UPSC NDA I 2024 Mathematics Paper

Concept:Use the inclusion-exclusion principle for three events and constraints from given probabilities to find the minimum possible value of $P(B \cap C)$.Explanation:Given: $P(A) = 0.6$, $P(B) = 0.4$, $P(C) = 0.5$, $P(A \cup B) = 0.8$, $P(A \cap C) = 0.3$, $P(A \cap B \cap C) = 0.2$, and $P(A \cup B \cup C) \geq 0.85$.First, find $P(A \cap B)$ using $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.$0.8 = 0.6 + 0.4 - P(A \cap B) \implies P(A \cap B) = 0.2$.Apply inclusion-exclusion for three events:$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$.Substitute known values:$P(A \cup B \cup C) = 0.6 + 0.4 + 0.5 - 0.2 - P(B \cap C) - 0.3 + 0.2 = 1.2 - P(B \cap C)$.Since $P(A \cup B \cup C) \geq 0.85$, we have $1.2 - P(B \cap C) \geq 0.85$.Solve: $\,- P(B \cap C) \geq -0.35 \implies P(B \cap C) \leq 0.35$. This gives the maximum possible value.Now, consider the overlap with $A$.The event $A \cap B \cap C$ is a subset of $B \cap C$. So $P(B \cap C) \geq P(A \cap B \cap C) = 0.2$.This gives a lower bound. Therefore, the minimum value of $P(B \cap C)$ is $0.2$.Answer:$0.2$, which corresponds to option B.