Suppose we have binary channel from which we are able to receive zeroes and ones. We also know apriory probability $p$ of receiving "1". Then we can calculate information amount of each digit $q$ we receive:

* q=0*: $I=-\log_2(1-p)$ bits;

* q=1*: $I=-\log_2(p)$ bits.

Now imagine that the channel is "fussy": instead of receiving exact digits we receiving probability $q$ that transferred digit is "1". Previous example of "unfussy" channel is when $q$ can take only two values: $q\in\{0,1\}$.

What will be the amount of information of receiving probability $q$ given probability $p$?

Thanks!