Out of sixty students, 14 are taking Economics and 29 are taking Calculus. Suppose 13 students are taking both Economics and Calculus. What is the probability that a randomly chosen student from this group is taking only the Calculus class?

Introduction / Context:
This question uses basic set theory and probability. Students may be taking Economics, Calculus, both or neither. We are given how many students are in each subject and how many students are in both subjects, and we must find the probability that a randomly selected student is taking only the Calculus class.

Given Data / Assumptions:

• Total number of students = 60.
• Students taking Economics = 14.
• Students taking Calculus = 29.
• Students taking both Economics and Calculus = 13.
• All students are equally likely to be picked at random.
• We seek P(student is taking Calculus and not Economics).

Concept / Approach:
We use the principle of inclusion exclusion for sets. The number of students taking only Calculus equals the total Calculus students minus those taking both subjects. The probability is then this count divided by the total number of students. No information about students who take neither subject is needed, as long as the total number of students is known.

Step-by-Step Solution:
Let E be the set of students taking Economics and C be the set of students taking Calculus.Given |E| = 14, |C| = 29 and |E ∩ C| = 13.Students taking only Calculus are in C but not in E. Their number is |C only| = |C| - |E ∩ C|.Compute |C only| = 29 - 13 = 16.Total students = 60.Probability that a randomly selected student is taking only Calculus = |C only| / 60 = 16 / 60.Simplify 16 / 60 by dividing numerator and denominator by 4 to get 4 / 15.

Verification / Alternative check:
We can also compute the number of students taking only Economics: |E only| = |E| - |E ∩ C| = 14 - 13 = 1. Then the number of students taking neither subject is 60 - (|E only| + |C only| + |E ∩ C|) = 60 - (1 + 16 + 13) = 30. All categories add up correctly: 1 only Economics, 16 only Calculus, 13 both, 30 neither, total 60. This consistency check supports the probability 4/15.

Why Other Options Are Wrong:
The value 8/15 would correspond to 32 students taking only Calculus, which is impossible since only 29 take Calculus in total. The value 7/15 corresponds to 28 students taking only Calculus, which would leave too few in the intersection. The fraction 1/15 corresponds to only 4 students taking only Calculus, which contradicts the given subject counts. Only 4/15 matches the correct count 16 out of 60.

Common Pitfalls:
Students sometimes misinterpret the phrase taking Calculus and double count those in both subjects when computing probabilities. Others forget to subtract the intersection from the total Calculus students to get the number taking only Calculus. Drawing a Venn diagram with two overlapping circles labeled Economics and Calculus is often very helpful to keep the counts clear.

Final Answer:
The probability that a randomly chosen student is taking only the Calculus class is 4/15.