Is there a way to proof to prove this for independence using the method of contradiction ?

Let $A_1, A_2,$ and $A_3$ be events, and let $B_i$ represent either $A_i$ or its complement $A^c _i$. Then there are eight possible choices for the triple $(B_1,B_2,B_3)$. Prove that the events $A_1, A_2, A_3$ are independent if and only if $P(B_1 \cap B_2 \cap B_3) = P(B_1)P(B_2)P(B_3) $, for all eight of the possible choices for the triple $(B_1, B_2, B_3)$.

The first thing I thought of is contradiction but I guess writing out $8 $ equations and showing that if they are independent, then $A_1, A_2, A_3$ must also be independent is also a good way. Is there any other way to prove this ?

Thanks in advance.


1 Answer 1


The following shows the "if" part.

You need to show $p(A_1\cap A_2\cap A_3)=p(A_1)p(A_2)p(A_3)$ and $p(A_i\cap A_j)=p(A_i)p(A_j)$ for all $i\not=j$. That's the definition of three events being independent, you have to show all four things.

The first one is basically given, just let $B_i=A_i$ for all $i$.

Now consider $p(A_1\cap A_2)$. Note that $A_1\cap A_2=A_1\cap A_2\cap(A_3\cup A_3^c)$. So

$p(A_1\cap A_2)=p(A_1\cap A_2\cap(A_3\cup A_3^c))$

$=p((A_1\cap A_2\cap A_3)\cup(A_1\cap A_2\cap A_3^c))$ (because intersection distributes over union)

$=p(A_1)p(A_2)p(A_3)+p(A_1)p(A_2)p(A_3^c)$ (by assumption)



This shows the "if" part, you should now be able to do the converse.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .