B) 0.2370 (C) 0.2560 (D) 0.3058 (E) 0.6942
20. The probability that at most 4 advisory members are university faculty members is closest to
(A) 0.0498 (B) 0.2370 (C) 0.2560 (D) 0.3058 (E) 0.6942
The more details the better. I really dont understand this question.
Explanation / Answer
Total faculty members =12
Business managers = 6
X to denote the number of advisory members who are university faculty members
17. The probability that exactly 5 advisory members are university faculty members= P(X=5) = 12C5 * 6C1 / 18C6 = 0.2559793
So ans = (C) 0.2560
18. The probability that all 6 advisory members are university faculty members = P(X=6) = 12C6/18C6 =0.04977376
So ans = (A) 0.0498
19. The probability that at least 5 advisory members are university faculty members = P(X>=5)
= P(X=5) + P(X=6) = 0.2560 + 0.0498 = 0.3058
So ans = (D) 0.3058
20. The probability that at most 4 advisory members are university faculty members = P(X<=4) = 1-P(X>=5) = 1- [P(X=5)+P(X=6)] = 1-(0.2560 + 0.0498) =0.6942
So ans = (E) 0.6942
PS: nCr = n! / r! * (n - r)!
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