In a classroom, $60\%$ of the students have brown hair, $30\%$ have brown eyes, and $10\%$ have both brown hair and eyes. A student is selected at random.
- If the student has brown hair, what is the probability that the student also has brown eyes?
- If the student does not have brown hair, what is the probability that the student does not have brown eyes?
Given: $60\%$ of the students have brown hair; $30\%$ have brown eyes; $10\%$ have both brown hair and brown eyes
Let $A$ be the event of selecting a student with brown hair
Then, $P \left(A\right) = \dfrac{60}{100} = \dfrac{6}{10}$
Let $B$ be the event of selecting a student with brown eyes
Then, $P \left(B\right) = \dfrac{30}{100} = \dfrac{3}{10}$
$\left(A \cap B\right)$ is the event of selecting a student with brown hair AND brown eyes
Then, $P \left(A \cap B\right) = \dfrac{10}{100} = \dfrac{1}{10}$
- P(if the student has brown hair then the student also has brown eyes)
$= P \left(B | A\right) = \dfrac{P \left(A \cap B\right)}{P \left(A\right)} = \dfrac{1 / 10}{6 / 10} = \dfrac{1}{6}$
- P(student does not have brown hair) $= P\left(\overline{A}\right) = 1 - P \left(A\right) = 1 - \dfrac{6}{10} = \dfrac{4}{10}$
P(if the student does not have brown hair then the student also does not have brown eyes)
$= P \left(\overline{B} | \overline{A}\right) = \dfrac{P \left(\overline{A} \cap \overline{B}\right)}{P \left(\overline{A}\right)}$
Now,
$\begin{aligned} P \left(\overline{A} \cap \overline{B}\right) & = P \left(\overline{A \cup B}\right) \\\\ & = 1 - P \left(A \cup B\right) \\\\ & = 1 - \left[P \left(A\right) + P \left(B\right) - P \left(A \cap B\right)\right] \\\\ & = 1 - \left[\dfrac{6}{10} + \dfrac{3}{10} - \dfrac{1}{10}\right] \\\\ & = 1 - \dfrac{8}{10} \\\\ & = \dfrac{2}{10} \end{aligned}$
$\therefore \;$ $P \left(\overline{B} | \overline{A}\right) = \dfrac{2 / 10}{4 / 10} = \dfrac{2}{4} = \dfrac{1}{2}$