JEE Mains · Maths · STD 12 - 13. probability
Let \(E_{1}\) and \(E_{2}\) be two events such that the conditional probabilities \(P \left( E _{1} \mid E _{2}\right)=\frac{1}{2}\), \(P \left( E _{2} \mid E _{1}\right)=\frac{3}{4}\) and \(P \left( E _{1} \cap E _{2}\right)=\frac{1}{8}\). Then
- A \(P \left( E _{1} \cap E _{2}\right)= P \left( E _{1}\right) \cdot P \left( E _{2}\right)\)
- B \(P \left( E _{1}^{\prime} \cap E _{2}^{\prime}\right)= P \left( E _{1}^{\prime}\right) \cdot P \left( E _{2}\right)\)
- C \(P \left( E _{1} \cap E _{2}^{\prime}\right)= P \left( E _{1}\right) \cdot P \left( E _{2}\right)\)
- D \(P \left( E _{1}^{\prime} \cap E _{2}\right)= P \left( E _{1}\right) \cdot P \left( E _{2}\right)\)
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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