If A and B are any two events such that P(A) | vedclass

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(B) Given $P(A) = \frac{2}{5} = \frac{8}{20}$ and $P(A \cap B) = \frac{3}{20}$.By De Morgan's Law,$A' \cup B' = (A \cap B)'$.Therefore,$P(A' \cup B')

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$5/17$

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(B) Given $P(A) = \frac{2}{5} = \frac{8}{20}$ and $P(A \cap B) = \frac{3}{20}$.By De Morgan's Law,$A' \cup B' = (A \cap B)'$.Therefore,$P(A' \cup B') = P((A \cap B)') = 1 - P(A \cap B) = 1 - \frac{3}{20} = \frac{17}{20}$.Now,we need to find $P(A | A' \cup B') = \frac{P(A \cap (A' \cup B'))}{P(A' \cup B')}$.Using the distributive law,$A \cap (A' \cup B') = (A \cap A') \cup (A \cap B') = \emptyset \cup (A \cap B') = A \cap B'$.Since $A = (A \cap B) \cup (A \cap B')$,we have $P(A \cap B') = P(A) - P(A \cap B) = \frac{8}{20} - \frac{3}{20} = \frac{5}{20}$.Thus,$P(A | A' \cup B') = \frac{5/20}{17/20} = \frac{5}{17}$.

$Face$	$1$	$2$	$3$	$4$	$5$	$6$
$P(F)$	$0.1$	$0.24$	$0.19$	$0.18$	$0.15$	$0.14$

If $A$ and $B$ are any two events such that $P(A) = \frac{2}{5}$ and $P(A \cap B) = \frac{3}{20}$,then the conditional probability $P(A | A' \cup B')$,where $A'$ denotes the complement of $A$,is equal to:

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