A, B are the events in a random experiment. I | vedclass

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(C) Given: $P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(A \cap B) = \frac{1}{4}$.First, we find $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{1/3} = \

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$\frac{11}{8}$

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(C) Given: $P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(A \cap B) = \frac{1}{4}$.First, we find $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/4}{1/3} = \frac{3}{4}$.Next, we find $P(A^c \cap B^c) = P((A \cup B)^c) = 1 - P(A \cup B)$.$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{2} + \frac{1}{3} - \frac{1}{4} = \frac{6+4-3}{12} = \frac{7}{12}$.So, $P(A^c \cap B^c) = 1 - \frac{7}{12} = \frac{5}{12}$.Also, $P(B^c) = 1 - P(B) = 1 - \frac{1}{3} = \frac{2}{3}$.Then, $P(A^c|B^c) = \frac{P(A^c \cap B^c)}{P(B^c)} = \frac{5/12}{2/3} = \frac{5}{12} 	imes \frac{3}{2} = \frac{5}{8}$.Finally, $P(A^c|B^c) + P(A|B) = \frac{5}{8} + \frac{3}{4} = \frac{5+6}{8} = \frac{11}{8}$.

$A, B$ are the events in a random experiment. If $P(A)=\frac{1}{2}, P(B)=\frac{1}{3}, P(A \cap B)=\frac{1}{4}$, then $P\left(\frac{A^{c}}{B^{c}}\right)+P\left(\frac{A}{B}\right)=$

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