I am asked to find $P(A^c | B^c)$ and I am given $P(A | B)$ = $\frac{1}{2}$ and $P(B | A) = \frac{1}{3}$.
I've tried to solve it but unfortunately, I couldn't complete it.
Here are my steps:
$P(A^c | B^c)$ = $\frac {1-P(A \bigcup B)}{1 - P(B)}$.
$P(A|B) = \frac{P(A \bigcap B)}{P(B)}$.
$\frac{1}{2} = \frac{P(A \bigcap B)}{P(B)}$.
$P(A \bigcap B) = \frac{1}{2} \times P(B)$
$P(B|A) = \frac{P(A \bigcap B)}{P(A)}$.
$P(A \bigcap B) = \frac{1}{3} \times P(A)$.
Then $\frac{P(A)}{P(B)} = \frac{3}{2}$.
$P(A) = \frac{3}{2} P(B)$.
$P(A \bigcup B) = \frac{3}{2} P(B) + P(B) - \frac{1}{2} P(B)$
then
$\frac {1-P(A \bigcup B)}{1 - P(B)} = \frac{1-2P(B)}{1 - P(B)}$, and here's my last step and I couldn't complete.
Update I misread the question statement
$A$ and $B$ are considering applying for a job. The probability that $A$ applies for the job is the probability that $A$ applies for the job given that $B$ applies for the job is $\frac{1}{2}$, and the probability that $B$ applies for the job given that $A$ applies for the job is $\frac{1}{3}$. What is the probability that $A$ doesn't apply for the job given that $B$ doesn't apply for the job?
I would starting with noticing that $\mathbb{P}(A) = 3\mathbb{P}(A\cap B)$ and $\mathbb{P}(B) = 2\mathbb{P}(A\cap B)$.
Based on such results, one arrives at the result: \begin{align*} \mathbb{P}(A^{c}|B^{c}) & = \frac{\mathbb{P}(A^{c}\cap B^{c})}{\mathbb{P}(B^{c})}\\\\ & = \frac{1 - \mathbb{P}(A\cup B)}{1 - \mathbb{P}(B)}\\\\ & = \frac{1 - \mathbb{P}(A) - \mathbb{P}(B) + \mathbb{P}(A\cap B)}{1 - \mathbb{P}(B)}\\\\ & = \frac{1 - 4\mathbb{P}(A\cap B)}{1 - 2\mathbb{P}(A\cap B)} \end{align*}
I guess this is as far as one can go based on the information provided since there are three unknowns and two constraints.
Hopefully this helps!