The problem is as follows:
You have a box of 6 coins.
2 of the coins are weighted so that they come up heads with probability 0.1. (Category 1)
3 are weighted so that it comes up heads with probability 0.6. (Category 2)
1 is weighted so that they come up heads with probability 0.7. (Category 3)
Suppose you take a coin out of the box and flip it and it is heads. What is the probability that the second flip of the same coin is also heads?
The way I approached this was, let $X_{1}, X_{2}$ be random variables representing the first and second flip. Let $C$ be the random variable representing the category that the coin is from. Then
$$ P(X_{2}=H|X_{1}=H)=P(X_{2}=H, C=1 | X_{1}=H) + P(X_{2}=H, C=2 | X_{1}=H) + P(X_{2}=H, C=3 | X_{1}=H). $$
Now for each of the summands on the right, I said something along the lines of
$$ P(X_{2}=H, C=1 | X_{1}=H) = \frac{P(X_{2}=H, C=1, X_{1}=H)}{P(X_{1}=H)} = \frac{P(X_{2}=H, X_{1}=H | C=1)P(C=1)}{P(X_{1}=H)}=\frac{P(X_{2}=H|C=1)P( X_{1}=H | C=1)P(C=1)}{P(X_{1}=H)}. $$
That last part is what makes me think that I am doing this wrong. Is it true that
$$ P(X_{2}=H, X_{1}=H | C=1) = P(X_{2}=H|C=1)P( X_{1}=H | C=1) $$
when $X_{1}, X_{2}$ are not independent? If they are independent then wouldn't I just say that $P(X_{2}=H|X_{1}=H)=P(X_{2}=H)=P(X_{1}=H)$? If they are not independent, then how do I go about solving this?
I want to argue that flipping the same coin a second time is independent of the first time but I am not certain.
I think you'd be better off conditioning on which category the first coin came from. If we know that the first coin was category 1, say then we know that 2 coins of category 1, 2 of category 2, and 1 of category 3 remain. Now it's easy to compute the probability that the second coin comes up heads under these conditions.