Q.Let $A$ and $B$ be two events such that $P\left(\overline{A\cup B}\right)=\dfrac{1}{6}$, $P(A\cap B)=\dfrac{1}{4}$ and $P(\overline{A})=\dfrac{1}{4}$. Then the events $A$ and $B$ are

$P(\overline{A}\cap B) = P(B)-P(A\cap B) = 0.8-0.24 = 0.56$.

First find $P(A\cap B)$ using the conditional probability formula: $P(B/A) = \dfrac{P(A\cap B)}{P(A)}$, so $P(A\cap B) = P(B/A)\cdot P(A) = 0.6\times0.4 = 0.24$.

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Solving for $P(X)=\frac12$ and $P(Y)=\frac13$ from the given conditionals, the addition rule gives $P(X\cup Y)=\frac23$.

From $P(X/Y)=\dfrac{P(X\cap Y)}{P(Y)}$: $\dfrac12 = \dfrac{1/6}{P(Y)} \Rightarrow P(Y) = \dfrac{1/6}{1/2} = \dfrac13$.

From $P(Y/X)=\dfrac{P(X\cap Y)}{P(X)}$: $\dfrac13 = \dfrac{1/6}{P(X)} \Rightarrow P(X) = \dfrac{1/6}{1/3} = \dfrac12$.

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First get $P(A\cap B)$ from $P(B|A)=P(A\cap B)/P(A)$, then use $P(A|B)=P(A\cap B)/P(B)$ to solve for $P(B)=1/3$.

Given $P(A)=\dfrac{1}{4}$, $P(A|B)=\dfrac{1}{2}$, $P(B|A)=\dfrac{2}{3}$.

From $P(B|A) = \dfrac{P(A\cap B)}{P(A)}$: $P(A\cap B) = P(B|A)\cdot P(A) = \dfrac{2}{3}\times\dfrac{1}{4} = \dfrac{1}{6}$.

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