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(C) Given that $A \subset B$,it implies that $A \cap B = A$. By the definition of conditional probability,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$. Su

Vedclass · Accepted Answer

$P(A \mid B) \geq P(A)$

Vedclass · Answer

(C) Given that $A \subset B$,it implies that $A \cap B = A$. By the definition of conditional probability,$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$. Substituting $A \cap B = A$,we get $P(A \mid B) = \frac{P(A)}{P(B)}$. Since $A \subset B$,we have $P(A) \leq P(B)$,which implies $\frac{1}{P(B)} \leq \frac{1}{P(A)}$. Multiplying both sides by $P(A)$ (assuming $P(A) > 0$),we get $\frac{P(A)}{P(B)} \leq 1$. Also,since $P(B) \leq 1$,we have $\frac{P(A)}{P(B)} \geq P(A)$. Thus,$P(A \mid B) \geq P(A)$.

If $A$ and $B$ are two events such that $A \subset B$ and $P(B) \neq 0$,then which of the following is correct?

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