# Probability based on a percentage

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?

• Hint: Converse probability: 1 minus the probability picking no man/5 woman. – callculus Apr 15 at 14:25
• 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? – sdds Apr 15 at 14:36
• @sds No, you just have to divide your result (product) by the number of ways to pick 5 people. See my answer. – callculus Apr 15 at 14:39
• 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. – sdds Apr 15 at 14:47
• 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 $\color{blue}5$. – callculus Apr 15 at 14:52

## 1 Answer

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:

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