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We have a group of 15 people, 7 men and 8 women.

Randomly selecting 5 people, what's the probability to pick 3 men and 2 women? What's the probability to pick at least 1 man?

I tried solving the first question like this: ${7 \choose 3} = \frac{7!}{(7-3)!3!} = 35 $ and ${8 \choose 2} = \frac{8!}{(8-2)!2!} = 28 $,

so the probability should be $\frac 1{980}$.

But I'm stuck on the second question, how should I proceed?

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  • $\begingroup$ Hint: Converse probability: 1 minus the probability picking no man/5 woman. $\endgroup$ – callculus Apr 15 at 14:25
  • $\begingroup$ Thank you, so if I'm not mistaken the probability to pick 5 women is (8!/[(8-5)!5!]) = 1/56, so the probability to pick at least one man is 55/56? $\endgroup$ – sdds Apr 15 at 14:36
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    $\begingroup$ @sds No, you just have to divide your result (product) by the number of ways to pick 5 people. See my answer. $\endgroup$ – callculus Apr 15 at 14:39
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    $\begingroup$ All right, I think I get it now. So I get to 56/3003 ≈1.86, which I then subtract from 100, getting ≈ 98.13% as the probability to pick at least one man. $\endgroup$ – sdds Apr 15 at 14:47
  • $\begingroup$ Your calculation is right. I have a different rounding. $ 0.98135...\approx 98.14\%$. Here is the rule: If the number you are rounding is followed by $\color{blue}5$, 6, 7, 8, or 9, round the number up. In your case 3 is followed by $\color{blue}5$. $\endgroup$ – callculus Apr 15 at 14:52
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You are right the number of ways to pick 3 men (m) and 2 women (w) are ${7 \choose 3}$ and ${8 \choose 2}$. And the number of ways to pick 5 people without any conditions is ${15 \choose 5}$. Therefore the probability to pick $3$ men (m) and $2$ women (w) is $\frac{{7 \choose 3}\cdot {8 \choose 2}}{{15 \choose 5}}=\frac{140}{429}\approx 32.63\%$

The formula is related to the Hypergeometric distribution.

Hint for the second question:

$P("\text{Picking at least one man}")=1-P("\text{Picking no man (5 woman)}")$

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