Negation of Bayes' theorem.

Question

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.

Answer 1

I wonder how long I've been doing completely the wrong thing and fudging the result with VERY italic handwriting.

Anyway, using the correct formula:

$$\mathbb{P}[A|B]=\frac{\mathbb{P}[A\text{ and }B]}{\mathbb{P}[B]}$$ It's trivial: $$1-\mathbb{P}[A|B]=1-\frac{\mathbb{P}[A\text{ and }B]}{\mathbb{P}[B]}=\frac{\mathbb{P}[B]-\mathbb{P}[A\text{ and } B]}{\mathbb{P}[B]}=\frac{\mathbb{P}[\text{not }A\text{ and }B]}{\mathbb{P}[B]}=\mathbb{P}[\text{not A|B]}$$

Answer 2

From the definition of conditional probability, $$\begin{align*} \Pr(A|B) =& \frac{\Pr(A\cap B)}{\Pr(B)}\\ =& \frac{\Pr(B|A)\Pr(A)}{\Pr(B|A)\Pr(A) + \Pr(B|A')\Pr(A')}\\ \Pr(A'|B) = 1- \Pr(A|B) =& \frac{\Pr(B|A')\Pr(A')}{\Pr(B|A)\Pr(A) + \Pr(B|A')\Pr(A')}\\ \end{align*}$$