Probability of Blood Type VS Probability of Choosing from a Class Scenario 1:
There are 12 boys and 8 girls in a class. Two students are chosen at random. Find the probability that both are boys.
We know that we have $\frac{12}{20} \times \frac{11}{19} = \frac{33}{95}$
Scenario 2:
The population in a city is divided as follows: 45% Blood O, 40% Blood B, 11% Blood AB+, and 4% Blood AB-.
Two people are chosen at random. Find the probability that both are Blood O.
Here, the working is $0.45 \times 0.45 = 0.2025$
My question is, why do we, in Scenario 2, consider that choosing a person with a particular blood type as Independent, whereas in Scenario 1, choosing a person with a particular gender is not considered to be Independent events?
Just as Blood A, B, AB and O are independent of each other, should it not be that Boys and Girls are therefore independent as well?
 A: Essentially, because the population of the city is a) very large, so that independence is a good approximation, and b) unknown, so you can't hope to do better than a good approximation.
Independence is sampling with replacement, so you can select the same person twice, whereas actually you are sampling without replacement. The only difference assuming independence makes comes from the possibility that you do choose the same person twice. This has probability $1/N$ if there are $N$ people, which makes a significant difference to the answer in the first case but not the second where we likely have $N\geq 200,000$.
Note also that in any real-life problem of this form the $45\%$, etc, are rounded anyway, and there is therefore an amount of uncertainty in the final answer coming from the precision of those figures which is vastly more significant than the difference between independent selections and selecting without replacement.
A: Scenario 1 is a "sampling from finite population" while Scenario 2 is a sampling from "infinite population"
(imagine to chose two guys at random from a population of 7,594,000,000)
