# How to prove P(A|B)>P(B)>P(A|complement of B)

In context, $A$ refers to the event of getting a positive in a cancer test and $B$ is the event of actually having cancer. i.e. Prove: True positive rate $\geq$positive rate $\geq$false positive rate

question b: Prove: For a test in which the false positive rate is $1$ in $n$, the posterior probability of $B$ given a positive test is at most a factor of $n$ times larger than the prior probability (incidence) of $A$.

question c: Given $P(B)\sim 0$ and $P(A|B)\sim 1$ prove $P(B|A)\sim P(B)/P(A/B')$ Note: $B'$ is complement of $B$.

• For (b) and (c) you will probably want to use $$P(A)=P(A \cap B) +P(A \cap B') = P(A\mid B)P(B) + P(A\mid B^\prime)P(B^\prime)$$ – Henry Sep 17 '15 at 6:07
• The first question is impossible without some explicit probabilities given. Since $\mathbb{P}( A \mid B ) > \mathbb{P}(A \mid B^C )$ cannot hold in general, as this would imply $\mathbb{P}( A \mid B^C) > \mathbb{P}( A \mid (B^C)^C ) = \mathbb{P}(A \mid B) > \mathbb{P}(A \mid B^C)$. – Hetebrij Sep 17 '15 at 7:17