A binary symmetric channel (BSC) has a transition probability of $\frac{1}{8}$. If the binary transmit symbol $X$ is such that $P(X=0)\:=\:\frac{9}{10}$, then the probability of error for an optimum receiver will be
- $\frac{7}{80}$
- $\frac{63}{80}$
- $\frac{9}{10}$
- $\frac{1}{10}$