Given $P(A) = .4, P(A\mid B) = .5$, and $P(A\mid B') = .2$, find $P(B\mid A)$

Question

I have $P(A\mid B) = P(A \cap B) / P(B) = .5$, and $P(A\mid B') = P(A \cap B') / P(B') = .2$

I am trying to find $P(B|A) = P(B \cap A) / P(A) = P(B \cap A) /.4$ but I cannot find $P(B \cap A)$.

I got $P(B \cap A) = P(B) \times .5$, and thus $P(B)\times.5 / .4$ gets $P(B)\times 0.2$. Just manipulation of the information but cannot find anything.

If anyone can help, I will be extremely appreciative and it will help a bunch!

Thanks!!

Answer 1

By the Theorem of Total Probability: $\mathsf P(A) = \mathsf P(A\mid B)\;\mathsf P(B) + \mathsf P(A\mid B')\;\mathsf P(B') \\ \therefore 0.4 = 0.5\; \mathsf P(B) + 0.2\; (1-\mathsf P(B))$

Solve for $\mathsf P(B)$ then use: $\mathsf P(B\mid A) = \mathsf P(A\mid B) \mathsf P(B) / \mathsf P(A)$

Answer 2

Let $x=P(A \cap B)$ and $y=P(A \cap B^c)$. Then from $P(A | B)=0.5$ and $P(A | B^c)=0.2$, we get $P(B)=x/0.5$ and $P(B^c)=5/0.2$. But $P(B)+P(B^c)=1$. Thus we get $$\frac{x}{0.5}+\frac{y}{0.2}=1.$$ Moreover $A=(A \cap B) \cup (A \cap B^c)$, thus we also get $P(A)=x+y$. Furthermore $$x+y=0.4.$$ Now solve for $x$ and $y$ and you will get your answer for $P(B | A)=x/P(A)=x/0.4$.

Answer 3

For these kinds of problems and getting a feel for what is going on, you might find a Venn diagram helps.

This then gives:

$P(A) = x + y = 2/5$

$P(B) = y+z$

$w+x+y+z = 1$

$P(A|B) = y/(y+z) = 1/2$

$P(A|B') = x/(x+w) = 1/5$

We can solve all these easily determine $w,x,y,z$ and then:

$P(B|A) = z/(y+z)$