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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.

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    $\begingroup$ But...$$\Pr(A|B) = \frac{\Pr(A\cap B)}{\Pr(B)}$$ $\endgroup$
    – peterwhy
    Nov 27, 2013 at 15:05
  • $\begingroup$ @peterwhy that explains a lot. What on earth was I thinking of? $\endgroup$
    – Alec Teal
    Nov 27, 2013 at 15:07
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    $\begingroup$ @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. $\endgroup$
    – Alec Teal
    Nov 27, 2013 at 15:18
  • $\begingroup$ I hope the dubious downvotes are not back. $\endgroup$
    – ToolPurger
    Sep 24, 2015 at 23:39

2 Answers 2

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

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

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  • $\begingroup$ Why did you answer this, I don't get what you're trying to show. $\endgroup$
    – Alec Teal
    Nov 28, 2013 at 21:10
  • $\begingroup$ The negation of Bayes' theorem, which is $$\Pr(A|B) =\frac{\Pr(B|A)\Pr(A)}{\Pr(B)}$$ $\endgroup$
    – peterwhy
    Nov 29, 2013 at 5:34

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