Question:
The probabilities of three events A, B and C are given by \(P(A)=0.6\), \(P(B)=0.4\) and \(P(C)=0.5\). If \(P(A\cup B)=0.8\), \(P(A\cap C)=0.3\), \(P(A\cap B\cap C)=0.2\), \(P(B\cap C)=\beta\) and \(P(A\cup B\cup C)=\alpha\), where \(0.85\le\alpha\le0.95\), then \(\beta\) lies in the interval :
The probabilities of three events A, B and C are given by \(P(A)=0.6\), \(P(B)=0.4\) and \(P(C)=0.5\). If \(P(A\cup B)=0.8\), \(P(A\cap C)=0.3\), \(P(A\cap B\cap C)=0.2\), \(P(B\cap C)=\beta\) and \(P(A\cup B\cup C)=\alpha\), where \(0.85\le\alpha\le0.95\), then \(\beta\) lies in the interval :
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Write \(P(A\cup B\cup C)\) by inclusion-exclusion in terms of \(\beta\) and use the range of \(\alpha\).
Updated On: Nov 5, 2026
- \([0.35,\,0.36]\)
- \([0.20,\,0.25]\)
- \([0.25,\,0.35]\)
- \([0.36,\,0.40]\)
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The Correct Option is C
Solution and Explanation
\(P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.6+0.4-0.8=0.2\).
\(\alpha=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)\)
\(=0.6+0.4+0.5-0.2-\beta-0.3+0.2=1.2-\beta\).
\(0.85\le1.2-\beta\le0.95 \Rightarrow 0.25\le\beta\le0.35\).
Answer: (3) [0.25, 0.35]
\(\alpha=P(A)+P(B)+P(C)-P(A\cap B)-P(B\cap C)-P(A\cap C)+P(A\cap B\cap C)\)
\(=0.6+0.4+0.5-0.2-\beta-0.3+0.2=1.2-\beta\).
\(0.85\le1.2-\beta\le0.95 \Rightarrow 0.25\le\beta\le0.35\).
Answer: (3) [0.25, 0.35]
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