# Probability of picking someone who opposes and favors

1. Apr 11, 2012

### MACHO-WIMP

Out of 25 people, 17 favor and 8 oppose something. What is the probability of picking two who favor and four who oppose out of six people?

25C6=177100 to find the # of ways to get 6 people.
Then find those who favor 17C2=136
Then find those who oppose 8C4=70
Multiply that together 136×70=9520
Then divide by total number of ways to pick 6 people 9520/177100=.0537549407
Did I do this right?

2. Apr 11, 2012

### Ray Vickson

Yes, you did just fine. If you want to look at it more generally, this problem involves a standard hypergeometric distribution, defined as follows. If we have a population of N1 things of type 1 and N2 things of type 2, and pick a sample of n things at random without replacement, the probability of getting k type 1 things in the sample is
$$P\{ k \text{ type 1 }\} = \frac{C(N_1,k) C(N_2,n-k)}{C(N_1 + N_2,n)},$$
where C(a,b) is the binomial coefficient
$$C(a,b) = \frac{a!}{b! (a-b)!} = \frac{1(a-1) \cdots (a-b+1)}{b!}.$$
In your case you have N1 = 17, N2 = 8, n = 6 and k = 2, so the probability is
$$p = \frac{C(17,2)C(8,4)}{C(25,6)} = \frac{17\cdot 16}{2 \cdot 1} \cdot \frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1} \cdot \frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{25 \cdot 24 \cdot 23 \cdot 22 \cdot 21} = \frac{68}{1265} \doteq 0.05375 .$$

RGV