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Let $A, B$ and $C$ be three events, such that $P(A∩C)$ and $P(B∩C)$ are both strictly between $0$ and $1$ and let $B^c$ denote the complement of $B$. Prove that

$P(B|A∩C) = P(A|B∩C)P(B|C)/P(A|C)$

Express $P(A|C)$ in terms of $P(A|B∩C)$, $P(A|B^c∩C)$ and $P(B|C)$.

Any help and explanation would be greatly appreciated.

Many thanks in advance.

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\begin{align} P(A|B∩C)P(B|C)/P(A|C)&=\dfrac{P(A\cap B\cap C)}{P(B\cap C)}\cdot\dfrac{P(B\cap C)}{P(C)}\cdot\dfrac{P(C)}{P(A\cap C)} \\ &=\dfrac{P(A\cap B\cap C)}{P(A\cap C)} \\ &=P(B|A\cap C) \end{align}

\begin{align} P(A|C)&=\dfrac{P(A\cap C)}{P(C)} \\ &=\dfrac{P(A\cap B\cap C)+P(A\cap B^c\cap C)}{P(C)} \\ &=\dfrac{P(A\cap B\cap C)}{P(B\cap C)}\cdot\dfrac{P(B\cap C)}{P(C)}+\dfrac{P(A\cap B^c\cap C)}{P(B^c\cap C)}\cdot\dfrac{P(B^c\cap C)}{P(C)} \\ &=P(A|B∩C)\cdot P(B|C)+P(A|B^c∩C)\cdot\dfrac{P(C)-P(B\cap C)}{P(C)} \\ &=P(A|B∩C)\cdot P(B|C)+P(A|B^c∩C)\cdot(1-P(B|C)) \end{align}

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  • $\begingroup$ Thanks! Have you any idea how I should approach the second part? Also, from the above statement "P(A∩C) and P(B∩C) are both strictly between 0 and 1" does this mean event A and C are independent of each other given there's no explicit relationship? $\endgroup$ – Devesh Jun 7 '15 at 22:23
  • $\begingroup$ $P(A∩C)$ and $P(B∩C)$ are both strictly between $0$ and $1$ means that they are $\ne0$. $\endgroup$ – Math Wizard Jun 7 '15 at 22:49
  • $\begingroup$ There is no need to assume independence of events for it is not used. Only thing needed are $P(A∩C), P(B∩C), P(B^c∩C), P(C)\ne0$. $\endgroup$ – Math Wizard Jun 7 '15 at 23:29
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You want to prove:

$$ \Pr(B|A \cap C) = \frac{\Pr(A|B\cap C) \Pr(B|C)}{\Pr(A|C)} $$

For $\Pr(C) \neq 0$, by definition for a probability space $(S, \mathcal{A}, P$) with $P(A) \neq 0$ and for every $B \in \mathcal{A}$ you get conditional probability of $B$ given $A$: $$ \begin{align} \Pr(B|A) &\equiv \frac{\Pr(B \cap A)}{\Pr(A)}\\ \end{align} $$

For independent $A, B$, by definition we have: $\Pr(A \cap B) = \Pr(A) \cdot \Pr(B)$. Then we can condition on C: $$ \begin{align} \Pr(B|AC) &= \frac{\Pr(AB | C)}{\Pr(A|C)}\\ \Pr(B|A\cap C) &= \frac{\Pr(A \cap B | C)}{\Pr(A|C)}\\ &= \frac{\Pr(A|B\cap C) \Pr(B|C)}{\Pr(A|C)} \end{align} $$

Answering the comment (and also consulting this answer): $$ \begin{align} \Pr(A|B^\prime \cap C) &= \frac{\Pr(A \cap B^\prime \cap C)}{\Pr(B^\prime \cap C)} \\ &= \frac{\overbrace{(1 - \Pr(B))}^{\Pr(B^\prime)}\Pr(C|B^\prime)\Pr(A|B^\prime \cap C)}{\underbrace{(1-\Pr(B))}_{\Pr(B^\prime)}\Pr(C|B^\prime)} \\ &= \frac{(1 - \Pr(B))\Pr(C|B^\prime)\Pr(A|B^\prime \cap C)}{(1-\Pr(B))\Pr(C|B^\prime)} \end{align} $$

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  • $\begingroup$ So we can assume independence i.e. Pr(A∩B)=Pr(A)⋅Pr(B) from the fact P(A ∩ B) was not defined to be strictly between 0 and 1? $\endgroup$ – Devesh Jun 7 '15 at 22:32
  • $\begingroup$ No. We have Independence if and only if $\Pr(A \cap B) = \Pr(A)\Pr(B)$: en.wikipedia.org/wiki/… where there is also a numerical example in the section "Pairwise and mutual independence". As a proof I like more the proof of math craze than mine. :) Also if you have time during the summer, I urge you to go further and work with probability using "probability density functions". It will help you a lot in the future! $\endgroup$ – Konstantinos Jun 7 '15 at 22:36
  • $\begingroup$ Hey thanks! genuinely appreciate the follow up! Have any ideas as to how I can complete the second half to express P(A|C) in terms of P(A|B∩C), P(A|B'∩C) and P(B|C)? $\endgroup$ – Devesh Jun 7 '15 at 22:38
  • $\begingroup$ I think you can derive this very easily using math craze's first equality, using the fact that $P(B) = 1 - P(B^\prime)$. $\endgroup$ – Konstantinos Jun 7 '15 at 22:46
  • $\begingroup$ P(A|Bc∩C) is not equivalent to 1 - P(A|B∩C) I'm assuming? $\endgroup$ – Devesh Jun 7 '15 at 22:56

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