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

Step 1. Apply the complement rule to $P\left(\overline{A\cup B}\right)=\dfrac16$: $P(A\cup B)=1-\dfrac16=\dfrac56$.

Step 2. Apply the complement rule to $P(\overline{A})=\dfrac14$: $P(A)=1-\dfrac14=\dfrac34$.

Step 3. Use the addition (inclusion-exclusion) rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ to solve for $P(B)$:

$\dfrac56=\dfrac34+P(B)-\dfrac14 \;\Rightarrow\; \dfrac56=\dfrac12+P(B) \;\Rightarrow\; P(B)=\dfrac56-\dfrac12=\dfrac13$. …

Complement rule to convert the given $P(\overline{A\cup B})$ and $P(\overline A)$ into $P(A\cup B)$ and $P(A)$, the addition rule to isolate $P(B)$, then the definitions o …

Misreading $P\left(\overline{A\cup B}\right)$ (the complement of the UNION) as $P(A\cup B)$ directly, i.e. dropping the overline - that misreading makes the data look impossible, since it would give $P(A\cup B)=\dfrac16 < P(A\cap B)=\dfrac14$, which cannot happen because $A\cap B\subseteq A\cup B$ always. Similarly, forgetting to complement $P(\overline A)$ to get $P(A) …