Question:
If A, B, C are mutually exclusive and exhaustive events such that \(P(A):P(B):P(C)=1:l:m\), then \(P(A\cup B)+P(B\cup C)+P(C\cup A)+P(A\cup B\cup C)=\):
If A, B, C are mutually exclusive and exhaustive events such that \(P(A):P(B):P(C)=1:l:m\), then \(P(A\cup B)+P(B\cup C)+P(C\cup A)+P(A\cup B\cup C)=\):
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For mutually exclusive events, unions reduce to simple sums.
Updated On: Jun 18, 2026
- \( \frac{1}{3} \)
- \( 3 \)
- \( \frac{3}{4} \)
- \( 1 \)
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The Correct Option is B
Solution and Explanation
Concept:
For mutually exclusive and exhaustive events:
\[
P(A)+P(B)+P(C)=1
\]
Step 1: Express probabilities.
Let: \[ P(A)=x,\;P(B)=lx,\;P(C)=mx \] \[ x(1+l+m)=1 \Rightarrow x=\frac{1}{1+l+m} \]
Step 2: Compute union expressions.
Since events are mutually exclusive: \[ P(A\cup B)=P(A)+P(B) \] Similarly: \[ P(A\cup B)+P(B\cup C)+P(C\cup A) =2(P(A)+P(B)+P(C))=2 \] Adding: \[ P(A\cup B\cup C)=1 \] \[ \Rightarrow \text{Total} = 3 \]
Step 1: Express probabilities.
Let: \[ P(A)=x,\;P(B)=lx,\;P(C)=mx \] \[ x(1+l+m)=1 \Rightarrow x=\frac{1}{1+l+m} \]
Step 2: Compute union expressions.
Since events are mutually exclusive: \[ P(A\cup B)=P(A)+P(B) \] Similarly: \[ P(A\cup B)+P(B\cup C)+P(C\cup A) =2(P(A)+P(B)+P(C))=2 \] Adding: \[ P(A\cup B\cup C)=1 \] \[ \Rightarrow \text{Total} = 3 \]
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