P(AuB) = .7
P(AuB')=.9
Find P(A)
So we know everything outside of B is .9
We know A or B =.7
We want to find A.
I am kind of stuck here.
I think I need to know the outside
of A and B, which is P(AuB)' which is = A' intersect B'
While I agree that the use of Venn diagrams, see reply #4, are useful, here is another approach.
We know that $\mathscr{P}(A)=\mathscr{P}(A\cap B)+\mathscr{P}(A\cap B')$
$ \begin{align*}\mathscr{P}(A\cup B)&=\mathscr{P}(A)+\mathscr{P}(B)-\mathscr{P}(A\cap B) \\\mathscr{P}(A\cup B')&=\mathscr{P}(A)+\mathscr{P}(B')-\mathscr{P}(A\cap B')\\\mathscr{P}(A\cup B)+\mathscr{P}(A\cup B')&=2\mathscr{P}(A)+\mathscr{P}(B)+( B')-\mathscr{P}(A\cap B)-\mathscr{P}(A\cap B')\\0.7+0.9&=2\mathscr{P}(A)+1-\mathscr{P}(A)\\\mathscr{P}(A)&=0.6 \end{align*}$