Thread: Set Theory: find P(A)

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'

Your basic problem is that the given information, p(AuB)= 0.7 and p(AuB')= 0.9, is impossible. Since B and B' are disjoint, AuB and AuB' are disjoint so p(A)= p(AuB)+ p(AuB') but that sum is greater than 1.

Originally Posted by HallsofIvy

Your basic problem is that the given information, p(AuB)= 0.7 and p(AuB')= 0.9, is impossible. Since B and B' are disjoint, AuB and AuB' are disjoint so p(A)= p(AuB)+ p(AuB') but that sum is greater than 1.

I think you mean AnB and AnB' are disjoint?? And P(A)=P(AnB)+P(AnB') ??

I find these easier to do using a Venn diagram. See attachment.

Originally Posted by math951

P(AuB) = .7
P(AuB')=.9
Find P(A)

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*}$