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### Exercise

A course in Economics is jointly taught to students from the Economics, Law, and Agriculture departments. Among the students enrolled in this course, 65% are from the Economics department, 20% are from the Law department, and 15% are from the Agriculture department. Of the enrolled students, 25% from the Economics department, 63% from the Law department, and 23% from the Agriculture department take the exam during the June session.

Calculate the following:

1. The probability that a randomly selected student will take the exam in June.

2. Given that a student takes the exam in June, calculate the probability that the student is from the Economics department.

3. Given that a student takes the exam in June, calculate the probability that the student is from the Law department.

4. Given that a student takes the exam in June, calculate the probability that the student is from the Agriculture department.

### Solution

Letâ€™s solve this step by step.

#### Step 1: Calculate the probability that a randomly selected student will take the exam in June.

Let \( P(E) \), \( P(L) \), and \( P(A) \) represent the probability that a randomly selected student is from the Economics, Law, and Agriculture departments, respectively. These are given as:

\[

P(E) = 0.65, \quad P(L) = 0.20, \quad P(A) = 0.15

\]

Let \( P(J|E) \), \( P(J|L) \), and \( P(J|A) \) represent the probability that a student from the Economics, Law, and Agriculture departments, respectively, will take the exam in June. These are given as:

\[

P(J|E) = 0.25, \quad P(J|L) = 0.63, \quad P(J|A) = 0.23

\]

The total probability that a randomly selected student will take the exam in June, \( P(J) \), is calculated using the law of total probability:

\[

P(J) = P(J|E)P(E) + P(J|L)P(L) + P(J|A)P(A)

\]

\[

P(J) = (0.25 \times 0.65) + (0.63 \times 0.20) + (0.23 \times 0.15)

\]

\[

P(J) = 0.1625 + 0.126 + 0.0345 = 0.323

\]

So, the probability that a randomly selected student will take the exam in June is 0.323.

#### Step 2: Calculate the probability that a student who takes the exam in June is from the Economics department.

We use Bayes’ theorem to find the conditional probability:

\[

P(E|J) = \frac{P(J|E)P(E)}{P(J)}

\]

\[

P(E|J) = \frac{0.25 \times 0.65}{0.323} = \frac{0.1625}{0.323} \approx 0.5031

\]

So, the probability that a student who takes the exam in June is from the Economics department is approximately 0.5031.

#### Step 3: Calculate the probability that a student who takes the exam in June is from the Law department.

Similarly, using Bayes’ theorem:

\[

P(L|J) = \frac{P(J|L)P(L)}{P(J)}

\]

\[

P(L|J) = \frac{0.63 \times 0.20}{0.323} = \frac{0.126}{0.323} \approx 0.3901

\]

So, the probability that a student who takes the exam in June is from the Law department is approximately 0.3901.

#### Step 4: Calculate the probability that a student who takes the exam in June is from the Agriculture department.

Again, using Bayes’ theorem:

\[

P(A|J) = \frac{P(J|A)P(A)}{P(J)}

\]

\[

P(A|J) = \frac{0.23 \times 0.15}{0.323} = \frac{0.0345}{0.323} \approx 0.1068

\]

So, the probability that a student who takes the exam in June is from the Agriculture department is approximately 0.1068.

### Summary of Answers:

1. The probability that a randomly selected student will take the exam in June: **0.3230**

2. The probability that a student who takes the exam in June is from the Economics department: **0.5031**

3. The probability that a student who takes the exam in June is from the Law department: **0.3901**

4. The probability that a student who takes the exam in June is from the Agriculture department: **0.1068**