UPSC NDA I 2024 Mathematics Paper

To determine the minimum value of $P(B \cap C)$, we can use the principle of inclusion-exclusion for three events $A, B$, and $C$. The principle of inclusion-exclusion states:$P(A \cup B \cup C)=P(A)+P(B)+P(C)-P(A \cap B)-P(A \cap C)-P(B \cap C)+P(A \cap B \cap C)$We are given the following probabilities:$P(A)=0.6$$P(B)=0.4$$P(C)=0.5$$P(A \cap C)=0.3$$P(A \cap B \cap C)=0.2$$P(A \cup B)=0.8 \; \text{ (this is not directly used in the minimum calculation but is given information) } \;$$P(A \cup B \cup C) \geq 0.85$Now, substitute the known values into the inclusion-exclusion principle formula:$0.85 \leq 0.6+0.4+0.5-P(A \cap B)-0.3-P(B \cap C)+0.2$Simplify the expression:$0.85 \leq 1.5-P(A \cap B)-P(B \cap C)-0.1$Combine like terms:$0.85 \leq 1.4-P(A \cap B)-P(B \cap C)$ We know from the given information that for $P(A \cup B)=0.8$ :$P(A \cup B)=P(A)+P(B)-P(A \cap B)=0.8$So we can solve for $P(A \cap B)$ :$0.8=0.6+0.4-P(A \cap B)$$P(A \cap B)=0.2$Now, substitute $P(A \cap B)=0.2$ back into the inequality:$0.85 \leq 1.4-0.2-P(B \cap C)$Combine terms again:$0.85 \leq 1.2-P(B \cap C)$Rearrange to solve for $P(B \cap C)$ :$P(B \cap C) \leq 0.35$