I got a tricky math conditional probability task: Find $$P(\overline{B} \mid \overline{A})$$ Having $$P(A) = a = 0.47$$ $$P(B) = b = 0.24$$ $$P(A \cap B) = c = 0.11$$
Solution: The thing is, that by definition the conditional probability of A given B is:$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$ Thus, moving into the calculation, it's not hard to swap variables and: $$P(\overline{B} \mid \overline{A}) = \frac{P(\overline{B} \cap \overline{A})}{P(\overline{A})}$$
The thing is, that we already got an input in the task: $$P(A \cap B) = c = 0.11$$
However, $$P(A \cap B) = P(A) * P(B) = 0.47*0.24 = 0.1128$$And this number is not rounded, what means, that by calculating the result by using only P(A) and P(B) variables would be different than implementing $P(A \cap B) = c$ variable.
My complete solution: Using Bayes' theorem, we can find the following $P(\overline{B} \mid \overline{A}) = \frac{P(\overline{B} \cap \overline{A})}{P(\overline{A})}$
$P(\overline{A}) = 1 - P(A) = 1 - 0.47 = 0.53$
Then $P(A \cap B) = P(B \mid A) * P(A)$
Therefore: $P(B \mid A) = \frac {P(A \cap B)}{P(A)} = \frac {0.11}{0.47} = 0.234$
Now we need to find $P(\overline{B} \cap \overline{A})$ by using a rule: $P(\overline{B} \cap \overline{A}) = 1 - P(A \cup B)$
Where $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.47 + 0.24 - 0.11 = 0.6$
Then: $P(\overline{B} \cap \overline{A}) = 1 - P(A \cup B) = 1 - 0.6 = 0.4$
Finally: $P(\overline{B} \mid \overline{A}) = \frac{0.4}{0.53} = 0.7547...$
Is my solution reasonable? The tricky thing is the dependency between the events A and B