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UPSC NDA I 2024 Mathematics Paper

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

A, B

and

C

are three events such that

P(A)=0.6, P(B)=0.4, P(C)=0.5, P(A \cup B)=

0.8, P(A \cap C)=0.3

and

P(A \cap B \cap C)=0.2

and

P(A \cup B \cup C) \ge 0.85

Question : 114 of 120

Marks: +1, -0

P(B \cap C)

Solution:

We know that

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

Substituting the given values, we get

0.85 \leq 0.6+0.4+0.5-P(A \cap B)-0.3-P(B \cap C)+0.2

Simplifying the inequality, we get

P(A \cap B)+P(B \cap C) \leq 0.55

We also know that

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

Substituting the given values, we get

0.8=0.6+0.4-P(A \cap B)

Solving for

P(A \cap B)

, we get

P(A \cap B)=0.2

Substituting this value into the inequality obtained earlier, we get

0.2+P(B \cap C) \leq 0.55

Solving for

P(B \cap C)

, we get

P(B \cap C) \leq 0.35

Therefore, the maximum value of

P(B \cap C)

is 0.35 .

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