Conditional probability

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?

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

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

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