In Barcelona, $$60\%$$ of the population are brunette, $$70\%$$ have brown eyes, and $$80\%$$ are brunette or have brown eyes.

We choose a person at random. If he or she is brunette: what is the probability to that he or she also has brown eyes?

### Development:

We are considering two events, $$C =$$ "to be brunette", $$O =$$"to have brown eyes". For the statement, we know that $$P(C)=\dfrac{6}{10}={3}{5}$$, $$P(O)=\dfrac{7}{10}$$, $$P(O\cup C)=\dfrac{8}{10}=\dfrac{4}{5}$$.

They ask us about the probability of having brown eyes, knowing that the person is brunette, this is $$P(O/C)$$.

Applying the formula of the conditional probability $$P(O/C)=\dfrac{P(O\cap C)}{P(C)}$$, yet we still do not know $$P(O\cap C)$$.

As we know the probability of the union, we can use the formula $$P(O\cup C)=P(O)+P(C)-P(O\cap C)$$.

By substituting, $$$\dfrac{4}{5}=\dfrac{7}{10}+\dfrac{3}{5}-P(O\cap C)$$$ and therefore, $$$P(O\cap C)=\dfrac{7}{10}+\dfrac{3}{5}-\dfrac{4}{5}=\dfrac{5}{10}=\dfrac{1}{2}$$$

And so, $$$P(O/C)=\dfrac{P(O\cap C)}{P(C)}=\dfrac{\dfrac{1}{2}}{\dfrac{3}{5}}=\dfrac{5}{6}$$$

### Solution:

$$P(O/C)=\dfrac{5}{6}$$.