Calculate Conditional probability 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, $P(A| B^c ) = 0.15 / 0.4 = 0.375$?
Thanks!
 A: $P(A\cup B)=P(A^c)$ 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.
A: 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 $P(A^c \cap B) \neq P(B)$.
