2
$\begingroup$

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?

  • 1
    probability-theory and set-theory tags are for more advanced topics. axiom-of-choice is not applicable here.2012-02-29

2 Answers 2

3

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.
  • 0
    Thanks Ilmari Karonen.I'have to review the theme.2012-02-29
1

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