Let $A_1$, $A_2$, $A_3$ be three events. Find the probability of the event $Z$ that exactly two of these events will occur. So, my solution so far is: 
Event $Z = A_1A_2$\ $A_3$ $ + $ $A_2A_3$\ $A_1$ $ + $ $A_1A_3$\ $A_2$
$P(Z) = P(A_1 \cap A_2$ \ $A_3 ) \cup P(A_2 \cap A_3$ \ $A_1) \cup P(A_1 \cap A_3$ \ $A_2)$ 
I am stuck here. I can not seem to progress to the part where I reduce $3P(A_1\cap A_2\cap A_3)$
Thank you
 A: You are right that 
$$
Z= (A_1 \cap A_2 \cap A_3^c)\cup(A_1 \cap A_2^c \cap A_3)\cup(A_1^c \cap A_2 \cap A_3)
$$
Since all above three events are disjoint, there is
$$
P(Z)=P(A_1 \cap A_2 \cap A_3^c)+P(A_1 \cap A_2^c \cap A_3)+P(A_1^c \cap A_2 \cap A_3)
$$
Also we can have another result. The event that exactly one of these events will occur is
$$
Y= (A_1 \cap A_2^c \cap A_3^c)\cup(A_1 \cap A_2^c \cap A_3^c)\cup(A_1^c \cap A_2 \cap A_3^c)
$$
The event that exactly three of these events will occur is
$$
X=A_1 \cap A_2 \cap A_3
$$
Since all above three events are disjoint and 
$$
X\cup Y \cup Z=A_1 \cup A_2 \cup A_3
$$
There is 
\begin{align}
Z&=(A_1 \cup A_2 \cup A_3)-(X\cup Y)
\\
&=((A_1 \cap (A_2 \cup A_3))\cup(A_2 \cap (A_1 \cup A_3))\cup(A_3 \cap (A_1 \cup A_2)))-(A_1 \cap A_2 \cap A_3)
\end{align}
Since 
$$
A_1 \cap A_2 \cap A_3\subset A_1 \cap (A_2 \cup A_3)
$$
$$
A_1 \cap A_2 \cap A_3\subset A_2 \cap (A_1 \cup A_3)
$$
$$A_1 \cap A_2 \cap A_3\subset A_3\cap (A_1 \cup A_2)
$$
We have
\begin{align}
P(Z)&=P((A_1 \cap (A_2 \cup A_3))\cup(A_2 \cap (A_1 \cup A_3))\cup(A_3 \cap (A_1 \cup A_2)))-P(A_1 \cap A_2 \cap A_3)
\end{align}
