Two competing clinics A and B treat one disease. After the treatment round in clinic A, 80% of patients recover. In clinic B, 70% of patients recover after one treatment round. Patients are directed for treatment by an insurance company to either clinic A or B, wherein clinic A is chosen twice as often as clinic B.
What's the probability that a client of the insurance company, who needs treatment will get to clinic A and recover after the treatment round?
I don't think I have solved this task correctly. Could you guide my thoughts to the right direction?). Below is my strange solution
P(R | A)=0.8 (After the treatment round in clinic A, 80% of patients recover)
P(R | B)=0.7 (In clinic B, 70% of patients recover after one treatment round)
P(A) = $\frac 23$ (clinic A is chosen twice as often as clinic B)
P(B) = $\frac 13$ (probability of choosing clinic B )
\begin{align} P(A \,|\, R) = \frac{P(R \,|\, A)\cdot P(A)}{P(A)} \end{align}
$$P(A | R) = \frac {0.8 \cdot 0.66}{0.66}$$
P(A | R) = P(R | A)???
What about this additional task?
Under the same conditions, What's the probability that a client of the insurance company, who needs treatment will get to clinic A and not recover after the treatment round?
Am I supposed to find $P(A \cap \overline R)$?
Then $P(R) = P(R \mid A)\cdot P(A)+P(R \mid B)\cdot P(B) = \frac 8 {15} + \frac 7 {30} = \frac {23} {30}$
$P(\overline R) = 1 - \frac {23} {30} = \frac {7} {30}$
$P(A \cap \overline R) = \frac {2} {3} \cdot \frac {7} {30} = \frac {14} {90} = \frac {7} {45}$