Let A and B be two events such that P(B mid A | vedclass

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(A) Given $P(B \mid A) = \frac{2}{5}$,$P(A \mid B) = \frac{1}{7}$,and $P(A \cap B) = \frac{1}{9}$.From $P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac

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Both $(S1)$ and $(S2)$ are true

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(A) Given $P(B \mid A) = \frac{2}{5}$,$P(A \mid B) = \frac{1}{7}$,and $P(A \cap B) = \frac{1}{9}$.From $P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{1}{7}$,we get $P(B) = 7 	imes P(A \cap B) = 7 	imes \frac{1}{9} = \frac{7}{9}$.From $P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{2}{5}$,we get $P(A) = \frac{5}{2} 	imes P(A \cap B) = \frac{5}{2} 	imes \frac{1}{9} = \frac{5}{18}$.For $(S1)$: $P(A' \cup B) = P(A') + P(B) - P(A' \cap B) = (1 - P(A)) + P(B) - (P(B) - P(A \cap B)) = 1 - P(A) + P(A \cap B) = 1 - \frac{5}{18} + \frac{1}{9} = 1 - \frac{5}{18} + \frac{2}{18} = 1 - \frac{3}{18} = 1 - \frac{1}{6} = \frac{5}{6}$. Thus,$(S1)$ is true.For $(S2)$: $P(A' \cap B') = 1 - P(A \cup B) = 1 - (P(A) + P(B) - P(A \cap B)) = 1 - (\frac{5}{18} + \frac{7}{9} - \frac{1}{9}) = 1 - (\frac{5}{18} + \frac{6}{9}) = 1 - (\frac{5}{18} + \frac{12}{18}) = 1 - \frac{17}{18} = \frac{1}{18}$. Thus,$(S2)$ is true.Therefore,both $(S1)$ and $(S2)$ are true.

Let $A$ and $B$ be two events such that $P(B \mid A) = \frac{2}{5}$,$P(A \mid B) = \frac{1}{7}$ and $P(A \cap B) = \frac{1}{9}$. Consider:
$(S1) P(A' \cup B) = \frac{5}{6}$
$(S2) P(A' \cap B') = \frac{1}{18}$.
Then:

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For two events $A$ and $B$,if $P(A) = P(A|B) = \frac{1}{4}$ and $P(B|A) = \frac{1}{2}$,then:

$A$ and $B$ are two events such that $P(A) \neq 0$. Find $P(B \mid A)$ if: $(i)$ $A \subset B$ (ii) $A \cap B = \phi$.

It is given that $A$ and $B$ are such that $P(A) = \frac{1}{4}$,$P(A|B) = \frac{1}{2}$,and $P(B|A) = \frac{2}{3}$. Then $P(B) = $?

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Let $A$ and $B$ be two events such that $P(B \mid A) = \frac{2}{5}$,$P(A \mid B) = \frac{1}{7}$ and $P(A \cap B) = \frac{1}{9}$. Consider:$(S1) P(A' \cup B) = \frac{5}{6}$$(S2) P(A' \cap B') = \frac{1}{18}$.Then: