Suppose two coins $C_1$ and $C_2$ are tossed and suppose further that $C_1$ was tossed $40$ times and $C_2$ only $20$ times. The outcome for $C_1$ is $8$ times heads and $32$ times tails. For $C_2$ we have once head and $19$ times tails. My question is, what test do I use in order to try to show that the probabilities for having heads are significantly different for $C_1$ and $C_2$?
In addition I'd be interested if anyone could explain the following confusion to me. If I estimate the probability for heads of $C_1$ by $p_1=1/5$ then the probability of the outcome of $C_2$ under the hypothesis that the probability is $1/5$ as well equals $$ \binom{20}{1}\frac{1}{5}\left(\frac{4}{5}\right)^{19}\approx 5.7\% $$ which is not significant (I would like to have a significance level of $<5\%$). On the other hand If I reverse the situation I obtain for the probability of the outcome of $C_1$ assuming a probability for heads of $p_2=1/20$ $$ \binom{40}{8}\frac{1}{20^8}\left(\frac{19}{20}\right)^{32}\approx 0.06\% $$ which is by far smaller than $5\%$. So one test tells me the probabilities are significantly different, while the other doesn't (this is the source of confusion) but in fact I would just like to know how I can tell that $p_1\ne p_2$ significantly.
