Q.If $E$ and $F$ are events such that $P(E) = \dfrac{1}{4}$, $P(F) = \dfrac{1}{2}$ and $P(E \text{ and } F) = \dfrac{1}{8}$ then the value of $P(\text{not } E \text{ and not } F)$ is

The Intuition: "Or" Means We Add — But Carefully

Imagine you have a bag of 20 marbles: 5 red, 3 blue, and 12 green. You pick one marble at random.

What's the probability that the marble is red or blue?

Your instinct might be: count the red ones (5), count the blue ones (3), add them up (8), and divide by total marbles (20). That gives $\frac{8}{20} = 0.4$.

That works perfectly here. Why? Because no marble is both red and blue. The events "red" and "blue" cannot happen at the same time — they are mutually exclusive.

Now change the problem. The bag has 20 marbles: 5 red, 3 blue, and 4 striped red-and-blue marbles (counted in both colours). The rest are plain green.

What's the probability of picking a marble that is red or blue?

If you just add red (5 + 4 striped = 9) and blue (3 + 4 striped = 7), you get 16. But that double-counts the 4 striped marbles — they are both red and blue. The correct count is: red-only (5) + blue-only (3) + striped (4) = 12. Probability = $\frac{12}{20} = 0.6$.

The simple addition overcounts when events can happen together. That's the core problem the Addition Rule solves.

The Precise Statement

Where:

• $P(A \cup B)$ = probability that A or B (or both) occur
• $P(A \cap B)$ = probability that both A and B occur together

The subtraction of $P(A \cap B)$ removes the double-counted overlap.

Two Special Cases

Case 1: Mutually exclusive events (can't happen together)

If $A$ and $B$ cannot both occur, then $P(A \cap B) = 0$, and the rule simplifies to:

This is the "red or blue marble" case — no overlap, so just add.

Case 2: Events that can overlap (general case)

You must subtract the overlap. This is the "striped marble" case.

Why It Works — A Visual Argument

Draw a rectangle for all possible outcomes. Inside, draw two overlapping circles — one for event A, one for event B. The overlap region is $A \cap B$.

• $P(A)$ counts everything in circle A.
• $P(B)$ counts everything in circle B.
• Adding them counts the overlap twice.
• Subtracting $P(A \cap B)$ once corrects that.

The result is exactly the area covered by either circle — which is $P(A \cup B)$.

Worked Example

A class has 30 students. 18 play cricket, 15 play football, and 8 play both. One student is chosen at random.

Question: What's the probability the student plays cricket or football?

Let $C$ = plays cricket, $F$ = plays football.

$P(C) = \frac{18}{30}$, $P(F) = \frac{15}{30}$, $P(C \cap F) = \frac{8}{30}$

Using the rule: …