Solving probability problem with given information that doesn't add up I hope you all are doing okay.
I have a quick question regarding a probability problem that I've been struggling to solve.
This is what we were given $$\Pr(A^c) = 2\Pr(A)\\ \Pr(A|B^c)=.2 \\ 2\Pr(B|A) = .2 $$
So I've started working on it.  First, trying to find $\Pr(A)$: $$1-\Pr(A) = 2\Pr(A) \\ 1 = 3\Pr(A) \\ \frac{1}{3} = \Pr(A)$$.
Then, trying to find $Pr(A\cap B)$: $$\Pr(B|A)=.1 \\ \frac{\Pr (A \cap B)}{\Pr(A)} = .1 \\ \Pr(A \cap B) = \frac{1}{30}$$
I only need $\Pr(B)$ now. Let's use infromation from $2\Pr(B|A)$: $$2\Pr(B|A) = .2 \\ 2[1-\Pr(B^c|A)] = .2 \\ 1-\Pr(B^c|A) = .1 \\ 1- \frac{\Pr(B^c \cap A)}{\Pr(A)} = .1 \\ 1 - \frac{\Pr(B^c \cap A}{\frac{1}{3}} = .1 \\ \Pr(B^c \cap A) = .3$$
Thus,  since $\Pr(A|B^c)=.2$, and $\Pr(A|B^c) = \frac{\Pr(A \cap B^c)}{\Pr(B^c)}$, we observe: $$ .2 = \frac{.3}{\Pr(B^c)} \\ \Pr(B^c) = \frac{.3}{.2} = 1.5$$
Which isn't right at all.  I can't really figure out what I'm doing wrong here and I would absolutely appreciate your help.
 A: Your calculations are correct. The problem is that the conditions are inconsistent, leading to a nonsensical result.
Another way to compute $P(B^c)$ is to rearrange the second two conditions:
$$
P(A\cap B^c) = .2 P(B^c)
$$
$$
P(B\cap A) = .1 P(A)
$$
Add the two equations, using $P(A) = P(A\cap B) + P(A\cap B^c)$:
$$
P(A) = .2 P(B^c) + .1 P(A)
$$
Given $P(A)=\frac13$, we solve for $P(B^c)$ and obtain $1.5$ again.
A: You may indeed obtain absurd results while using valid reasoning but inconsistent facts.
These facts are inconsistent.

Let us take $\Pr(A^\complement)=2\Pr(A)$ and $2\Pr(B\mid A)=0.2$ as given, and derive $\Pr(A\mid B^\complement)$ to compare.
We have $\Pr(A)=1/3$ as you obtained, $\Pr(B\mid A)=1/10$ as is clear, and so $\Pr(A\cap B)=1/30$ by definition: $\Pr(A\cap B)=\Pr(A)\Pr(B\mid A)$
Likewise by definition, $\Pr (B\mid A)=\Pr(A\cap B)\div\Pr(B)$.
Therefore: $~\Pr(B)~{=\Pr(A\cap B)\div\Pr(B\mid A)\\=\tfrac 1{30}\div\tfrac 1{10}\\=\tfrac 13}\\\Pr(B^\complement)=\tfrac 23$

By Law of Total Probability: $\Pr(A\cap B^\complement)~{=\Pr(A)-\Pr(A\cap B)\\=\tfrac 13-\tfrac 1{30}\\=\tfrac{3}{10}}$

Therefore $\Pr(A\mid B^\complement)~{=\dfrac{3/10}{2/3}\\=1/20\\=0.05}$
Not $0.2$ as claimed.  Therefore the three facts, which you were provided, are inconsistent.

You will obtain different results for $\Pr(B^\complement)$ depending on which two of the three 'facts' you choose to derive it.
