Calculate the probability of sampling the same number of real and fake gold coins Julien is invited to open one of two treasure chests. He is twice as likely to open treasure chest $A$ than treasure chest $B$.
In treasure chest $A$, there are $15$ gold coins. However, $60\%$ of them are fake. In treasure chest $B$, there are $8$ gold coins. However, $25\%$ of them are fake. After selecting a treasure chest to open, Julien randomly samples four coins.
Calculate the probability that Julien samples an equal number of real and fake gold coins from the random sample of four.
I tried to do a binomial for each
$P(\text{ choosing real gold coins from A)}={15\choose 2}(.4)^2(.6)^{13}$
$P(\text{ choosing real gold coins from B)}={8\choose 2}(.75)^2(.25)^6$
And the fakes are just the probabilities switched so like $(.4)^{13}(.6)^2$ for A, now the probability of choosing $A$ is $2\over 3$ and $B$ is $1\over 3$ so here is where I get stuck. I'm not sure if I should do $$P(A)P(2 \text{ real from A) $P($$2$ fake from A})+P(B)P(2 \text{ real from B) $P($$2$ fake from B})$$
or not.
 A: So in chest A there are 6 reals and 9 fakes, and in chest B there are 6 reals and 2 fakes.
Let $R$ be the number of reals sampled, and $F$ be the number of fakes sampled.
Presumably, you sample without replacement since it (it doesn't make sense to put the coins back in the chest). So the probabilities are not constant among the draws and therefore you cannot model $R$ and $F$ as binomial random variables. Instead, you model them as hypergeometric random variables.
So let $A$ be the event that he chooses chest A and let $B$ be the event that he chooses chest B. Notice that if we draw $R = 2$, then $F = 2$ since we draw $n = 4$ coins and there are only two kinds of coins. Then
\begin{align*}
P(R = 2) &= P(A, R = 2) +P(B, R = 2)\\
&=P(R = 2|A)P(A)+P(R=2|B)P(B)\\
&=\frac{\binom{6}{2}\binom{9}{2}}{\binom{15}{4}}\cdot\frac{2}{3}+\frac{\binom{6}{2}\binom{2}{2}}{\binom{8}{4}}\cdot \frac{1}{3}\\
&=\frac{61}{182}\\
&\approx 0.335.
\end{align*}

If for some strange reason you sample with replacement, then yes you can model $R$ and $F$ as binomial random variables. However, notice again that in four draws, if $R=2$, then automatically $F = 2$ since there are only two kinds of coins. So you only have to find the chance that $R=2$ in each case. Then
\begin{align*}
P(R = 2) &= P(A, R = 2) +P(B, R = 2)\\
&=P(R = 2|A)P(A)+P(R=2|B)P(B)\\
&=\binom{4}{2}(.40)^2(1-.40)^{4-2}\cdot \frac{2}{3}+\binom{4}{2}(.75)^2(1-.75)^{4-2}\cdot \frac{1}{3}\\
&\approx 0.301.
\end{align*}
