Let A, B and C be three events,which are pair | vedclass

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(C) We need to find $P[(\bar{A} \cap \bar{B})|C]$.By the definition of conditional probability,$P[(\bar{A} \cap \bar{B})|C] = \frac{P(\bar{A} \cap \ba

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$P(\bar{A}) - P(B)$

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(C) We need to find $P[(\bar{A} \cap \bar{B})|C]$.By the definition of conditional probability,$P[(\bar{A} \cap \bar{B})|C] = \frac{P(\bar{A} \cap \bar{B} \cap C)}{P(C)}$.Using De Morgan's Law,$\bar{A} \cap \bar{B} = \overline{A \cup B}$.Thus,$\bar{A} \cap \bar{B} \cap C = C \setminus ((A \cap C) \cup (B \cap C))$.Therefore,$P(\bar{A} \cap \bar{B} \cap C) = P(C) - P((A \cap C) \cup (B \cap C))$.Using the inclusion-exclusion principle,$P((A \cap C) \cup (B \cap C)) = P(A \cap C) + P(B \cap C) - P(A \cap B \cap C)$.Since $A, B, C$ are pairwise independent,$P(A \cap C) = P(A)P(C)$ and $P(B \cap C) = P(B)P(C)$.Given $P(A \cap B \cap C) = 0$,we have $P((A \cap C) \cup (B \cap C)) = P(A)P(C) + P(B)P(C) - 0 = P(C)(P(A) + P(B))$.Substituting this back,$P(\bar{A} \cap \bar{B} \cap C) = P(C) - P(C)(P(A) + P(B)) = P(C)(1 - P(A) - P(B))$.Finally,$P[(\bar{A} \cap \bar{B})|C] = \frac{P(C)(1 - P(A) - P(B))}{P(C)} = 1 - P(A) - P(B)$.Since $1 - P(A) = P(\bar{A})$,the expression becomes $P(\bar{A}) - P(B)$.

Let $A, B$ and $C$ be three events,which are pairwise independent and $\bar{E}$ denotes the complement of an event $E$. If $P(A \cap B \cap C) = 0$ and $P(C) > 0$,then $P[(\bar{A} \cap \bar{B})|C]$ is equal to

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