# Simple Random Sampling

• Dec 6th 2009, 01:27 AM
Dredito
Simple Random Sampling
If you have a population of size n and a sample of size m what is the probability that two values x and y from the population both appear in the sample? So far, I have that each member of the population has probability m/n of being in the sample. Therefore I would guess that the probability that both x and y appear in the sample is (m/n)*(m/n)=m^2/n^2. However the answers show that the probability is (m^2)/n(n-1) but I don't understand why?
• Dec 8th 2009, 04:06 PM
awkward
Quote:

Originally Posted by Dredito
If you have a population of size n and a sample of size m what is the probability that two values x and y from the population both appear in the sample? So far, I have that each member of the population has probability m/n of being in the sample. Therefore I would guess that the probability that both x and y appear in the sample is (m/n)*(m/n)=m^2/n^2. However the answers show that the probability is (m^2)/n(n-1) but I don't understand why?

I'm not sure I understand your problem, so let me state a problem which I think is equivalent. You have a bowl containing n balls, 2 of which are red, the remainder white. You draw a sample of m balls without replacement.

What is the probability that both red balls are in the sample?
$\frac{\binom{2}{2} \binom{n-2}{m-2}}{\binom{n}{m}} = \frac{m(m-1)}{n(n-1)}$