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?

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 $P(G_1 \mid E) = \frac{3}{4}$.

The post Find the Conditional Probability About Math Exam Experiment first appeared on Problems in Mathematics.

A certain model of smartphone is manufactured by three factories A, B, and C. Factories A, B, and C produce $60\%$, $25\%$, and $15\%$ of the smartphones, respectively.

Suppose that their defective rates are $5\%$, $2\%$, and $7\%$, respectively. Determine the overall fraction of defective smartphones of this model.

Solution.

Let $E$ be the event that a smartphone of this model is defective. Let $F_A$ be the event that a smartphone is manufactured by factory A. Similarly for $F_B$ and $F_C$.

Then the overall fraction of defective smartphones of this model can be found as follows.
\begin{align*}
P(E) &= P(F_A \cap E) + P(F_B \cap E) + P(F_C \cap E)\\
&= P(F_A)\cdot P(E \mid F_A) + P(F_B)\cdot P(E \mid F_B) + P(F_C)\cdot P(E \mid F_C)\\
&= (0.6)(0.05) + (0.25)(0.02) + (0.15)(0.07)\\
&= 0.0455.
\end{align*}
Thus, the overall defective rate is $4.55\%$.

Further Question

In the context of the above problem, if a smartphone of this model is found out to be detective, find the probability that this smartphone was manufactured in factory C.

The solution is available in the post If a Smartphone is Defective, Which Factory Made It?

The post Overall Fraction of Defective Smartphones of Three Factories first appeared on Problems in Mathematics.