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UPSC NDA I 2024 Mathematics Paper

Consider the following for the next two ( 02 ) items that follow:

Let

A

and

B

be two events such that

P(A \cup B) \ge 0.75

and

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

Question : 112 of 120

Marks: +1, -0

P(A)+P(B)

Solution:

We know that:

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

Therefore:

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

To maximize

P(A)+P(B)

, we need to maximize both

P(A \cup B)

and

P(A \cap B)

The maximum value of

P(A \cup B)

is given as 0.75 . The maximum value of

P(A \cap B)

is 0.375 .

Therefore, the maximum value of

P(A)+P(B)

is:

0.75+0.375=1.125

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