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(C) Given that $A \subset B$,we have $A \cap B = A$. By the definition of conditional probability,$P(A | B) = \frac{P(A \cap B)}{P(B)}$. Substituting

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

Vedclass · Answer

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

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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