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Question #41087

Difficulty: 8/10 | 🟢 Free

The Question

Let $A, B,$ and $C$ be three events in a probability space. It is given that $A$ and $B$ are mutually exclusive, while $B$ and $C$ are independent, and $A$ and $C$ are independent. If $P(A \cup B) = \frac{2}{3}$ , $P(B \cup C) = \frac{3}{4}$ , and $P(A \cup C) = \frac{5}{8}$ , then the value of $P(A \cup B \cup C)$ is

Options

A.
$\frac{49}{48}$
B.
$\frac{25}{32}$
C.
$\frac{27}{32}$
D.
$\frac{11}{16}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Inclusion-Exclusion Principle|Independent vs Disjoint Events|Mutually Exclusive vs Independent

Why This Question Is Tricky

Common Mistakes

This question targets the critical distinction between mutually exclusive (disjoint) and independent events within the Inclusion-Exclusion Principle.

A common trap for students is to conflate these properties, either assuming $P(A \cap B) = P(A)P(B)$ for disjoint events or $P(A \cap B) = 0$ for independent events.

By explicitly mixing mutually exclusive ( $A, B$ ) and independent ( $A, C$ and $B, C$ ) pairs, the problem forces the correct application of $P(A \cup B \cup C)$ .

Furthermore, it rewards algebraic insight: recognizing that $P(A \cap B \cap C) = 0$ (since $A \cap B = \emptyset$ ) allows the entire expression to be factored in terms of $P(A)+P(B)$ and $P(C)$ , bypassing the need to solve for $P(A)$ and $P(B)$ individually.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Inclusion-Exclusion Principle
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free