Does it make sense to use the complement of both events when finding a probability using Bayes' Theorem?
For ex: say that 60% of 911 calls concern domestic disputes, and only 3% of those domestic abuse calls end with an arrest being made. Furthermore suppose that 92% of all police calls don't end in an arrest.
What is the probability that a police call that does not concern a domestic dispute does not result in an arrest.
My attempt at this would be to find P(A'|D'): the probability there's no arrest given that it's not a domestic dispute.
I did that and got 0.368, but I'm not sure if that number makes sense given the scenario.
Best Answer
That kind of question can be answered without using Bayes Rule: A call to 911 can lead be related to domestic violence or not, there is no other posibilities. Also, they are disjoint so they form a partition. Then: $$ P(A^{C}) = P(A^{C} \cap D) + P(A^{C} \cap D^{C}) = P(A^{C} | D) P(D) + P(A^{C} | D^{C}) P(D^{C}) $$ And now solving for $P(A^{C} | D^{C})$ $$ P(A^{C} | D^{C}) = \frac{P(A^{C}) - P(A^{C} | D) P(D)}{P(D^{C})} $$
However, regarding your original question, yes, you can apply Bayes rule to any complement of sets, furthermore, you could apply to any union and intersections of sets too. For example: $$ P(A^{C} | D^{C}) = \frac{P(D^{C} | A^{C}) P(A^{C})}{P(D^{C})} $$ Which using the complement turns out to be the expression above. But also for an intersection for example: $$ P(A \cap B | C) = \frac{P(C | A \cap B) P(A \cap B)}{P(C)} $$