conditional probability disjoint events

Question

If $D_i$ are disjoint events and $P(C|D_i) = p$ for any i, then :

How to prove: $ P(C|\bigcup_{i}D_i) = p$?

Answer 1

By Bayes' Theorem: $$P(C|\bigcup_iD_{i}) = \frac{P(C\cap(\bigcup_iD_{i}))}{P(\bigcup_iD_i)}$$ and distributing gives $$= \frac{P(\bigcup_i({CD}_{i}))}{P(\bigcup D_i)}$$ and since $D_i$'s disjoint, ${CD}_i$'s disjoint, so probability of the unions is sum of the probabilities. Hence $$= \frac{\sum_iP(C\cap D_i)}{\sum_i P(D_i)}$$ and using Bayes' Theorem on numerator gives $$= \frac{\sum_iP(C|D_i)P(D_i)}{\sum_i P(D_i)}$$ But $P(C|D_i)=p$, a constant, by assumption so we have $$=\frac{\sum_ipP(D_i)}{\sum_i P(D_i)} = \frac{p\sum_iP(D_i)}{\sum_i P(D_i)} = p$$

An intuitive way to view this is that since $D_i$'s are disjoint, $P(C|\bigcup_i D_i)$, the probability that C occurs given that any of $D_i$'s occur, is really just the probability that C occurs given that a particular $D_i$ occurs (only one can occur as they are disjoint, so the union having occurred means that exactly one occurred). But regardless of which $D_i$ occurred, the probability that $C$ now occurs is p, since $P(C|D_i)=p$ for any $i$.