A magnetic tape storing information in binary form has been corrupted,so it can only be read with bit errors. The probability that you correctly detect a 0 is 0.9, while the probability that you correctly detect a 1 is 0.85. Each digit is a 1 or a 0 with equal probability. Given that you read a 1, what is the probability that this is a correct reading?
I define $C=$ {correct reading}, so $P(zero|C)=.9$ and $P(one|C)=.8$. Also, $P(one)=P(zero)=.5$
Using Bayes Rule, $P(C|one)=\frac{P(one|C)P(C)}{P(one)}$.
My problem is that I can't figure out what $P(C)$ is. I would appreciate if someone could guide me on the right direction to finish the problem.
Don't make things so complicated.
Think in terms of "actual" and "read".
$\displaystyle p(Actual\ One) = p(Actual\ Zero) = 0.5$
$\displaystyle p(Read \ One |Actual\ One) = 0.85$
$\displaystyle p(Read \ One |Actual\ Zero) = 1-0.9 = 0.1$
By Bayes' Theorem,
$\displaystyle p(Actual \ One|Read \ One) =\frac{p(Read \ One |Actual\ One) \cdot p(Actual\ One) }{p(Read \ One |Actual\ One) \cdot p(Actual\ One) + p(Read \ One |Actual\ Zero) \cdot p(Actual\ Zero) } = \frac{(0.85)(0.5)}{(0.85)(0.5)+(0.1)(0.5)} \approx 0.895 $