Question:
Let \(E^C\) denote the complement of an event E. Let \(E_1\), \(E_2\) and \(E_3\) be any pairwise independent events with \(P(E_1) > 0\) and \(P(E_1 \cap E_2 \cap E_3) = 0\).
Then \(P\left(E_2^C \cap E_3^C / E_1\right)\) is equal to :
Let \(E^C\) denote the complement of an event E. Let \(E_1\), \(E_2\) and \(E_3\) be any pairwise independent events with \(P(E_1) > 0\) and \(P(E_1 \cap E_2 \cap E_3) = 0\).
Then \(P\left(E_2^C \cap E_3^C / E_1\right)\) is equal to :
Then \(P\left(E_2^C \cap E_3^C / E_1\right)\) is equal to :
Show Hint
Use \(E_2^C\cap E_3^C = (E_2\cup E_3)^C\) and inclusion-exclusion inside the conditioning on \(E_1\); pairwise independence gives \(P(E_1\cap E_i)=P(E_1)P(E_i)\).
Updated On: Nov 5, 2026
- \(P\left(E_3^C\right) - P(E_2)\)
- \(P\left(E_2^C\right) + P(E_3)\)
- \(P\left(E_3^C\right) - P\left(E_2^C\right)\)
- \(P(E_3) - P\left(E_2^C\right)\)
Show Solution
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The Correct Option is A
Solution and Explanation
\(P(E_2^C\cap E_3^C\mid E_1) = 1 - P(E_2\cup E_3\mid E_1) = 1-P(E_2\mid E_1)-P(E_3\mid E_1)+P(E_2\cap E_3\mid E_1)\).
Pairwise independence: \(P(E_2\mid E_1)=P(E_2)\), \(P(E_3\mid E_1)=P(E_3)\). Also \(P(E_2\cap E_3\mid E_1)=\frac{P(E_1\cap E_2\cap E_3)}{P(E_1)}=0\).
So the value is \(1-P(E_2)-P(E_3) = \left(1-P(E_3)\right)-P(E_2) = P\left(E_3^C\right)-P(E_2)\).
Answer: (1) \(P\left(E_3^C\right) - P(E_2)\)
Pairwise independence: \(P(E_2\mid E_1)=P(E_2)\), \(P(E_3\mid E_1)=P(E_3)\). Also \(P(E_2\cap E_3\mid E_1)=\frac{P(E_1\cap E_2\cap E_3)}{P(E_1)}=0\).
So the value is \(1-P(E_2)-P(E_3) = \left(1-P(E_3)\right)-P(E_2) = P\left(E_3^C\right)-P(E_2)\).
Answer: (1) \(P\left(E_3^C\right) - P(E_2)\)
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