The Big Idea: Does One Event Affect Another?#
In Grade 10, you calculated the probability of single events. Grade 11 asks a deeper question: when two events happen together, does the first one change the chances of the second?
This single question — “does A affect B?” — is the foundation of everything on this page.
1. Independent vs Dependent Events#
| Type | Meaning | Real-life example |
|---|---|---|
| Independent | The outcome of A has no effect on the probability of B | Flipping a coin AND rolling a die — the coin doesn’t care what the die does |
| Dependent | The outcome of A changes the probability of B | Drawing cards WITHOUT replacement — what you draw first changes what’s left |
The Product Rule for Independent Events#
If A and B are independent:
$$\boxed{P(A \text{ and } B) = P(A) \times P(B)}$$This only works when the events are truly independent. If they’re dependent, you need a different approach (see Section 3).
The Mathematical Test for Independence#
You’re given $P(A)$, $P(B)$, and $P(A \text{ and } B)$. To test whether the events are independent:
Calculate $P(A) \times P(B)$ and compare it to $P(A \text{ and } B)$.
| Result | Conclusion |
|---|---|
| $P(A \text{ and } B) = P(A) \times P(B)$ | Independent |
| $P(A \text{ and } B) \neq P(A) \times P(B)$ | Dependent |
Worked Example 1 — Independent#
$P(A) = 0.4$, $P(B) = 0.5$, $P(A \text{ and } B) = 0.2$
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 2 — Dependent#
$P(A) = 0.3$, $P(B) = 0.6$, $P(A \text{ and } B) = 0.1$
Test: $P(A) \times P(B) = 0.3 \times 0.6 = 0.18$
$P(A \text{ and } B) = 0.1 \neq 0.18$ → Dependent
Worked Example 3 — Finding Missing Probabilities Using Independence#
Events A and B are independent. $P(A) = 0.3$ and $P(A \text{ or } B) = 0.72$. Find $P(B)$.
Since A and B are independent: $P(A \text{ and } B) = P(A) \times P(B) = 0.3 \times P(B)$
Using the addition rule:
$P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$
$0.72 = 0.3 + P(B) - 0.3 \times P(B)$
$0.42 = P(B)(1 - 0.3) = 0.7 \times P(B)$
$$P(B) = \frac{0.42}{0.7} = 0.6$$2. Conditional Probability — “Given That”#
Conditional probability answers the question: “What is the probability of B, given that A has already happened?”
$$\boxed{P(B \mid A) = \frac{P(A \text{ and } B)}{P(A)}}$$The vertical bar “$\mid$” is read as “given” or “given that.”
Why This Matters#
When A has already happened, the sample space shrinks. You’re no longer looking at all possible outcomes — only those where A occurred.
Worked Example 4#
In a class of 40 learners: 25 take Maths, 18 take Science, and 10 take both. A learner is chosen at random. Given that the learner takes Maths, what is the probability they also take Science?
$P(\text{Science} \mid \text{Maths}) = \frac{P(\text{Science and Maths})}{P(\text{Maths})} = \frac{10/40}{25/40} = \frac{10}{25} = 0.4$
Interpretation: Of the 25 Maths learners, 10 also take Science. That’s a 40% chance.
The Connection to Independence#
If A and B are independent, then knowing A happened doesn’t change the probability of B:
$$P(B \mid A) = P(B)$$This gives us another way to test for independence: check whether $P(B \mid A) = P(B)$.
3. Dependent Events — Without Replacement#
The most common source of dependent events in exams is drawing without replacement. Each draw changes the total number of items and potentially the number of “success” items.
The General Rule for Dependent Events#
$$P(A \text{ and } B) = P(A) \times P(B \mid A)$$This is the multiplication rule — it always works, whether events are independent or dependent.
Worked Example 5 — Drawing Without Replacement#
A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement. Find the probability of drawing two red balls.
First draw: $P(R_1) = \frac{5}{8}$
Second draw (given first was red): One red ball removed → 4 red and 3 blue left (7 total)
$P(R_2 \mid R_1) = \frac{4}{7}$
$$P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$$Worked Example 6 — With Replacement (Comparison)#
Same bag (5 red, 3 blue), but now with replacement.
First draw: $P(R_1) = \frac{5}{8}$
Second draw (ball replaced): Still 5 red and 3 blue (8 total)
$P(R_2) = \frac{5}{8}$
$$P(\text{both red}) = \frac{5}{8} \times \frac{5}{8} = \frac{25}{64}$$Compare: Without replacement gives $\frac{5}{14} \approx 0.357$. With replacement gives $\frac{25}{64} \approx 0.391$. Replacing makes getting two reds slightly more likely because you haven’t reduced the supply.
