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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

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

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