You pick 4 marbles from a bag of 5 red and 5 blue, with no replacement. What is the probability of getting exactly 2 marbles of each color?
I think it's either 3/8 or 10/21 but I'm not sure which.
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$\begingroup$I like to add another perspective to this as well. If we instead didn't care about the order. For example, RRBB and RBRB are indistinguishable. So what are the number of possible 4 marbles I can get? It can be all blue, all red, or in between. It is
\begin{align*} {10 \choose 4} \end{align*}
Now, we want to know how many ways I can get $2$ Red marbles with no regards to ordering. It is
\begin{align*} {5 \choose 2} \end{align*}
Same thing goes for the blue marbles as well. So ${5 \choose 2}{5 \choose 2}$ gives you the number of $2$ red and $2$ blue marbles you can get without worrying about ordering. So the answer is
\begin{align*} \frac{{5 \choose 2}{5 \choose 2}}{10 \choose 4} \end{align*}
$\endgroup$ $\begingroup$In simple cases, it is easy to visualize picking up balls by drawing a tree diagram. I'll skip that part.
To pick a red marble first, you have a probability $\frac{5}{10}$. To pick another red marble, now you will have probability $\frac{4}{9}$ (since no replacement).
To pick a blue marble next, you have probability $\frac{5}{8}$, and for another $\frac{4}{7}$.
Note that this is not the only way to pick up the required balls, you can have other arrangements as $RRBB,RBRB,BRBR,BBRR,BRRB,RBBR$, in total $\binom{4}{2}=6$ ways.
So your final probablity is $$6.\frac{5}{10}.\frac{4}{9}.\frac{5}{8}.\frac{4}{7}$$
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