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

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(B) Given $P(A) = \frac{2}{3}$, $P(B) = \frac{1}{2}$, and $P(A | B) = \frac{2}{3}$.Since $P(A | B) = \frac{P(A \cap B)}{P(B)}$, we have $P(A \cap B) =

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(B) Given $P(A) = \frac{2}{3}$, $P(B) = \frac{1}{2}$, and $P(A | B) = \frac{2}{3}$.Since $P(A | B) = \frac{P(A \cap B)}{P(B)}$, we have $P(A \cap B) = P(A | B) \cdot P(B) = \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}$.Note that $P(A \cap B) = P(A)$, which implies $A \subset B$.We need to find $P(A' \cup B) + P(A \cup B')$.Using $P(X \cup Y) = P(X) + P(Y) - P(X \cap Y)$:$P(A' \cup B) = P(A') + P(B) - P(A' \cap B) = (1 - P(A)) + P(B) - (P(B) - P(A \cap B)) = 1 - P(A) + P(A \cap B) = 1 - \frac{2}{3} + \frac{1}{3} = \frac{2}{3}$.$P(A \cup B') = P(A) + P(B') - P(A \cap B') = P(A) + (1 - P(B)) - (P(A) - P(A \cap B)) = 1 - P(B) + P(A \cap B) = 1 - \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$.Sum $= \frac{2}{3} + \frac{5}{6} = \frac{4+5}{6} = \frac{9}{6} = \frac{3}{2}$.Wait, re-evaluating: $P(A' \cup B) = P(A' \cup B) = 1 - P(A \cap B') = 1 - (P(A) - P(A \cap B)) = 1 - (\frac{2}{3} - \frac{1}{3}) = 1 - \frac{1}{3} = \frac{2}{3}$.$P(A \cup B') = 1 - P(A' \cap B) = 1 - (P(B) - P(A \cap B)) = 1 - (\frac{1}{2} - \frac{1}{3}) = 1 - \frac{1}{6} = \frac{5}{6}$.Sum $= \frac{2}{3} + \frac{5}{6} = \frac{9}{6} = 1.5$. Given the options, let's re-check the expression. If $A \subset B$, then $P(A' \cup B) = P(S) = 1$ and $P(A \cup B') = P(B') = 1 - 1/2 = 1/2$. Sum $= 1.5$. The provided options seem inconsistent with the calculation.

If $A$ and $B$ are two events such that $P(A) = \frac{2}{3}$, $P(B) = \frac{1}{2}$ and $P(A | B) = \frac{2}{3}$, then $P(A' \cup B) + P(A \cup B') = $

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