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JEE Mains · Maths · STD 12 - 13. probability

माना \(E _1\) तथा \(E _2\) दो घटनायें इस प्रकार है कि प्रतिबन्धित प्रायिकता \(P \left( E _1 \mid E _2\right)=\frac{1}{2}\), \(P \left( E _2 \mid E _1\right)=\frac{3}{4}\) तथा \(P \left( E _1 \cap E _2\right)=\frac{1}{8}\) है तो

  1. A \(P \left( E _{1} \cap E _{2}\right)= P \left( E _{1}\right) \cdot P \left( E _{2}\right)\)
  2. B \(P \left( E _{1}^{\prime} \cap E _{2}^{\prime}\right)= P \left( E _{1}^{\prime}\right) \cdot P \left( E _{2}\right)\)
  3. C \(P \left( E _{1} \cap E _{2}^{\prime}\right)= P \left( E _{1}\right) \cdot P \left( E _{2}\right)\)
  4. D \(P \left( E _{1}^{\prime} \cap E _{2}\right)= P \left( E _{1}\right) \cdot P \left( E _{2}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(P \left( E _{1} \cap E _{2}^{\prime}\right)= P \left( E _{1}\right) \cdot P \left( E _{2}\right)\)

Step-by-step Solution

Detailed explanation

(A) \(P \left( E _{1}\right) \cdot P \left( E _{2}\right)=\frac{1}{6} \cdot \frac{1}{4}=\frac{1}{24} \neq P \left( E _{1} \cap E _{2}\right)\) (B) \(P \left( E _{1}^{\prime} \cap E _{2}^{\prime}\right)=1- P \left( E _{1} \cup E _{2}\right)\)…
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