Bayes with multiple conditions

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Doctor $B$ and $B^c$ can make 2 types of surgery: type $C$ and $C^c$.

The probability of receiving the surgery from $B$ = $B^c$ = $0.5$

Both Doctors have in their curriculums a total of 100 surgeries.

Doctor B has 90 surgeries of type $C$ of which 70 were successful and 20 failed, and 10 surgeries of type $C^c$ that were all successful.

Doctor $B^c$ has 10 surgeries of type $C$ of which 2 were successful and 8 failed, and 90 surgeries of type $C^c$ that were 81 successful and 9 failed.

What is the probability that the surgery was successful given that it was a type C surgery made by Doctor B?

or, what is $P(A|B,C)$ ?


My approach:

for what I know

$P(A|B,C) = \frac{P(A)P(B,C|A)}{P(B,C)}$ where $P(B,C|A)= P(B|A)*P(C|A)$;

$P(B|A)=\frac{P(B)P(A|B)}{P(A)}$ and $P(C|A)=\frac{P(C)P(A|C)}{P(A)}$

$P(B,C) = P(B)*P(C)$

Calculating:

$P(A) = \frac{163}{200}$

$P(B,C) = 0.45$

$P(B) = 0.5$

$P(A|B) = \frac{7}{9}\frac{9}{10}+\frac{1}{10}=\frac{8}{10}$

$P(C) = \frac{1}{2}\frac{9}{10}+\frac{1}{2}\frac{1}{10}=0.5$

$P(A|C) = \frac{7}{9}+\frac{1}{5}=\frac{44}{45}$

$P(B|A) = \frac{80}{163}$

$P(C|A) = \frac{880}{1467}$

Finally,

$P(B,C|A) \approx0.2944$

$P(A|B,C) \approx 0,5332$

WHat do you think?