If P(B)=p
P(Ac|B)=q
P(Ac∩Bc)=r
Find P(A∩Bc)
P(A)
and
P(B|A)
Please and THANKS. and also the steps? ive been trying this forever!
You would be well advised to draw a diagram with A and B: what you know is $\displaystyle P(B)=p$, $\displaystyle P(A^c\cap B^c)=r$ (this is the points outside A and B) and, using the definition of conditional expectation, $\displaystyle P(A^c\cap B)=qp$ (this is the points inside B but not inside A). Write these numbers on your diagram, and fill in the gaps using the fact that the probability of the whole space is 1. First, you find the probability of $\displaystyle A\cap B$, then that of $\displaystyle A\cap B^c$ (the points of A not in B), then that of A, and you can compute anything you want from that point on.
Or, formally (but draw your sketch first), if I try to answer the question in the order they are asked:
First $\displaystyle P(B^c)=1-P(B)=1-p$.
Then $\displaystyle P(A\cap B^c)=P(B^c)-P(A^c\cap B^c)$ (because $\displaystyle B^c=(B^c\cap A)\cup(B^c\cap A^c)$ where the union is disjoint), which gives you $\displaystyle P(A\cap B^c)=1-p-r$.
Now, you know that $\displaystyle q=P(A^c|B)=\frac{P(A^c\cap B)}{P(B)}=\frac{P(A^c\cap B)}{p}$, so that $\displaystyle P(A^c\cap B)=qp$. By the same trick as before, $\displaystyle P(A\cap B)=P(B)-P(A^c\cap B)=p-qp$. You deduce (the same way) $\displaystyle P(A)=P(A\cap B)+P(A \cap B^c)=p-qp+1-p-r=1-qp-r$.
Finally, $\displaystyle P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{p-qp}{1-qp-r}$.