Q:

Hypergeometric distribution excercise

There are 3 red and 4 blue balls in a box. Someone picks two balls from this box randomly. 
What is the probability that if we pick 2 balls from the remaining 5 balls, we pick exactly 2 red.
I calculated the probabilites when we have 1,2 and 3 red balls remaining, then I added these up. Is that correct? I got $P=0,4$ but it seems a lot to me.

A:

Imagine that the balls all have ID numbers. Now imagine taking a ball from the box, then another, then another, and so on, until we have taken out all seven. All sequences of balls are equally likely. 
The probability the third and fourth balls are red is the same as the probability that the first and second are red, that is, $\frac{3}{7}\cdot\frac{2}{6}$.
Remark: As an exercise, I suggest also doing this problem the harder way. Alicia picks first. Calculate the probabilities $p_0$, $p_1$, $p_2$ that she gets $0$ red, $1$ red, $2$ red.
Beti picks next. Calculate the conditional probabilities $x$, $y$, $z$, she got $2$ given that Alicia $0$ red, $1$ red, $2$ red respectively ($z$ is $0$).
Then our required probability is $p_0x+p_1y+p_2z$. Calculate. After a while, you will get the same number as the one we got using the simple calculation that we used to find the answer. 

