(3) The Human Resources department of a big company is trying to fill 3 vacant p

Question

(3) The Human Resources department of a big company is trying to fill 3 vacant positions by randomly picking 3 individuals (all at once) from among a pool of 10 equally qualified candidates, 6 of whom are men and 4 are women. What are the chances that, among the 3 selected individuals, there are more women than men? If the 3 positions being filled are those of general manager, regional manager and assistant regional manager, what are the chances that a female candidate gets the general manager's position?

Explanation / Answer

there are 10 equally qualified candidates 6 of whom are men and 4 are women

there are 3 vacant positions

so 3 candidates are to be selected from 10 candidates in 10C3 ways which is the all possible ways of selecting the candidates

now there would be more women than men if 3 women from 4 women and 0 men from 6 men are selected in 6C0*4C3 ways

or if 2 women from 4 women and 1 man from 6 men are selected in 6C1*4C2 ways

so the required probability is   (6C1*4C2+6C0*4C3)/10C3=(36+4)/120=1/3

now if 3 women are selected then the general manager's position to be filled by a woman is a sure event

if 2 women and 1 man is selected then the general manager's position to be filled by a woman in 2C1/3C1 ways

so the required probability is (6C0*4C3*1+6C1*4C2*2/3)/10C3=28/120=7/30

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