In a hospital, patients are treated by doctors $A, B,$ and $C$. Doctor $A$ treats $50\%$ of the patients, doctor $B$ treats $20\%$ and doctor $C$ treats $30\%$. $3\%$ of doctor $A'$s diagnoses are false, $2\%$ of doctor $B'$s diagnoses are false and $5\%$ of doctor $C'$s diagnoses are false.
To improve the diagnosis process, doctor $B$ has recently been double-checking the work of his colleagues $A$ and $C$ and finding $80\%$ of all their false diagnoses and correcting them. The diagnoses of doctor $B$ remain unchecked. What is the probability of being misdiagnosed at this hospital?
Here are my thoughts:
Let $A$ be the event of being treated by doctor $A$, $B$ by doctor $B$, $C$ by doctor $C$ and $F$ the event of being misdiagnosed before doctor $B$ started double-checking the other doctors' work. It follows:
$P(A) = 0.5$, $P(B) = 0.2$, $P(C) = 0.3$.
$P(F) = P(A) \cdot P(F | A) + P(B)\cdot P(F | B) + P(C) \cdot P(F | C) = 0.5 \cdot 0.03 + 0.2 \cdot 0.02 + 0.3 \cdot 0.05 = 0.034.$
Let $F'$ be the event of being misdiagnosed after $B$ started double-checking the other doctors' work. Then:
$P(F') = (P(A) \cdot P(F | A) + P(C) \cdot P(F | C)) \cdot 0.8 + P(B)\cdot P(F | B) = (0.5 \cdot 0.03 + 0.3 \cdot 0.05) \cdot 0.8 + 0.2 \cdot 0.02 = 0.028.$
Which is as expected less than $P(F).$
However, my textbook gives $0.01$ as the solution. Where's my mistake?
Since we're looking for the probability of being misdiagnosed:
$$P(F') = (P(A) \cdot P(F | A) + P(C) \cdot P(F | C)) \cdot 0.\color{red}{2} + P(B)\cdot P(F | B) = 0.01.$$