# 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

• Are the three events $A_1,A_2,A_3$ independent? – Parcly Taxel Sep 11 '16 at 16:32
• It does not say, but I guess they are independent. @ParclyTaxel – zeeks Sep 11 '16 at 16:40
• Then you can merely multiply the probabilities together and add: $Z=A_1A_2(1-A_3)+A_1A_3(1-A_2)+A_2A_3(1-A_1)$. – Parcly Taxel Sep 11 '16 at 16:42

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}