Venn Diagrams, Addition Rule & Event Relationships#
In Grade 10, you used basic Venn diagrams and the sample space. In Grade 11, we formalise the rules and learn to classify events as mutually exclusive, complementary, or independent — and understand why these are all different things.
1. The Addition Rule — The Master Formula#
$$\boxed{P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)}$$This works for every pair of events, always.
Why subtract the intersection?#
$P(A)$ counts everything in the A circle. $P(B)$ counts everything in the B circle. But the overlap region ($A \text{ and } B$) gets counted twice — once in $P(A)$ and once in $P(B)$. Subtracting it once fixes the double-count.
Special Cases#
| Condition | Simplification |
|---|---|
| $A$ and $B$ are mutually exclusive ($P(A \text{ and } B) = 0$) | $P(A \text{ or } B) = P(A) + P(B)$ |
| $A$ and $B$ are independent | $P(A \text{ or } B) = P(A) + P(B) - P(A) \times P(B)$ |
Rearranging the Addition Rule#
The addition rule can be rearranged to find any one of its four values:
$$P(A \text{ and } B) = P(A) + P(B) - P(A \text{ or } B)$$$$P(A) = P(A \text{ or } B) - P(B) + P(A \text{ and } B)$$Exam tip: If the question gives you three of the four values, rearrange to find the fourth.
2. Filling a Venn Diagram — Two Events#
Golden rule: Always start from the INSIDE and work OUTWARD.
Worked Example 1#
In a class of 40 learners: 25 play soccer (S), 18 play cricket (C), and 10 play both.
Step 1 — Fill the intersection first: $n(S \text{ and } C) = 10$
Step 2 — S only: $25 - 10 = 15$
Step 3 — C only: $18 - 10 = 8$
Step 4 — Neither: $40 - 15 - 10 - 8 = 7$
| Region | Count | Probability |
|---|---|---|
| S only | 15 | $\frac{15}{40}$ |
| S and C | 10 | $\frac{10}{40}$ |
| C only | 8 | $\frac{8}{40}$ |
| Neither | 7 | $\frac{7}{40}$ |
| Total | 40 | 1 |
Verify with addition rule: $P(S \text{ or } C) = \frac{25}{40} + \frac{18}{40} - \frac{10}{40} = \frac{33}{40}$
Check: $\frac{15 + 10 + 8}{40} = \frac{33}{40}$ ✓
Worked Example 2 — Finding the Intersection#
$P(A) = 0.6$, $P(B) = 0.5$, $P(A \text{ or } B) = 0.8$. Find $P(A \text{ and } B)$.
$P(A \text{ and } B) = P(A) + P(B) - P(A \text{ or } B) = 0.6 + 0.5 - 0.8 = 0.3$
Worked Example 3 — Using “Neither”#
In a survey of 100 people: 60 like coffee, 45 like tea, and 15 like neither. How many like both?
$n(\text{coffee or tea}) = 100 - 15 = 85$
$n(\text{both}) = n(\text{coffee}) + n(\text{tea}) - n(\text{coffee or tea}) = 60 + 45 - 85 = 20$
3. The Complement Rule#
$$\boxed{P(\text{not } A) = 1 - P(A)}$$Also written as $P(A') = 1 - P(A)$.
Why It’s So Useful#
Either A happens or it doesn’t — there’s no third option. So:
$$P(A) + P(A') = 1$$The Power of the Complement: “At Least One”#
Whenever a problem asks for “at least one”, it’s almost always easier to calculate:
$$P(\text{at least one}) = 1 - P(\text{none})$$This avoids having to list every possible case.
Worked Example 4#
$P(A) = 0.7$. Find $P(\text{not } A)$.
$P(A') = 1 - 0.7 = 0.3$
Worked Example 5#
$P(A \text{ or } B) = 0.85$. Find the probability that neither A nor B occurs.
$P(\text{neither}) = P((A \text{ or } B)') = 1 - 0.85 = 0.15$
4. Three Types of Event Relationships#
Type 1: Mutually Exclusive Events#
Definition: Events that cannot happen at the same time. Their Venn circles don’t overlap.
$$P(A \text{ and } B) = 0$$Example: Rolling a 3 and rolling a 5 on the same die — you can’t get both.
Consequence for the addition rule: $P(A \text{ or } B) = P(A) + P(B)$ (no overlap to subtract).
Type 2: Complementary Events#
Definition: Two events where exactly one MUST happen. They are mutually exclusive AND exhaustive (they cover the entire sample space with no gaps).
$$P(A) + P(A') = 1$$Example: Passing or failing a test — exactly one of these will happen.
