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JEE Mains · Maths · STD 11 - 14. probability

ત્રણ ઘટનાઓ \(A , B\) અને \(C\) ની સંભાવના અનુક્રમે \(P ( A )=0.6, P ( B )=0.4\) અને \(P ( C )=0.5\) આપેલ છે જો \(P ( A \cup B )=0.8, P ( A \cap C )=0.3, P ( A \cap B \cap\) \(C)=0.2, P(B \cap C)=\beta\) અને \(P(A \cup B \cup C)=\alpha\) જ્યાં \(0.85 \leq \alpha \leq 0.95,\) હોય તો \(\beta\) ની કિમત ........ અંતરાલમાં રહે છે 

  1. A \([0.36,0.40]\)
  2. B \([0.35,0.36]\)
  3. C \([0.25,0.35]\)
  4. D \([0.20,0.25]\)
Verified Solution

Answer & Solution

Correct Answer

(C) \([0.25,0.35]\)

Step-by-step Solution

Detailed explanation

\(P ( A \cup B )= P ( A )+ P ( B )- P ( A \cap B )\) \(0.8=0.6+0.4- P ( A \cap B )\) \(P ( A \cap B )=0.2\) \(P ( A \cup B \cup C )=\Sigma P ( A )-\Sigma P ( A \cap B )+ P ( A \cap B \cap C )\) \(\alpha=1.5-(0.2+0.3+\beta)+0.2\) \(\alpha=1.2-\beta \in[0.85,0.95]\) (where…
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