UPSC NDA I 2024 Mathematics Paper

To find the minimum value of $P(A)+P(B)$, we can use the principle of addition for probabilities and the given constraints. The principle of addition for two events $A$ and $B$ states:$P(A \cup B)=P(A)+P(B)-P(A \cap B)$Given that $P(A \cup B) \geq 0.75$ and $0.125 \leq P(A \cap B) \leq 0.375$, we can rewrite the equation as:$P(A \cup B)=P(A)+P(B)-P(A \cap B) \geq 0.75$To minimize $P(A)+P(B)$, we need to consider the maximum possible value of $P(A \cap B)$ within the given range, which is 0.375 . This is because a higher $P(A \cap B)$ would reduce the required sum $P(A)+P(B)$ due to the subtraction in the equation. Substituting $P(A \cap B)=0.375$ into the equation, we get:$P(A \cup B) \geq P(A)+P(B)-0.375 \geq 0.75$Simplifying this, we get:$P(A)+P(B)-0.375 \geq 0.75$Adding 0.375 to both sides, we get:$P(A)+P(B) \geq 1.125$However, because the probability of an event cannot exceed 1, the maximum value of the sum $P(A \cup B)=0.75$ and $P(A \cap B)=0.125$, we re-evaluate with the other bound:$P(A \cup B)=P(A)+P(B)-0.125 \geq 0.75$ Here, we want the minimum sum, so we try substracting:$P(A)+P(B)=x$Substitute:$x-0.125 \geq 0.75$Simplifying:$x \geq 0.875$Thus, the minimum value of $P(A)+P(B)$ required to satisfy the conditions is 0.875 .