💡 All complementary events are mutually exclusive, but NOT all mutually exclusive events are complementary. Rolling a 1 and rolling a 2 are mutually exclusive, but they’re not complementary (you could roll a 3, 4, 5, or 6 instead).
Type 3: Independent Events#
Definition: One event does NOT affect the probability of the other.
$$P(A \text{ and } B) = P(A) \times P(B)$$Example: Flipping heads on a coin AND rolling a 6 on a die — the coin doesn’t care what the die does.
5. The Critical Distinction: Independent ≠ Mutually Exclusive#
This is the most commonly confused concept in Grade 11 probability. Students assume these are the same thing, but they are almost opposites:
| Mutually Exclusive | Independent | |
|---|---|---|
| Meaning | Cannot happen together | Don’t affect each other |
| $P(A \text{ and } B)$ | $= 0$ | $= P(A) \times P(B)$ |
| Venn diagram | No overlap (circles separate) | Circles DO overlap |
| Dependent? | YES — very dependent! | No — no influence |
| Can both happen? | No | Yes |
⚠️ If events are mutually exclusive (with non-zero probabilities), they are ALWAYS DEPENDENT. Think about it: if A happens, then B definitely cannot happen — that’s a very strong dependency! The occurrence of A changes $P(B)$ from some positive value to exactly 0.
Why This Matters#
An exam might ask: “Events A and B are mutually exclusive. Are they independent?” The answer is NO (unless one of them has probability 0).
Proof: If $P(A) > 0$ and $P(B) > 0$, and $A$ and $B$ are mutually exclusive:
$P(A \text{ and } B) = 0$
$P(A) \times P(B) > 0$
$0 \neq P(A) \times P(B)$ → Not independent
6. Testing for Independence — Worked Examples#
Worked Example 6#
$P(A) = 0.4$, $P(B) = 0.5$, $P(A \text{ and } B) = 0.2$. Are A and B independent?
Test: $P(A) \times P(B) = 0.4 \times 0.5 = 0.2$
$P(A \text{ and } B) = 0.2 = P(A) \times P(B)$ ✓ → Independent
Worked Example 7#
$P(A) = 0.3$, $P(B) = 0.4$, $P(A \text{ or } B) = 0.58$. Are A and B independent?
First find $P(A \text{ and } B)$:
$P(A \text{ and } B) = 0.3 + 0.4 - 0.58 = 0.12$
Test: $P(A) \times P(B) = 0.3 \times 0.4 = 0.12$
$0.12 = 0.12$ ✓ → Independent
Worked Example 8#
$P(A) = 0.5$, $P(B) = 0.6$, $P(A \text{ and } B) = 0.2$. Are A and B independent?
Test: $P(A) \times P(B) = 0.5 \times 0.6 = 0.3$
$0.2 \neq 0.3$ → Dependent
7. Algebraic Probability Problems#
These are the hardest Venn diagram questions. You’re given probabilities in terms of a variable and must solve for it.
Worked Example 9#
$P(A) = 0.5$, $P(B) = 0.4$, and A and B are independent. Find $P(A \text{ or } B)$.
$P(A \text{ and } B) = P(A) \times P(B) = 0.5 \times 0.4 = 0.2$ (independent)
$P(A \text{ or } B) = 0.5 + 0.4 - 0.2 = 0.7$
Worked Example 10#
$P(A) = 0.3$ and $P(A \text{ or } B) = 0.6$. If A and B are mutually exclusive, find $P(B)$.
Mutually exclusive → $P(A \text{ or } B) = P(A) + P(B)$
$0.6 = 0.3 + P(B)$
$P(B) = 0.3$
Worked Example 11#
$P(A) = x$, $P(B) = 0.4$, $P(A \text{ and } B) = 0.12$. If A and B are independent, find $x$.
$P(A) \times P(B) = P(A \text{ and } B)$
$x \times 0.4 = 0.12$
$x = 0.3$
Worked Example 12#
In a group of 200 learners, $x$ play basketball, 80 play netball, 30 play both, and 70 play neither. Find $x$.
