I have two events $A$ and $B$ satisfying: \begin{align} P(A) = 0.25, P(B) = 0.6, P(A \cup B) = 0.75 \end{align}

I have calculated:

$$P(A^c) = 0.75, P(A \cap B) = 0.1.$$

I now need $P(A^c | B)$. Is it right to use the standard conditional probability formula and assume that:

$P(A^c \cap B) = P(B) = 0.6$

and calculate $P (A^c \cap B) / P(B) = 1$. This seems strange.

Also the same way, $P(A| B^c ) = 0.15 / 0.4 = 0.375$?


  • $\begingroup$ We have $\Pr(A^c\cap B)=0.5$. Now you can use the defining formula for conditional probability. $\endgroup$ – André Nicolas Dec 2 '15 at 18:56
  • $\begingroup$ any feedback for the answers? $\endgroup$ – Carlos Mendoza Dec 2 '15 at 23:57
  • $\begingroup$ Hi, Yes I think I understood , I made a mistake in my initial question. thanks $\endgroup$ – user290335 Dec 3 '15 at 0:24

$P(A\cup B)=P(A^c)$ thus $P(A^c\cap B^c)=P(A)=0.25$ thus $P(A^c\cap B)=0.5$.

So no, it's not right.

To complete the answer, $P(A^c\mid B)=\frac{P(A^c\cap B)}{P(B)}=\frac{0.5}{0.6}\approx0.833\dots$ and $P(A\mid B^c)=\frac{P(A\cap B^c)}{P(B^c)}=\frac{0.15}{0.4}=0.375$ as you said.


Let me start by suggesting another approach that avoids the direct computation of the probabilities of an intersection of events that are the complement of $A$ and/or $B$. For the first probability is easy:

\begin{align} P(A^c \mid B) &= 1 - P(A \mid B)\\ &=1 - \frac{P(A \cap B)}{P(B)} \end{align}

For the second probability it's a little trickier, and may be it's not worth from a practical point of view, but it is nice anyway to see De Morgan's law in action!

\begin{align} P(A \mid B^c) &= 1- P(A^c \mid B^c)\\ &= 1-\frac{P(A^c \cap B^c)}{P(B^c)}\\ &= 1-\frac{P((A \cup B)^c)}{P(B^c)}\\ &= 1-\frac{1-P(A\cup B)}{P(B^c)} \end{align}

Now, in regards to your reasoning, the total probability theorem tell us that

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

From this we get

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

which show us that indeed $P(A^c \cap B) \neq P(B)$.


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