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∪B)=P(A)+P(B)−P(A∩B)
Given that P(A∪B)≥0.75 and 0.125≤P(A∩B)≤0.375, we can rewrite the equation as:
P(A∪B)=P(A)+P(B)−P(A∩B)≥0.75
To minimize P(A)+P(B), we need to consider the maximum possible value of P(A∩B) within the given range, which is 0.375 . This is because a higher P(A∩B) would reduce the required sum P(A)+P(B) due to the subtraction in the equation. Substituting P(A∩B)=0.375 into the equation, we get:
P(A∪B)≥P(A)+P(B)−0.375≥0.75
Simplifying this, we get:
P(A)+P(B)−0.375≥0.75
Adding 0.375 to both sides, we get:
P(A)+P(B)≥1.125
However, because the probability of an event cannot exceed 1, the maximum value of the sum P(A∪B)=0.75 and P(A∩B)=0.125, we re-evaluate with the other bound:
P(A∪B)=P(A)+P(B)−0.125≥0.75

Here, we want the minimum sum, so we try substracting:
P(A)+P(B)=x
Substitute:
x−0.125≥0.75
Simplifying:
x≥0.875
Thus, the minimum value of P(A)+P(B) required to satisfy the conditions is 0.875 .