$n(\text{B or N}) = 200 - 70 = 130$
$n(\text{B}) + n(\text{N}) - n(\text{B and N}) = n(\text{B or N})$
$x + 80 - 30 = 130$
$x = 80$
8. Three-Event Venn Diagrams#
With three events (A, B, C), the diagram has 8 regions (including “neither”). The filling strategy is the same — work from the inside out:
The Method#
- Start with the triple intersection: $n(A \text{ and } B \text{ and } C)$
- Fill pairwise intersections: Subtract the triple from each pair
- $n(A \text{ and } B \text{ only}) = n(A \text{ and } B) - n(A \text{ and } B \text{ and } C)$
- Similarly for $n(A \text{ and } C \text{ only})$ and $n(B \text{ and } C \text{ only})$
- Fill “only” regions: Subtract all overlaps from each event’s total
- Fill “neither”: Total minus the sum of all 7 regions inside the circles
Worked Example 13#
100 learners: 50 take Maths (M), 40 take Science (S), 30 take English (E). Also: 15 take M and S, 10 take M and E, 8 take S and E, and 5 take all three. Find the number in each region and the number taking none of these subjects.
Step 1 — Triple intersection: $n(M \cap S \cap E) = 5$
Step 2 — Pairwise only:
- M and S only: $15 - 5 = 10$
- M and E only: $10 - 5 = 5$
- S and E only: $8 - 5 = 3$
Step 3 — Only one subject:
- M only: $50 - 10 - 5 - 5 = 30$
- S only: $40 - 10 - 3 - 5 = 22$
- E only: $30 - 5 - 3 - 5 = 17$
Step 4 — Neither: $100 - (30 + 10 + 5 + 22 + 3 + 17 + 5) = 100 - 92 = 8$
Check: All 8 regions sum to 100 ✓
Questions you can now answer:
- $P(\text{exactly one subject}) = \frac{30 + 22 + 17}{100} = 0.69$
- $P(\text{at least two subjects}) = \frac{10 + 5 + 3 + 5}{100} = 0.23$
- $P(\text{M or S}) = \frac{30 + 10 + 5 + 22 + 3 + 5}{100} = 0.75$
9. Summary: Choosing the Right Tool#
| Given information | Strategy |
|---|---|
| Counts/numbers and asked to fill a diagram | Fill Venn diagram (inside → outside) |
| Probabilities and asked to find overlap | Use addition rule: $P(A \cap B) = P(A) + P(B) - P(A \cup B)$ |
| Asked “are they independent?” | Test: $P(A \cap B) = P(A) \times P(B)$? |
| Asked “are they mutually exclusive?” | Test: $P(A \cap B) = 0$? |
| “At least one” question | Use complement: $1 - P(\text{none})$ |
| Given “neither” count | Subtract from total to get $n(A \cup B)$ |
| Variable in the problem | Set up equation using addition rule, solve algebraically |
🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Forgetting “neither” | After filling the circles, there are often items outside both circles | Calculate: Total $-$ sum of all regions inside circles |
| Double counting | If 25 play soccer and 10 play both, soccer ONLY = $25 - 10 = 15$ | Don’t put the total count (25) in the “only” region |
| Confusing independent and mutually exclusive | They are almost opposites — see the table in Section 5 | ME: can’t happen together. Independent: don’t affect each other |
| Not starting from the middle | Going outside-in leads to incorrect overlap calculations | Always fill the intersection FIRST, then work outward |
| Forgetting the minus in the addition rule | It’s $P(A) + P(B) \mathbf{-} P(A \text{ and } B)$ | The minus prevents double-counting the overlap |
| Saying “mutually exclusive means independent” | ME events are dependent (if A happens, B definitely can’t) | Know the proof: $P(A \cap B) = 0 \neq P(A) \times P(B) > 0$ |
| Three-event diagrams: using raw pair values | $n(A \text{ and } B) = 15$ includes the triple intersection | Subtract the triple: $n(A \text{ and } B \text{ only}) = 15 - n(A \cap B \cap C)$ |
💡 Pro Tips for Exams#
1. The “Algebra” Approach#
If you’re given three of the four addition rule values, rearrange to find the fourth. This is faster than drawing a full Venn diagram.
2. The Independence Test — Show Your Working#
Always write:
“$P(A) \times P(B) = \ldots$” and “$P(A \text{ and } B) = \ldots$”
Then conclude: “Since [equal/not equal], the events [are/are not] independent.”
Markers award marks for the comparison, not just the conclusion.
3. Three-Event Venn — The Counting Check#
After filling all 8 regions, add them up. The sum must equal the total. If it doesn’t, you have an error — find it before moving on.
4. “Or” vs “And” in Exam Questions#
- “or” → addition rule (union, $\cup$, larger region)
- “and” → intersection ($\cap$, overlap region)
- “not” → complement ($1 - P$)
🔗 Related Grade 11 topics:
- Combined Events & Tree Diagrams — applying independence and dependence to multi-step experiments
- Variance & Standard Deviation — contingency tables bridge statistics and probability
📌 Grade 10 foundation: Probability Basics and Venn Diagrams