# 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?

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probability-theory and set-theory tags are for more advanced topics. axiom-of-choice is not applicable here. –  Aryabhata Feb 29 '12 at 21:55

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.
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I get the answer as $\frac{1}{8}$. Ah, I see, you have put 0.50 instead of 0.35. –  Aryabhata Feb 29 '12 at 22:07
Ah, great. I was about to ask "how did you calculate the four possibilities". –  Gigili Feb 29 '12 at 22:10
@Aryabhata: Oops, well spotted! Thanks. –  Ilmari Karonen Feb 29 '12 at 22:12
Thanks Ilmari Karonen.I'have to review the theme. –  João Feb 29 '12 at 22:35

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

$(1)$:

• $P(A \cap B)= P(B|A)P(A)=\frac{2}{9} \cdot 0.45=0.1$
• $P(\bar A \cap \bar B) = 1 - P(A) - P(\bar A \cap B)=1-0.45-0.5=0.05$

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

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