100 people comprised of 75 women and 25 men have applied for promotions at local postal distribution facility. only 6 of these 100 can be promoted. also, since each of these 100 people are equally qualified to be promoted, management has decided to implement a lottery system so that each person has a equal chance to be promoted. (a) compute the probability that no woman is promoted. simplify completely - you’ll need a calculator. (b) compute the probability that exactly 1 woman is promoted. simplify completely - you’ll need a calculator. (c) if you observed at most one woman being promoted, would you suspect foul-play by management? why?

(a) The probability of no women being promoted is 17%.

(b) The probability of only one women being promoted is 6%.

(c) The probability of having 2 women or more in the promotions is 32%, so it is not suspicious.

Step-by-step explanation:

This is a case of combination (no replacement and order doesn't matter).

(a) The probability that no woman is promoted is equal to the probability of a man being choose for every promotion:

The probability of no women being promoted is 17%.

(b) The probability of only one woman being promoted is

The probability of only one women being promoted is 6%.

(c) The expected number of women in the 6 promotions can be calculated as .

This expected value, as approximated by a binomial distribution with p=0.25 (chances of picking a woman) and n=6, is

If we compute the z-value for 2 woman, and approximating by the central limit theorem, we can calculate the probability of this event.

Then P(z>-.47)=0.32 or 32%, what can be interpreted as the probability of having 2 or more women in the promotions.

We can conclude that it is not suspicious to have 2 women selected for promotions.

f(x) = y

f(-3) = -5

step-by-step explanation:

the answer is 99.