Let E^{C} denote the complement of an event E | vedclass

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Question

(A) Given that $E_{1}, E_{2}, E_{3}$ are pairwise independent events,we have $P(E_{1} \cap E_{2}) = P(E_{1})P(E_{2})$,$P(E_{2} \cap E_{3}) = P(E_{2})P

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$P(E_{3}^{C}) - P(E_{2})$

Vedclass · Answer

(A) Given that $E_{1}, E_{2}, E_{3}$ are pairwise independent events,we have $P(E_{1} \cap E_{2}) = P(E_{1})P(E_{2})$,$P(E_{2} \cap E_{3}) = P(E_{2})P(E_{3})$,and $P(E_{3} \cap E_{1}) = P(E_{3})P(E_{1})$.Also,$P(E_{1} \cap E_{2} \cap E_{3}) = 0$.We need to find $P(E_{2}^{C} \cap E_{3}^{C} | E_{1})$.Using the definition of conditional probability,$P(E_{2}^{C} \cap E_{3}^{C} | E_{1}) = \frac{P(E_{1} \cap E_{2}^{C} \cap E_{3}^{C})}{P(E_{1})}$.By De Morgan's Law,$E_{2}^{C} \cap E_{3}^{C} = (E_{2} \cup E_{3})^{C}$.Thus,$E_{1} \cap (E_{2} \cup E_{3})^{C} = E_{1} \setminus (E_{1} \cap (E_{2} \cup E_{3})) = E_{1} \setminus ((E_{1} \cap E_{2}) \cup (E_{1} \cap E_{3}))$.Using the inclusion-exclusion principle for probability,$P(E_{1} \cap (E_{2} \cup E_{3})) = P(E_{1} \cap E_{2}) + P(E_{1} \cap E_{3}) - P(E_{1} \cap E_{2} \cap E_{3})$.Substituting the given values,$P(E_{1} \cap (E_{2} \cup E_{3})) = P(E_{1})P(E_{2}) + P(E_{1})P(E_{3}) - 0 = P(E_{1})(P(E_{2}) + P(E_{3}))$.Therefore,$P(E_{1} \cap E_{2}^{C} \cap E_{3}^{C}) = P(E_{1}) - P(E_{1} \cap (E_{2} \cup E_{3})) = P(E_{1}) - P(E_{1})(P(E_{2}) + P(E_{3})) = P(E_{1})(1 - P(E_{2}) - P(E_{3}))$.Finally,$P(E_{2}^{C} \cap E_{3}^{C} | E_{1}) = \frac{P(E_{1})(1 - P(E_{2}) - P(E_{3}))}{P(E_{1})} = 1 - P(E_{2}) - P(E_{3}) = (1 - P(E_{3})) - P(E_{2}) = P(E_{3}^{C}) - P(E_{2})$.

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 $P(E_{2}^{C} \cap E_{3}^{C} / E_{1})$ is equal to

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