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 ^{n}C_{x} = n! / [(n-x)!*x!]

____________________________________________________________________

Q1) (a) Choose 5 out of 11 in ^{11}C_{5} =
**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 ^{8}C_{5} =
**56
ways**

(d) Favourable outcomes = 56

Total Outcomes = 462

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

**___________________________________________________________________________**

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

Therefore Number of groups of 6 = ^{12}C_{6} =
**924**
**ways**

(b) There are 7 women. Therefore different groups of 6 out of 7
women in ^{7}C_{6} = **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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