Two fake coins of the same weight are on the same box as eight real coins. The weight of each of the fake coins is different as the real ones. We randomly select a pair of coins without replacement out of the ten.
Then we select another pair out of the remaining eight.
What is the probability that the four coins we got are real, knowing that the combined weight of the first pair is the same as the second pair?
I named the following events as:
A = "the four coins are real".
B = "the weight of the first pair is the same as the weight of the second pair"
C = "both pairs containing a fake coin and a real coin"
D = "both pairs contain only real coins"
Of course, $B = C \cup D$.
So, $$P(A |C \cup D)= \frac{P(A \cap (B \cup C))}{P(C \cup D)}=\frac{P(D)}{P(C \cup D)}$$
Now
- $P(D)=P(\text{"I get two real coins on the first pair"})P(\text{"I get two real coins on the second pair"})$
To calculate this, I used a random variable $X \sim \text{hypergeometric}(10, 8, 2)$
So $P(\text{"I get two real coins on the first pair"})=P(X=2)=\frac{8 \choose 2}{10 \choose 2}=\frac{28}{45}$
And $Y \sim \text{hypergeometric}(8, 6, 2)$
So $P(\text{"I get two real coins on the second pair"})=P(X=2)=\frac{6 \choose 2}{8 \choose 2}=\frac{15}{28}$
$\Rightarrow \boxed{ P(D)=\frac{28}{45} \cdot \frac{15}{28} = \frac{1}{3}}$
$P(C \cup D)=P(C)+P(D)-P(C \cap D)$ with $P(C \cap D)=0$
So $\boxed{P(C)=\frac{2}{10} \cdot \frac{8}{9} \cdot \frac{1}{8} \cdot \frac{7}{7}=\frac{1}{45}}$ (First I select a fake coin (so I have a probability of $\frac{2}{10}$), then I select a real one (I have a probability of $\frac{8}{9}$ as I already took a coin), and so on).
Finally,
$$\boxed{P(A | C \cup D) = \frac{1/3}{1/45 + 1/3}=\frac{15}{16}}$$
I don't know where is my mistake, the correct answer is under the spoiler markup.
The correct answer is $\frac{15}{19}$
For $C$ you don't have to pick the coins in the order fake-real-fake-real. There are actually four possible orders of fake and real. So $P(C)=\frac{4}{45}$. This should point you toward the right answer.