UPSC NDA I 2024 Mathematics Solved Paper

To determine the minimum value of P(B∩C), we can use the principle of inclusion-exclusion for three events A,B, and C. The principle of inclusion-exclusion states:
P(A∪B∪C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C)
We are given the following probabilities:
P(A)=0.6
P(B)=0.4
P(C)=0.5
P(A∩C)=0.3
P(A∩B∩C)=0.2
P(A∪B)=0.8‌ (this is not directly used in the minimum calculation but is given information) ‌
P(A∪B∪C)≥0.85
Now, substitute the known values into the inclusion-exclusion principle formula:
0.85≤0.6+0.4+0.5−P(A∩B)−0.3−P(B∩C)+0.2
Simplify the expression:
0.85≤1.5−P(A∩B)−P(B∩C)−0.1
Combine like terms:
0.85≤1.4−P(A∩B)−P(B∩C)

We know from the given information that for P(A∪B)=0.8 :
P(A∪B)=P(A)+P(B)−P(A∩B)=0.8
So we can solve for P(A∩B) :
0.8=0.6+0.4−P(A∩B)
P(A∩B)=0.2
Now, substitute P(A∩B)=0.2 back into the inequality:
0.85≤1.4−0.2−P(B∩C)
Combine terms again:
0.85≤1.2−P(B∩C)
Rearrange to solve for P(B∩C) :
P(B∩C)≤0.35