# Thread: Probability of choosing workers

1. ## Probability of choosing workers

How do I do this problem its driving me insane!!!

There are 200 applicants for 5 positions. Company narrows it down to 17 candidates. Ten are men. Seven are women. 5 workers are chosen at random from the 17. Find:

four of the five hired are women
only two of the males are hired
at least one one of those hired is male
exactly three of the males are hired

thank you so much I can not figure this out for my life!

2. Originally Posted by skeltonjoe
How do I do this problem its driving me insane!!!

There are 200 applicants for 5 positions. Company narrows it down to 17 candidates. Ten are men. Seven are women. 5 workers are chosen at random from the 17. Find:

four of the five hired are women
only two of the males are hired
at least one one of those hired is male
exactly three of the males are hired

thank you so much I can not figure this out for my life!
1
4 women

There are $\displaystyle \binom{7}{4}$ ways to choose 4 women from 7,
$\displaystyle \binom{10}{1}$ ways to choose a man to go with each female group, $\displaystyle \binom{17}{5}$ ways to choose 5 people from 17.

Hence P(4W) = $\displaystyle \frac{\binom{7}{4}\binom{10}{1}}{\binom{17}{5}}$

2
2 men

P(2M) = $\displaystyle \frac{\binom{10}{2}\binom{7}{3}}{\binom{17}{5}}$

3
at least 1 man

This is the probability of no men subtracted from 1,
since all probabilities sum to 1.

P(at least 1 man) = 1-P(no men)=1-P(all women)= $\displaystyle 1-\frac{\binom{7}{5}}{\binom{17}{5}}$

4
3 men

P(3M) = $\displaystyle \frac{\binom{10}{3}\binom{7}{2}}{\binom{17}{5}}$