Test Index

UPSC NDA I 2024 Mathematics Paper

Question Numbers: 113-114

Consider the following for the 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 ∪ B) = 0.8, P(A ∩ C) = 0.3 and P(A ∩ B ∩ C) = 0.2 and P(A ∪ B ∪ C) ≥ 0.85.

Question : 114 of 120

Marks: +1, -0

What is the maximum value of P(B ∩ C) ?

0.1
0.2
0.35
0.45

Validate

Solution:

Concept:

The probability of the union of three events is given by the inclusion-exclusion principle:

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

Explanation:

We are given

P(A) = 0.6

P(B) = 0.4

P(C) = 0.5

P(A \cup B) = 0.8

P(A \cap C) = 0.3

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

, and

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

First, find

P(A \cap B)

using

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

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

gives

P(A \cap B) = 0.2

Substitute all known values into the inclusion-exclusion formula:

P(A \cup B \cup C) = 0.6 + 0.4 + 0.5 - 0.2 - P(B \cap C) - 0.3 + 0.2 = 1.2 - P(B \cap C)

Since

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

, we have

1.2 - P(B \cap C) \ge 0.85

Rearrange:

-P(B \cap C) \ge -0.35

, so

P(B \cap C) \le 0.35

Therefore, the maximum possible value of

P(B \cap C)

0.35

Answer:

Option C:

0.35

Go to Question:

Prev Question

Next Question