What is the probability that the next toss of that coin is also a heads? 
A jar has 1000 coins, of which 999 are fair and 1 is double headed. Pick a coin at random, and toss it 10 times. Given that you see 10 heads, what is the probability that the next toss of that coin is also a heads?

I personally do not really like this question, since it does not state whether when we pick a random coin is with replacement or not. If if it is with replacement then the fact that we flipped 10 heads is not relevant here. Thus we would only need to figure out the probability that we flip a head which would be
$$\frac{1}{1000}\times 1 + \frac{999}{1000}\times 0.5$$
The answer is suppose to be  $.753$.
Note if you answer this question please refrain from long verbal explanations just use concise notation where you define events.
 A: Here is some notation for an approach. Let:
$X_n$ denotes the result of flip $n$. Note by nature of the problem, the $X_i$'s are independent of one another.
$A$ be the event that the coin picked was a fair coin.
$B$ be the event that the coin picked was the double-heads coin.
We are asked $$P(X_{11} = H\mid X_{10}, X_9,\dots X_1 = H) = \frac{P(X_1, X_2, X_3, \dots X_{11} = H)}{P(X_1, X_2, X_3, \dots X_{10} = H)}$$
See that $A$ and $B$ are complementary events. Therefore, $P(X_1,X_2, \dots X_n = H) = P(X_1,X_2, \dots X_n = H\mid A) P(A) + P( X_1,X_2, \dots X_n = H\mid B) P(B)$
Use this identity to arrive at the answer.
A: 
I personally do not really like this question, since it does not state whether when we pick a random coin is with replacement or not.

No, it does and it is neither of those.


Pick a coin at random, and toss it 10 times.


It specified that only a single coin selection is ever made; and that this coin is then flipped multiple times.   So the results of each successive flip is conditionally independent of the others, when given which type of coin was selected to flip (fair or double-headed).
  Let $E$ be the event of the evidence (10 heads shown), $F$ be the event of selecting a fair coin, and $N$ be the event that the next toss of that coin is a head.  You can now determine:
$$\mathsf P(F), \mathsf P(F^{\small\complement}),\\\mathsf P(E\mid F),\mathsf P(E\mid F^{\small\complement}),\\\mathsf P(N\cap E\mid F),\mathsf P(N\cap E\mid F^{\small\complement})$$
Then the solution is thus an application of the law of total probability and Bayes' Rule.

$$\begin{align}\mathsf P(N\mid E)&=\dfrac{\mathsf P(N\cap E)}{\mathsf P(E)}\\[2ex]&=\dfrac{\mathsf P(N\cap E\mid F)\,\mathsf P(F)+\mathsf P(N\cap E\mid F^{\small\complement})\,\mathsf P(F^{\small\complement})}{\mathsf P(E\mid F)\,\mathsf P(F)+\mathsf P(E\mid F^{\small\complement})\,\mathsf P(F^{\small\complement})}\end{align}$$

