Help with rearranging conditional probability

Question

Given $\mathbb{P}(A^c|B^c) = \frac{1}{2}$, $\mathbb{P}(B^c|A^c) = \frac{1}{4}$, $\frac{\mathbb{P}(A)}{\mathbb{P}(B^c)} = \frac{1}{3}$.

Calculate $\mathbb{P}(B)$.

I don't get how to rearrange $\mathbb{P}(B)$ to solve it. Maybe there is some trick I am missing but the furthest I got is still nowhere near a computable solution: $$\mathbb{P}(B) = \frac{\mathbb{P}(B|A^C)\mathbb{P}(A^c)}{1-\mathbb{P}(A|B)}$$

Answer 1

$$ P(A^c|B^c) = \frac{1}{2} \Rightarrow \frac{P(A^c \cap B^c)}{P(B^c)} = \frac{1}{2} \Rightarrow P(A^c \cap B^c) = \frac{P(B^c)}{2} \tag{1} $$

$$ P(B^c|A^c) = \frac{1}{4} \Rightarrow \frac{P(A^c \cap B^c)}{P(A^c)} = \frac{1}{4} \Rightarrow \frac{P(B^c)}{2P(A^c)} = \frac{1}{4} \Rightarrow P(B^c) =\frac{P(A^c)}{2}\tag{2} $$

$$ \frac{P(A)}{P(B^c)} = \frac{1}{3} \Rightarrow P(B^c) = 3P(A) \Rightarrow \frac{P(A^c)}{2} = 3(1 - P(A^c)) \Rightarrow P(A^c) = \frac{6}{7} \tag{3} $$

Therefore from $(2)$ and $(3)$ we get $$ P(B^c) = \frac{3}{7} \Rightarrow P(B) = 1 - P(B^c) \Rightarrow P(B) = \frac{4}{7} $$

Answer 2

The equalities:

• $2P\left(B^{\complement}\right)=4P\left(A^{\complement}\cap B^{\complement}\right)=P\left(A^{\complement}\right)$

• $P\left(B^{\complement}\right)=3P\left(A\right)=3-3P\left(A^{\complement}\right)$

enable you to find $P(B^{\complement})$

Answer 3

$$\frac{P(A^C|B^C)}{P(B^C|A^C)}=\frac{P(A^C)}{P(B^C)}=2$$which gives $P(A^C)=2P(B^C)$ and hence $P(A)=1-2P(B^C)$. You are also given $3P(A)=P(B^C)$. Can you take it from here?