Find the Conditional Probability About Math Exam Experiment

Probability problems

Problem 740

A researcher conducted the following experiment. Students were grouped into two groups. The students in the first group had more than 6 hours of sleep and took a math exam. The students in the second group had less than 6 hours of sleep and took the same math exam.

The pass rate of the first group was twice as big as the second group. Suppose that $60\%$ of the students were in the first group. What is the probability that a randomly selected student belongs to the first group if the student passed the exam?

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Solution.

Let $E$ be the event that a student passes the exam.
Let $G_i$ be the event that a student belongs to the group $i$ for $i= 1, 2$. Then the desired probability is $P(G_1 \mid E)$. ($G_1$ is the first group and $G_2$ is the second group.)

Using these notation, the pass rate of the first group is expressed as $P(E \mid G_1)$. Similarly, the pass rate of the second group is $P(E \mid G_2)$. By assumption, the pass rate of the first group is twice as big as the second group. Hence, we have
\[P(E \mid G_1) = 2\ P(E \mid G_2).\]

The required probability that a randomly selected student belongs to the first group given that the student passes the exam is expressed as $P(G_1 \mid E)$.

Using the above equality and Bayes’ rule, we get
\begin{align*}
&P(G_1 \mid E)\\[6pt] &= \frac{P(G_1) P(E \mid G_1)}{P(E)} & \text{(by Bayes’ rule)}\\[6pt] &= \frac{P(G_1)P(E \mid G_1)}{P(E \mid G_1)P(G_1) + P(E \mid G_2)P(G_2)} & \text{(rule of total probability)}\\[6pt] &= \frac{P(G_1)\cdot 2P(E \mid G_2)}{2P(E \mid G_2)P(G_1) + P(E \mid G_2)P(G_2)} \\[6pt] &= \frac{P(G_1)\cdot 2}{2P(G_1) + P(G_2)}\\[6pt] &= \frac{0.6 \cdot 2}{2\cdot 0.6 + 0.4}\\[6pt] &= \frac{3}{4}.
\end{align*}

Therefore, the required probability is $P(G_1 \mid E) = \frac{3}{4}$.


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