If $A_{n}$ are disjoint and $P (B\mid A_{n}) \geq c$ for every n, then $P(B\mid \cup A_{n})\geq c$

Question

how can I prove this:

If $A_{n}$ are disjoint, and $P (B\mid A_{n}) \geq c$ for every n, then $P(B\mid | \cup A_{n})\geq c$.

Any help guys?

Answer 1

Hint: $$P(B \mid \bigcup_n A_n) = \frac{P(B \cap \bigcup_n A_n)}{P(\bigcup_n A_n)} = \frac{\sum_n P(B \mid A_n) P(A_n)}{\sum_n P(A_n)}.$$ Can you finish the rest?

$$\frac{\sum_n P(B \mid A_n) P(A_n)}{\sum_n P(A_n)} \ge \frac{c \sum_n P(A_n)}{\sum_n P(A_n)} = c.$$