Probabilities involving $2018$ biased coins Consider a group of $2018$ biased coins such that for the $n^{th}$ coin the 
probability $P_{heads} = \frac{1}{p_n}$ where $p_n$ is the $n^{th}$ prime number. Thus the first coin will be unbiased while the second coin will give head $\frac{1}{3}$ times and so on.....

First Question

You toss the $2018$ coins one by one (starting with the first coin and ending with the last) and record the number of head you get. Let the total number of heads to get be equal to $H$. Find the probability $P$ such that $H$ is even.



Second Question

Now the first coin (which is unbiased) is removed and another coin is added at the end. So now the $n^{th}$ coin will have $P_{heads} = \frac{1}{p_{n+1}}$. 
Solve the first question again for this case.



Third Question

Does it really matter whether there are $2018$ coins? I believe that the answers should not depend on the total number of coins. Am I correct?



This is not homework. A friend asked me this question. I have a feeling that the answer to both questions must be $\frac{1}{2}$ but I am unable to prove it. Can someone help me?
(Expanding on comments)

Q1 Use induction (I got the idea from your Q3 whose answer is affirmative for Q1)

$$P(\sum_{n=1}^{k+1} 1_{H_n} \ \text{is even})$$

$$= P(\sum_{n=1}^{k} 1_{H_n} + 1_{H_{k+1}} \ \text{is even})$$

$$= P(\sum_{n=1}^{k} 1_{H_n} \ \text{is even} \cap 1_{H_{k+1}} = 0) + P(\sum_{n=1}^{k} 1_{H_n} \ \text{is odd} \cap 1_{H_{k+1}} = 1)$$

$$= P(\sum_{n=1}^{k} 1_{H_n} \ \text{is even}) P(1_{H_{k+1}} = 0) + P(\sum_{n=1}^{k} 1_{H_n} \ \text{is odd}) P(1_{H_{k+1}} = 1) \tag{independent?}$$

$$= \frac12 \frac1{p_{k+1}} + \frac12 (1-\frac1{p_{k+1}}) = \frac12 \tag{inductive hypothesis}$$

Now, $H = \sum_{n=1}^{2018} 1_{H_n}$. As it turns out, not only is 2018 irrelevant but also is $p_k$ irrelevant: as long as it is not $0$. It doesn't matter if it's prime, composite or 1.

Q2 

(Edit: wait they're not independent.  LOL. I can try again next time I guess. Or someone may edit.)

$$P(\sum_{n=1}^{2019} 1_{H_n} - 1_{H_1} \ \text{is even})$$

$$ = P(\sum_{n=1}^{2019} 1_{H_n} \ \text{is even})P(H_1^C) + P(\sum_{n=1}^{2019} 1_{H_n} \ \text{is odd})P(H_1) \tag{again, independent?}$$

$$ = \frac12 \frac12 + \frac12 (1-\frac12) \tag{by Q1}$$

$$ = \frac12$$
If $p_k$ is the probability to get heads on the $k$th throw and $P_k$ is the probability that the number of heads after $k$ throws is even, we have a recurrence relation
$$P_k = (1 - p_k) P_{k-1} + p_k (1 - P_{k-1}), \quad P_0 = 1.$$
If $p_k$ is the reciprocal of the $(k + 1)$st prime, $P_{2018}$ is a huge fraction approximately equal to $.50435$. The limit $P_\infty$ appears to be $1/2$. For other sequences the limit will be different.