# Negation of Bayes' theorem.

This is self-learning.

This is very hard to find, there are examples with numbers but none with ven diagrams. This is not homework, I'm studying Markov Chains and have little confidence with conditional probability.

We all know:

$$\mathbb{P}[A|B]=\frac{\mathbb{P}[A\ \text{and}\ B]}{\mathbb{P}[B|A]}$$ I wish to start with: $$1-\mathbb{P}[A|B]$$ and get to a result.

I have tried this but I keep going in circles (getting back to what I started with) and I'm not sure what the right result actually is. I believe:

$$1-\mathbb{P}[A|B]=\mathbb{P}[\text{not }A|B]$$

Using ven-diagrams, or common sense, P(not A and B) (I take not to have high precedence than and, so this is (not A) and B, not (pardon the pun) not (A and B).

Anyway P(not A and B) = P(B) - P(A and B) =P(B)-P(A|B)P(B|A) which feels like the closest I have got.

• But...$$\Pr(A|B) = \frac{\Pr(A\cap B)}{\Pr(B)}$$ Nov 27, 2013 at 15:05
• @peterwhy that explains a lot. What on earth was I thinking of? Nov 27, 2013 at 15:07
• @peterwhy thank you, you have just changed my life. For something like 3 years I've been doing stuff totally wrong and hated conditional probability because I could never get it right, you have fixed all that. Thank you. Nov 27, 2013 at 15:18
• I hope the dubious downvotes are not back. Sep 24, 2015 at 23:39

$$\mathbb{P}[A|B]=\frac{\mathbb{P}[A\text{ and }B]}{\mathbb{P}[B]}$$ It's trivial: $$1-\mathbb{P}[A|B]=1-\frac{\mathbb{P}[A\text{ and }B]}{\mathbb{P}[B]}=\frac{\mathbb{P}[B]-\mathbb{P}[A\text{ and } B]}{\mathbb{P}[B]}=\frac{\mathbb{P}[\text{not }A\text{ and }B]}{\mathbb{P}[B]}=\mathbb{P}[\text{not A|B]}$$
From the definition of conditional probability, \begin{align*} \Pr(A|B) =& \frac{\Pr(A\cap B)}{\Pr(B)}\\ =& \frac{\Pr(B|A)\Pr(A)}{\Pr(B|A)\Pr(A) + \Pr(B|A')\Pr(A')}\\ \Pr(A'|B) = 1- \Pr(A|B) =& \frac{\Pr(B|A')\Pr(A')}{\Pr(B|A)\Pr(A) + \Pr(B|A')\Pr(A')}\\ \end{align*}
• The negation of Bayes' theorem, which is $$\Pr(A|B) =\frac{\Pr(B|A)\Pr(A)}{\Pr(B)}$$ Nov 29, 2013 at 5:34