For a student and three subjects (Mathematics M, Physics P, Chemistry C), suppose: P(M ∪ P ∪ C) = 0.75 (pass at least one), P(at least two) = 0.50, and P(exactly two) = 0.40. Which relation is correct among the options (interpreting pmc as P(M ∩ P ∩ C))?

Introduction / Context:
The statements refer to probabilities of passing subjects M, P, and C. The options mix expressions for sums and a term denoted “pmc.” Under the usual contest shorthand, “pmc” here is best read as P(M ∩ P ∩ C), the probability a student passes all three subjects, not the product p*m*c. We use standard set identities to find that value.

Given Data / Assumptions:

• P(M ∪ P ∪ C) = 0.75.
• P(at least two) = 0.50.
• P(exactly two) = 0.40.
• Let S1 = P(M) + P(P) + P(C); S2 = sum of pairwise intersections; S3 = P(M ∩ P ∩ C).

Concept / Approach:
Recall: P(M ∪ P ∪ C) = S1 − S2 + S3. Also, P(exactly two) = S2 − 3S3, and P(at least two) = P(exactly two) + S3.

Step-by-Step Solution:
From P(at least two) = 0.50 and P(exactly two) = 0.40, we get S3 = 0.50 − 0.40 = 0.10.Thus pmc = P(M ∩ P ∩ C) = 0.10 = 1/10.Additionally, S2 = P(exactly two) + 3S3 = 0.40 + 0.30 = 0.70.Then 0.75 = S1 − 0.70 + 0.10 ⇒ S1 = 1.35 (not among the listed numeric options except by implication that pmc is 1/10).

Verification / Alternative check:
Check consistency: P(at least one) = 0.75 and P(at least two) = 0.50 implies a plausible triple-intersection S3 = 0.10.

Why Other Options Are Wrong:
The equalities for S1 (e.g., 19/20 or 17/20) contradict the deduced S1 = 27/20; pmc = 1/4 is too large and inconsistent with the given “exactly two.”

Common Pitfalls:
Misreading “pmc” as the product p*m*c; here it naturally denotes the three-way intersection probability.

Final Answer:
pmc = 1/10