# How to find $P(A^c | B^c)$ given that $P(A | B)$ and $P(B | A)$?

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?

• It is not possible to find $P(A^{c}|B^{c})$ from the given information. Whatever you have done is correct. Jun 23, 2022 at 23:28
• @geetha290krm That's right, I misread the statement, $P(A) = P(A | B)$, my bad :( Jun 23, 2022 at 23:36

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*}