I'm a little conflicted about a probability question and would like you help.
Suppose we have 5 different coins, with different probabilities for heads: $$p_i =\frac{i-1}{4}\in \{0,0.25,0.5,0.75,1\}$$ Let's define the event $H_i:$ The $i-th$ toss was heads.
There's an even probability to choose either one of the coins, therefore we also define $C_k:$ We chose the $i-th$ coin. $\forall k\in1,...,5: C_k=\frac{1}{5}$.
In order to help myself I calculated $\mathbb{P}(H1)$, using the Law of Total Probability:
$$\mathbb{P}(H_1) = \sum_{i=1}^{5}{\mathbb{P}(H_1\mid C_i)\mathbb{P}(C_i)}=\frac{0.25+0.5+0.75+1}{5}=\frac{1}{2}$$
Now since the second toss is a general event, and the probability for heads depends on which coin we toss, I claim that given $C_i$ we can tell that $H_1$ and $H_2$ are independent, because if I already know which I'm tossing the first toss doesnt give me any more information about the second.
$$\mathbb{P}(H_2\mid H_1) = \frac{\mathbb{P}(H_1 \cap H_2)}{\mathbb{P}(H_1)}=\frac{\sum_{i=1}^{5}{\mathbb{P}(H_1\cap H_2\mid C_i)\mathbb{P}(C_i)}}{0.5}=\frac{0.2 \sum_{i=1}^{5}{p_i^2}}{0.5}$$
When calculated with the given probabilities I get: $$\mathbb{P}(H_2\mid H_1) =0.75$$
A final answer that I saw somewhere points that I should get $\frac{3}{16}$. I'd like to know if there's a mistake I made along the way, or was the final answer I've seen was incorrect. I could not think of a way to get such result.
Any help or hints would be much appreciated.