Question

one professor grades homework by randomly choosing 5 out of 11homework problems to grade.

a.) how many different groups of 5 problems can be chosen fromthe 11 problems?

b.) probability extension: jerry did only 5 problems of one assignment. what is the probability that the problems he didcomprised the group that was selected to be graded?

c.) silvia did 8 problems. how many different groups of 5 did she complete?

d.) what is the probability that one of the groups of 5 she completed comprised the group selected to be graded?

The qualified applicant pool for 6 management trainee positions consists of 7 women and 5 men.

(a) How many different groups of applicants can be selected for the positions?


(b) How many different groups of trainees would consist entirely of women?


(c) Probability extension: If the applicants are equally qualified and the trainee positions are selected by drawing the names at random so that all groups of 6 are equally likely, what is the probability that the trainee class will consist entirely of women? (Round your answer to 3 decimal places.)

Answer 1

P(Event) = Number of favourable outcomes/Total Number of outcomes

Please note ⁿC_x = n! / [(n-x)!*x!]

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Q1) (a) Choose 5 out of 11 in ¹¹C₅ = 462 ways

(b) Favourable outcomes = 1 (the set of 5 chosen by the professor, and that done by jerry are the same)

Total Outcomes = 462

Therefore the required probability = 1/462 = 0.002

(c) Choose 5 out of 8 in ⁸C₅ = 56 ways

(d) Favourable outcomes = 56

Total Outcomes = 462

Therefore the required probability = 56/462 = 4/33 = 0.121

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Q2) (a) Total = 7 W + 5M = 12

Therefore Number of groups of 6 = ¹²C₆ = 924 ways

(b) There are 7 women. Therefore different groups of 6 out of 7 women in ⁷C₆ = 7 ways

(c) Favourable outcomes = 7 different groups of 6 consisting of only women.

Total outcomes = 924 groups of 6.

Therefore the required probability = 7/924 = 1/132 = 0.008

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