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UPSC NDA I 2024 Mathematics Paper

Question Numbers: 111-112

Consider the following for the items that follow:

Let A and B be two events such that P(A ∪ B) ≥ 0.75 and 0.125 ≤ P(A ∩ B) ≤ 0.375.

Question : 111 of 120

Marks: +1, -0

What is the minimum value of P(A) + P(B) ?

Solution:

Concept:

Use the inclusion-exclusion principle for probability.

Explanation:

Given:

P(A \cup B) \ge 0.75

and

0.125 \le P(A \cap B) \le 0.375

From

P(A \cup B) = P(A) + P(B) - P(A \cap B)

, we get

P(A) + P(B) = P(A \cup B) + P(A \cap B)

To minimize

P(A) + P(B)

, take the smallest possible values of

P(A \cup B)

and

P(A \cap B)

Minimum of

P(A \cup B)

0.75

and minimum of

P(A \cap B)

0.125

Thus,

P(A) + P(B) \ge 0.75 + 0.125 = 0.875

Therefore, the minimum value is

0.875

Answer:

0.875

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