0
$\begingroup$

Given two events $A$ and $B$ such that $P(A)=P(B)=P(A \mid B^c)=\frac{1}{3}$, are $A$ and $B$ independent?

I'm having trouble figuring out how I can manipulate these equations to see if they're independent. Of course, I just need to check to see if $P(A \cap B)=\frac{1}{9}$ but most of the formulas I can think of involving the intersection of two events assumes independence. I was thinking about how $P(A)=P(A \mid B^c)$ implies that $A$ and $B^c$ are independent, but that isn't enough to make the leap.

$\endgroup$
3
$\begingroup$

$P(A\cap B)=P(A)-P(A\cap B^{c})=P(A)- P(A|B^{c}) (1-P(B))=\frac 1 3 -\frac 2 9=\frac 1 9=P(A)P(B)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.