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Suppose I have two events $B$ and $A^c$ and I wish to compute the probability of their intersection. I just want to ensure that the following proof holds (i.e., is correct -- I'm a little rusty). Updated! Assume the events are independent.

\begin{gather*} P(A^c) = 1- P(A) \\ \end{gather*} so \begin{align*} P(B \cap A^c) &= P(B) \times \Big(1-P(A)\Big) \\ &= P(B) - P(B)P(A) \\ &= P(B) - P(B \cap A) \end{align*}

Another way to look at it, $A$ and $B$ are two events in some sample space $\mathcal{F}$, i.e., $A,B \in \mathcal{F}$. This means:

\begin{align*} B = (A\cap B) \cup (B\cap A^c) \end{align*}


\begin{align*} P(B) &= P(A\cap B) + P(B \cap A^c) \\ P(B \cap A^c) &= P(B) - P(A \cap B) \end{align*}

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Which probabilities are given to start with? – Zen Jun 24 '14 at 16:26
up vote 4 down vote accepted

You can not write $A^c = 1-A$ because these are events, not probabilities. So to prove that $P(B\cap A^c)=P(B)-P(B\cap A)$ you'll have to use different tactics. One of them is graphical - Venn diagrams. For analytical solution you can go the following way:

Recall that for two not mutually exclusive events we have:

$$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$


$$P(B \cap A^c) = P(A^c) + P(B) - P(A^c \cup B)$$

so you need to show that $P(A^c)-P(A^c \cup B)=P(A \cap B)$. This is true when you think about it, but can you show it why it is the case? Say, assume $x \in P(A \cap B)$. Could you show that $x$ also belongs to $P(A^c)-P(A^c \cup B)$? And if you assume that $x$ belongs to $P(A^c)-P(A^c \cup B)$. Could you show that it also belongs $P(A \cap B)$? Then you would be done. This could be achieved if you think about these events as sets.

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Thanks Sarunas! I've updated my question to reflect that I'm working with probabilities. What do you think now? – user13317 Jun 24 '14 at 17:09
now it seems coherent! good luck :) – Sarunas Jun 25 '14 at 11:48

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