Consider two events A and B such that P(A) = | vedclass

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(A) Given $P(A) = \frac{1}{4}$,$P(B/A) = \frac{1}{2}$,and $P(A/B) = \frac{1}{4}$.First,find $P(A \cap B)$:$P(A \cap B) = P(A) 	imes P(B/A) = \frac{1}

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$I$ only

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(A) Given $P(A) = \frac{1}{4}$,$P(B/A) = \frac{1}{2}$,and $P(A/B) = \frac{1}{4}$.First,find $P(A \cap B)$:$P(A \cap B) = P(A) 	imes P(B/A) = \frac{1}{4} 	imes \frac{1}{2} = \frac{1}{8}$.Since $P(A \cap B) 
eq 0$,the events $A$ and $B$ are not mutually exclusive. Thus,statement $II$ is incorrect.Next,find $P(B)$:$P(A/B) = \frac{P(A \cap B)}{P(B)} \Rightarrow \frac{1}{4} = \frac{1/8}{P(B)} \Rightarrow P(B) = \frac{1}{2}$.Since $P(A \cap B) = \frac{1}{8} = P(A) 	imes P(B)$,events $A$ and $B$ are independent.Check statement $I$:$P(A^c/B^c) = \frac{P(A^c \cap B^c)}{P(B^c)} = \frac{P(A^c)P(B^c)}{P(B^c)} = P(A^c) = 1 - P(A) = 1 - \frac{1}{4} = \frac{3}{4}$.Thus,statement $I$ is correct.Check statement $III$:$P(A/B) + P(A/B^c) = \frac{1}{4} + \frac{P(A \cap B^c)}{P(B^c)} = \frac{1}{4} + \frac{P(A) - P(A \cap B)}{1 - P(B)} = \frac{1}{4} + \frac{1/4 - 1/8}{1 - 1/2} = \frac{1}{4} + \frac{1/8}{1/2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.Since $\frac{1}{2} 
eq 1$,statement $III$ is incorrect.Therefore,only statement $I$ is true.

Consider two events $A$ and $B$ such that $P(A) = \frac{1}{4}$,$P(B/A) = \frac{1}{2}$,$P(A/B) = \frac{1}{4}$. For each of the following statements,which is true?
$I.$ $P(A^c/B^c) = \frac{3}{4}$
$II.$ The events $A$ and $B$ are mutually exclusive
$III.$ $P(A/B) + P(A/B^c) = 1$

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Consider two events $A$ and $B$ such that $P(A) = \frac{1}{4}$,$P(B/A) = \frac{1}{2}$,$P(A/B) = \frac{1}{4}$. For each of the following statements,which is true?$I.$ $P(A^c/B^c) = \frac{3}{4}$$II.$ The events $A$ and $B$ are mutually exclusive$III.$ $P(A/B) + P(A/B^c) = 1$