4. Tree Diagrams#
A tree diagram maps every possible path through a multi-step experiment. It is the most powerful tool for combined events.
The Three Rules of Tree Diagrams#
- Each set of branches from a node sums to 1 (something must happen)
- Multiply along a path to get that specific outcome’s probability
- Add separate paths to find the total probability of an event
Worked Example 7 — Without Replacement (Full Tree)#
Bag: 4 red, 6 blue. Two balls drawn without replacement.
First draw:
$P(R) = \frac{4}{10} = \frac{2}{5}$, $P(B) = \frac{6}{10} = \frac{3}{5}$
Second draw (if 1st was R): 3 red, 6 blue left (9 total)
$P(R \mid R) = \frac{3}{9} = \frac{1}{3}$, $P(B \mid R) = \frac{6}{9} = \frac{2}{3}$
Second draw (if 1st was B): 4 red, 5 blue left (9 total)
$P(R \mid B) = \frac{4}{9}$, $P(B \mid B) = \frac{5}{9}$
| Path | Calculation | Probability |
|---|---|---|
| RR | $\frac{4}{10} \times \frac{3}{9}$ | $\frac{12}{90}$ |
| RB | $\frac{4}{10} \times \frac{6}{9}$ | $\frac{24}{90}$ |
| BR | $\frac{6}{10} \times \frac{4}{9}$ | $\frac{24}{90}$ |
| BB | $\frac{6}{10} \times \frac{5}{9}$ | $\frac{30}{90}$ |
Check: $\frac{12 + 24 + 24 + 30}{90} = \frac{90}{90} = 1$ ✓
Questions you can answer from this tree:
$P(\text{both same colour}) = P(RR) + P(BB) = \frac{12 + 30}{90} = \frac{42}{90} = \frac{7}{15}$
$P(\text{exactly one red}) = P(RB) + P(BR) = \frac{24 + 24}{90} = \frac{48}{90} = \frac{8}{15}$
$P(\text{at least one red}) = 1 - P(BB) = 1 - \frac{30}{90} = \frac{60}{90} = \frac{2}{3}$
Worked Example 8 — Three-Step Tree Diagram#
A coin is flipped three times. Find the probability of getting exactly 2 heads.
Each flip: $P(H) = \frac{1}{2}$, $P(T) = \frac{1}{2}$ (independent events — with replacement, effectively)
The paths with exactly 2 heads:
| Path | Probability |
|---|---|
| HHT | $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$ |
| HTH | $\frac{1}{8}$ |
| THH | $\frac{1}{8}$ |
$P(\text{exactly 2 heads}) = 3 \times \frac{1}{8} = \frac{3}{8}$
Why 3 paths? There are 3 ways to choose which of the 3 flips is the tail: the 1st, the 2nd, or the 3rd.
5. Contingency Tables (Two-Way Tables)#
A contingency table organises data about two categorical variables. Every Grade 11 contingency table question asks: are these events independent?
How to Read a Contingency Table#
| Category X | Category Y | Total | |
|---|---|---|---|
| Group A | Both A and X | Both A and Y | Total in A |
| Group B | Both B and X | Both B and Y | Total in B |
| Total | Total in X | Total in Y | Grand total |
Every probability comes from dividing a cell by the grand total.
Worked Example 9 — Testing Independence#
| Maths | Science | Total | |
|---|---|---|---|
| Boys | 30 | 20 | 50 |
| Girls | 25 | 25 | 50 |
| Total | 55 | 45 | 100 |
$P(\text{Boy}) = \frac{50}{100} = 0.5$
$P(\text{Maths}) = \frac{55}{100} = 0.55$
$P(\text{Boy and Maths}) = \frac{30}{100} = 0.3$
Test: $P(\text{Boy}) \times P(\text{Maths}) = 0.5 \times 0.55 = 0.275$
$0.3 \neq 0.275$ → Not independent
Being a boy slightly increases the probability of taking Maths (30% actual vs 27.5% expected if independent).
Worked Example 10 — Completing a Contingency Table#
200 learners were surveyed about sport preference. 120 are boys. 90 play rugby, of whom 70 are boys. Determine if gender and sport preference are independent.
Step 1 — Complete the table:
| Rugby | Not Rugby | Total | |
|---|---|---|---|
| Boys | 70 | 50 | 120 |
| Girls | 20 | 60 | 80 |
| Total | 90 | 110 | 200 |
Step 2 — Test:
$P(\text{Boy}) = \frac{120}{200} = 0.6$
$P(\text{Rugby}) = \frac{90}{200} = 0.45$
$P(\text{Boy}) \times P(\text{Rugby}) = 0.6 \times 0.45 = 0.27$
$P(\text{Boy and Rugby}) = \frac{70}{200} = 0.35$
$0.35 \neq 0.27$ → Not independent. Boys are more likely to play rugby than expected.
