Boolean algebra probability not coming out right Assuming A,B,C,D are mutually independent.
$P[(A\cup\overline{B}\cup C)\cap(A\cup C \cup \overline{D})]$
I get $(P(A) + 1 - P(B) + P(C))(P(A) + P(C) + 1 - P(D))$
But when I plug in the numbers, I get a result that's larger than $1$, which makes no sense... what am I doing wrong?
 A: If you have taken a course in Boolean algebra, you might remember the result: 
$(x+y)(x+z) = x + yz$.  Similarly, we have that
$$\begin{align*}
F &= (A\cup\overline{B}\cup C)\cap(A\cup C \cup \overline{D})\\
&= ((A\cup C)\cup\overline{B})\cap((A\cup C) \cup \overline{D})\\
&= (A \cup C) \cup (\overline{B}\cap\overline{D}) = G \cup H
\end{align*}$$
where $G= A \cup C$ and $H = \overline{B}\cap\overline{D}$.  (If you have not had Boolean algebra, drawing a Venn diagram will
help you work out this equality).  Now, we can write
$$
\begin{align*}
P(F) &= P(G \cup H) = P(G) + P(H) - P(G\cap H)\\
&= P(A\cup C) + P(\overline{B}\cap\overline{D}) - P((A\cup C)\cap(\overline{B}\cap\overline{D}))
\end{align*}
$$
where 


*

*$P(A \cup C) = P(A) + P(C) - P(A \cap C) = P(A) + P(C) - P(A)P(C)$ because 
$A$ and $C$ are independent events,

*$P(\overline{B}\cap\overline{D}) = P(\overline{B})P(\overline{D})$ because 
$\overline{B}$ and $\overline{D}$ are independent events,

*$P((A\cup C)\cap(\overline{B}\cap\overline{D})) = P(A\cup C)P(\overline{B}\cap\overline{D})$ because 
$A\cup C$ and $\overline{B}\cap\overline{D}$ are independent events as Didier Piau has noted in his answer.
If you don't believe the claim of independence in the last bulleted item, note that 
$A \cup C$ is the union of the mutually exclusive events $A$ and 
$\overline{A} \cap C$ and so
$$
\begin{align*}
P((A\cup C)\cap(\overline{B}\cap\overline{D})) 
&= P((A\cup (\overline{A} \cap C))\cap(\overline{B}\cap\overline{D}))\\
&= P(A\cap\overline{B}\cap\overline{D}) +
P(\overline{A} \cap C \cap \overline{B}\cap\overline{D})\\
&= P(A)P(\overline{B})P(\overline{D}) +
P(\overline{A})P(C)P(\overline{B})P(\overline{D})\\
&= [P(A) +
P(\overline{A})P(C)]P(\overline{B})P(\overline{D})\\
&=[P(A) +
(1 - P(A))P(C)]P(\overline{B})P(\overline{D})\\
&= P(A\cup C)P(\overline{B}\cap\overline{D}) 
\end{align*}
$$
A: The problem is that $P(X \cap Y) \neq P(X) P(Y)$ if $X$ and $Y$ are not independent.  Here, both your $X = A \cup \bar{B} \cup C$ and your $Y = A \cup C \cup \bar{D}$ have $A\cup C$ in them -- they are not independent.
Also $P(X \cup Y) = 1 - P(\bar{X} \cap \bar{Y}) = 1 - (1-P(X)) (1-P(Y)) \neq P(X) + P(Y)$ for $X,Y$ independent.
