The Logic of Probability Rules#
Before using the Counting Principle (Grade 12’s main new content), you need to be rock-solid on the probability identities from Grade 10–11. These rules appear in almost every probability question.
1. Basic Probability#
$$ P(A) = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}} = \frac{n(A)}{n(S)} $$- $P(A)$ is always between $0$ and $1$ (inclusive).
- $P(A) = 0$ means the event is impossible.
- $P(A) = 1$ means the event is certain.
2. The Complementary Rule#
$$ P(\text{not } A) = P(A') = 1 - P(A) $$When to use it: When it’s easier to calculate the probability of something NOT happening.
Example: The probability of rolling at least one 6 in four rolls of a die.
- Hard: Calculate P(one 6) + P(two 6s) + P(three 6s) + P(four 6s).
- Easy: $P(\text{at least one 6}) = 1 - P(\text{no 6s}) = 1 - \left(\frac{5}{6}\right)^4 = 1 - \frac{625}{1296} = \frac{671}{1296}$
3. The Addition Rule (OR)#
$$ P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) $$Logic: If you just add $P(A) + P(B)$, you double-count the outcomes that are in BOTH events. Subtracting $P(A \text{ and } B)$ corrects this.
Venn Diagram: $P(A \text{ or } B)$ is the entire shaded region covering both circles.
Example#
In a class of 30 learners: 18 play soccer, 12 play cricket, 5 play both.
$$ P(\text{soccer or cricket}) = \frac{18}{30} + \frac{12}{30} - \frac{5}{30} = \frac{25}{30} = \frac{5}{6} $$4. Mutually Exclusive Events#
Two events are mutually exclusive if they cannot happen at the same time: $P(A \text{ and } B) = 0$.
The simplified Addition Rule:
$$ P(A \text{ or } B) = P(A) + P(B) $$Example: Rolling a die. Let $A$ = rolling a 2, $B$ = rolling a 5. You can’t roll a 2 AND a 5 at the same time, so:
$$ P(A \text{ or } B) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} $$How to test: If $P(A \text{ and } B) = 0$, the events are mutually exclusive.
5. Independent Events#
Two events are independent if one happening does NOT affect the probability of the other:
$$ P(A \text{ and } B) = P(A) \times P(B) $$The Product Rule: For independent events, “AND” means multiply.
Example#
Rolling a die and flipping a coin. Let $A$ = rolling a 6, $B$ = getting heads.
$$ P(A \text{ and } B) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12} $$How to Test for Independence#
Calculate $P(A) \times P(B)$ and compare it to $P(A \text{ and } B)$.
- If they’re equal → Independent.
- If they’re not equal → Dependent (also called “not independent”).
Example: Testing Independence#
From a survey: $P(A) = 0.4$, $P(B) = 0.5$, $P(A \text{ and } B) = 0.2$.
$P(A) \times P(B) = 0.4 \times 0.5 = 0.2 = P(A \text{ and } B)$ ✓
Conclusion: $A$ and $B$ are independent.
6. Dependent Events and Conditional Probability#
If events are dependent, the probability of the second event changes depending on what happened first.
$$ P(A \text{ and } B) = P(A) \times P(B|A) $$Where $P(B|A)$ means “the probability of $B$ given that $A$ has already happened.”
Example: Drawing Cards Without Replacement#
From a standard 52-card deck, you draw 2 cards without replacement. $P(\text{both aces}) = P(\text{1st ace}) \times P(\text{2nd ace | 1st was ace})$
$$ = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221} $$The second probability changed because one ace was already removed.
7. Tree Diagrams#
Tree diagrams are the best visual tool for multi-step probability problems.
The Rules#
- Each branch represents an outcome.
- Write the probability on each branch.
- To find $P(\text{path})$: multiply along the branches.
- To find $P(\text{event})$: add the probabilities of all paths that lead to that event.
Example: Two Dice Problem#
A bag contains 3 red and 2 blue balls. Two balls are drawn without replacement.
First draw: $P(R) = \frac{3}{5}$, $P(B) = \frac{2}{5}$
Second draw (given first was Red): $P(R) = \frac{2}{4}$, $P(B) = \frac{2}{4}$
Second draw (given first was Blue): $P(R) = \frac{3}{4}$, $P(B) = \frac{1}{4}$
$P(\text{both red}) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$
$P(\text{one of each}) = \frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4} = \frac{6}{20} + \frac{6}{20} = \frac{12}{20} = \frac{3}{5}$
8. Venn Diagram Calculations#
For two events $A$ and $B$ with a sample space $S$:
| Region | Formula |
|---|---|
| Only $A$ (not $B$) | $P(A) - P(A \text{ and } B)$ |
| Only $B$ (not $A$) | $P(B) - P(A \text{ and } B)$ |
| Both $A$ and $B$ | $P(A \text{ and } B)$ |
| Neither $A$ nor $B$ | $1 - P(A \text{ or } B)$ |
Example#
$P(A) = 0.6$, $P(B) = 0.5$, $P(A \text{ or } B) = 0.8$.
Find $P(A \text{ and } B)$:
$$ 0.8 = 0.6 + 0.5 - P(A \text{ and } B) $$$$ P(A \text{ and } B) = 0.3 $$Find $P(\text{neither})$:
$$ P(\text{neither}) = 1 - 0.8 = 0.2 $$Test independence: $P(A) \times P(B) = 0.6 \times 0.5 = 0.3 = P(A \text{ and } B)$ → Independent ✓
🚨 Common Mistakes#
- Confusing mutually exclusive with independent: These are DIFFERENT concepts.
- Mutually exclusive: $P(A \text{ and } B) = 0$ (can’t happen together).
- Independent: $P(A \text{ and } B) = P(A) \times P(B)$ (don’t affect each other).
- If events are mutually exclusive, they are almost never independent (unless one has probability 0).
- Forgetting “without replacement”: If balls/cards are NOT put back, the total and favourables change for the second draw.
- Adding instead of multiplying for “AND”: “OR” = add (after correcting for overlap). “AND” = multiply (for independent events).
- Not reading the Venn diagram correctly: “Only A” means the part of A that does NOT overlap with B.
💡 Pro Tip: The “At Least One” Shortcut#
Whenever a question says “at least one”, use the complement:
$$ P(\text{at least one}) = 1 - P(\text{none}) $$This is almost always much simpler than calculating each case individually.