Bits are sent through an information channel, the probability of incorrectly receiving a $1$ is 0.02, while the probability of incorrectly receiving a 0 is 0.01. What is the probability of receiving a correct bit?
Since both have equal probability of being sent, I thought I need to consider the converse probability of a mistake for each of these, namely $\frac{1}{2}99.9+\frac{1}{2}99,8$. But the solution tells me to use Bayes' theorem. Is my reasoning incorrect?
We do not need to use Bayes' theorem. By the law of total probability, \begin{multline*} P(\text{correct bit})=P(\text{correct bit}\mid\text{$1$ was sent})P(\text{$1$ was sent})\\+P(\text{correct bit}\mid \text{$0$ was sent})P(\text{$0$ was sent}). \end{multline*} Hence, $$ P(\text{correct bit})=0.98\cdot0.5+0.99\cdot0.5=0.985. $$