Biased coin hypothesis 
Let's assume, we threw a coin $110$ times and in $85$ tosses it was head. What is the probability that the coin is biased towards head?

We can use chi squared test to test, whether the coin is biased, but using this test we only find out, that the coin is biased towards heads or tails and there seems to be no one tailed chi squared test.
The same problem seems to appear when using z-test approach.
What is the correct way to solve this problem?
 A: Let's assume that you have a fair coin $p=.5$.  You can approximate a binomial distribution with a normal distribution.  In this case we'd use a normal distribution with mean $110p=55$ and standard deviation $\sqrt{110p(1-p)}\approx5.244$.  So getting 85 heads is a $(85-55)/5.244\approx5.72$ standard deviation event.  And looking this value up on a table (if your table goes out that far, lol) you can see that the probability of getting 85 heads or more is $5.3\times10^{-9}$.  An extremely unlikely event.  That is approximately the probability you have a fair coin.
A: I actually think I found a solution, but it is very ugly one. Let's have similar problem:

We have population of $1 000 000$ people. We take $110$ random samples - $85$ females, $25$ males. What is the probability, that there are more females than males?

This means that we need to know, what is the probability, there are $500 001$ females in the population or more. The solution is following:
$$\frac {\sum_{i=1}^{500000} \binom {25} {500000-i}  \binom {85} {500000+i}} {\binom {110} {1000000} }$$
So for the general solution, it would be 
$$\lim_x^\inf \dfrac {\sum_{i=1}^{x/2} \binom {25} {x-i}  \binom {85} {x+i} } {\binom {110} {x} }$$
This is pretty ugly solution and I hope there is better solution to this problem.
