How to find this probability? 
In a company, it's known that $45 \%$ of the employees have a graduate (university), a half are not graduate and make less than $\$1500$ per month.Those who are graduate, $\frac{2}{9}$ make less than $\$1500$

What is the probability of one employee not be graduate, knowing that makes more or equal than $\$1500$?
Fist I set two events:


*

*$A$ - "To be graduate"

*$B$ - "To make less than $\$1500$"
I know that the question is about $P(\bar{A}|\bar{B})$.From the introdution I know that: 
$P(A)=0{,}45$
$P(\bar{A}\cap B)=0{,}5$
$P(B|A)=\frac{2}{9}$
But I don't know how to find the probability asked.Can you explain me how to do it?
 A: With a problem like this, a straightforward approach is to enumerate the possibilities:
$$\begin{aligned}
P(A \cap B) &= 0.10 \\
P(A \cap \bar B) &= 0.35 \\
P(\bar A \cap B) &= 0.50 \\
P(\bar A \cap \bar B) &= 0.05 \\
\end{aligned}$$
Now we can calculate:
$$P(\bar A \,|\, \bar B)
 = \frac{P(\bar A \cap \bar B)}{P(\bar B)}
 = \frac{P(\bar A \cap \bar B)}{P(\bar A \cap \bar B) + P(A \cap \bar B)}
 = \frac{0.05}{0.05 + 0.35} = \frac{1}{8} = 12.5\%
$$

Ps. How did I calculate the four base probabilities above?  Well...


*

*$P(\bar A \cap B)$ we already know

*$P(\bar A \cap \bar B) = 1 - P(A) - P(\bar A \cap B)$

*$P(A \cap B) = P(A) \; P(B|A)$

*and $P(A \cap \bar B)$ is what's left over.

A: $$P(\bar A \,|\, \bar B)
 = \frac{P(\bar A \cap \bar B)}{P(\bar B)}$$
 $$=  \tag{1}\frac{1-P( A \cup B)}{1-P( B) }$$
 $$= \frac{0.05}{1-0.6} = \frac{1}{8} $$
$(1)$:


*

*$P(A \cap B)= P(B|A)P(A)=\frac{2}{9} \cdot 0.45=0.1$

*$P(\bar A \cap \bar B) = 1 - P(A) - P(\bar A \cap B)=1-0.45-0.5=0.05$


Then  we have $P(B)$ from $P(A \cup B)=P(A)+P(B)-P(A \cap B)$ 
