Q.If A and B are two events such that $P(A) = \frac{1}{4}$, $P(B) = \frac{1}{2}$ and $P(A \cap B) = \frac{1}{8}$, find P (not A and not B).

Concept: Complement of a union (De Morgan's law) / Independence check

We want $P(\text{not A and not B}) = P(A' \cap B')$. By De Morgan's law, $A' \cap B' = (A \cup B)'$, so $P(A' \cap B') = 1 - P(A \cup B)$.

Step 1: Addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{1}{4} + \frac{1}{2} - \frac{1}{8} = \frac{2}{8} + \frac{4}{8} - \frac{1}{8} = \frac{5}{8}$.

Step 2: $P(A' \cap B') = 1 - \frac{5}{8} = \frac{3}{8}$. …

The phrase "not A and not B" means both events fail to occur — the complement of $A$ and the complement of $B$ happen together. In set notation this is $A' \cap B'$.

Step 1 — Turn it into a union using De Morgan's law.

The complement of "both fail to occur" is "at least one occurs", which is exactly $A \cup B$. De Morgan's law states $A' \cap B' = (A \cup B)'$, so:

Step 2 — Find $P(A \cup B)$ with the addition rule.

For any two events,

Substitute the given values $P(A) = \frac{1}{4}$, $P(B) = \frac{1}{2}$, $P(A \cap B) = \frac{1}{8}$:

1. Convert to eighths: $\frac{1}{4} = \frac{2}{8}$, $\frac{1}{2} = \frac{4}{8}$.
2. Add: $\frac{2}{8} + \frac{4}{8} = \frac{6}{8}$.
3. Subtract: $\frac{6}{8} - \frac{1}{8} = \frac{5}{8}$.

So $P(A \cup B) = \frac{5}{8}$.

Step 3 — Get the required probability. …

Method: Probability of "neither" via De Morgan and the addition rule

Use this whenever a question asks for the chance that both events fail — "neither A nor B", "not A and not B", $P(A' \cap B')$ — given $P(A)$, $P(B)$ and the overlap.

Steps

Step 1: Rewrite "neither" as the complement of a union.

By De Morgan's law, both failing is the same as "at least one occurring" not happening:

This is the key move — a direct product $P(A')\,P(B')$ is only valid under independence, which you should not assume.

Step 2: Find $P(A \cup B)$ with the general addition rule. …

Common Mistakes

Mistake 1: Reading "not $A$ and not $B$" as $1-P(A)-P(B)$.

Why it's wrong: that drops the overlap correction. Correct approach: use De Morgan, $A'\cap B'=(A\cup B)'$, so $P(A'\cap B')=1-P(A\cup B)$ with $P(A\cup B)=P(A)+P(B)-P(A\cap B)=\tfrac58$, giving $\tfrac38$.

Mistake 2: Applying De Morgan the wrong way, e.g. $A'\cap B'=(A\cap B)'$. …