Q.In a class of 60 students, 30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that —

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: …