Need a pointer on conditional probability of a fair coin toss 
A fair coin is thrown $n$ times. Show that the conditional probability of a head on any specified trial, given a total of $k$ heads over the $n$ trials, is $\frac{k}{n}$ ($k > 0$).

I immediately think of Bernoulli trials, which would result in the following:
$$\binom{n}{k}\cdot\left(\frac{1}{2}\right)^4$$
But I don't see how to fit that in to conditional probability, something like:
$$\mathbb{P}\left[H|K\right]=\frac{\mathbb{P}\left[H\cap K\right]}{\mathbb{P}\left[K\right]}$$
And that's where I'm stuck. K is just the number of heads, so I'm not quite sure how to move forward from here...
Would $\mathbb{P}\left[K\right]=\binom{n}{k}\cdot\left(\frac{1}{2}\right)^4$?

By symmetry, the conditional probability is the same for all trials. By linearity of expectation, the conditional expectation value for the number of heads is the sum over the conditional probabilities of heads for all trials, which is just $n$ times one of these conditional probabilities, since they're all the same. Since the conditional expectation value for the number of heads given $k$ heads is $k$, it follows that the conditional probabilities are $k/n$.