The Quick Independence Check for Contingency Tables#
Events are independent if every cell satisfies:
$$\text{Expected cell value} = \frac{\text{row total} \times \text{column total}}{\text{grand total}}$$For the Boys–Rugby cell: $\frac{120 \times 90}{200} = 54$. The actual value is 70 — significantly higher. Not independent.
6. “At Least One” Problems — The Complement Strategy#
“At least one” is the most efficient application of the complement rule:
$$\boxed{P(\text{at least one}) = 1 - P(\text{none})}$$Worked Example 11#
A die is rolled 3 times. Find the probability of getting at least one 6.
$P(\text{not a 6}) = \frac{5}{6}$ per roll.
$P(\text{no sixes in 3 rolls}) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$
$$P(\text{at least one 6}) = 1 - \frac{125}{216} = \frac{91}{216} \approx 0.421$$Worked Example 12#
In a factory, Machine A has a 5% defect rate and Machine B has a 3% defect rate (independent). Find the probability that at least one machine produces a defective item.
$P(\text{A not defective}) = 0.95$, $P(\text{B not defective}) = 0.97$
$P(\text{neither defective}) = 0.95 \times 0.97 = 0.9215$
$$P(\text{at least one defective}) = 1 - 0.9215 = 0.0785 \approx 7.85\%$$7. Solving for Unknown Probabilities#
Exam questions often give partial information and ask you to find a missing probability.
Worked Example 13#
$P(A) = 0.4$ and $P(B) = 0.25$. A and B are independent. Find $P(A \text{ or } B)$.
$P(A \text{ and } B) = 0.4 \times 0.25 = 0.1$ (independent)
$P(A \text{ or } B) = 0.4 + 0.25 - 0.1 = 0.55$
Worked Example 14#
$P(A) = 0.5$, $P(A \text{ or } B) = 0.8$, and A and B are independent. Find $P(B)$.
Let $P(B) = b$.
$0.8 = 0.5 + b - 0.5b$
$0.3 = b(1 - 0.5) = 0.5b$
$b = 0.6$
$$P(B) = 0.6$$🚨 Common Mistakes#
| Mistake | Why it’s wrong | Fix |
|---|---|---|
| Confusing independent and mutually exclusive | Mutually exclusive: $P(A \text{ and } B) = 0$. This means they’re actually very dependent — if A happens, B definitely can’t! | See the table in Venn Diagrams & Logic |
| Not adjusting denominator for “without replacement” | After removing one item, the total drops by 1 and the count of the drawn type also drops by 1 | Always recalculate both numerator and denominator after each draw |
| Tree diagram branches not summing to 1 | Each set of branches from a single node must represent all possible outcomes | Add a “check: do these probabilities sum to 1?” step after each node |
| Using grand total as denominator for conditional probability | $P(B \mid A)$ uses $P(A)$ as the denominator, NOT the grand total | Use the formula $P(B \mid A) = \frac{P(A \text{ and } B)}{P(A)}$ |
| Forgetting “at least one = 1 − none” | Calculating “at least one” directly requires listing many paths; the complement is far simpler | Always use $1 - P(\text{none})$ for “at least one” problems |
| Contingency table — testing the wrong cell | You must test $P(A \text{ and } B) = P(A) \times P(B)$ using the same pair of events | Pick any one cell; if it fails, the events are dependent (one cell is enough) |
| Saying “independent” without showing the calculation | The conclusion must be supported by the mathematical test | Always write: “$P(A) \times P(B) = ... \neq P(A \text{ and } B) = ...$, therefore dependent/independent” |
💡 Pro Tips for Exams#
1. The “Replacement” Keyword#
With replacement → independent (probabilities don’t change).
Without replacement → dependent (probabilities change after each draw).
If the question doesn’t say, assume without replacement (this is the default in most exam contexts).
2. Tree Diagram Layout#
Always label:
- Each branch with its probability
- Each endpoint with the combined outcome (e.g., RR, RB, BR, BB)
- The product of each path at the end
Then answer the specific question by adding the relevant paths.
3. Contingency Table Strategy#
- Complete the table first — fill in all row and column totals
- Choose one cell to test for independence
- Calculate $P(\text{row event}) \times P(\text{column event})$ and compare to $P(\text{cell event})$
- Write a conclusion with the values
🔗 Related Grade 11 topics:
- Venn Diagrams & Logic — the addition rule and event relationships
- Variance & Standard Deviation — contingency tables bridge statistics and probability
- Quadratic Equations — probability equations sometimes need algebraic solving