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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

1. $\frac{7}{80}$
2. $\frac{63}{80}$
3. $\frac{9}{10}$
4. $\frac{1}{10}$