Grade 11 Mathematics
Grade 11 · Term 4Mathematics

Probability

We extend Grade 10 probability to include conditional probability, the multiplication rule, and independent events.

Week 3

10.1 Conditional Probability & Independent Events

  • Apply conditional probability: $P(A|B)=\frac{P(A\cap B)}{P(B)}$
  • Apply the multiplication rule
  • Test for independence: $P(A\cap B)=P(A)\cdot P(B)$
  • Use contingency tables and tree diagrams
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Real-World Connection

Medical testing uses conditional probability constantly. If a disease affects 1% of the population and a test is 99% accurate, the probability that a positive test result means you actually have the disease is NOT 99% — conditional probability gives the true answer, which surprises most people.

Conditional Probability

P(AB)=P(AB)P(B)P(A|B) = \frac{P(A\cap B)}{P(B)}

$P(A|B)$ = probability of $A$ given $B$ has occurred; $P(B)>0$

Multiplication Rule

P(AB)=P(AB)P(B)=P(BA)P(A)P(A\cap B) = P(A|B)\cdot P(B) = P(B|A)\cdot P(A)

Rearrangement of conditional probability

Property / Rule

Independent Events

Events AA and BB are independent if knowing BB occurred does NOT affect the probability of AA. Test: P(AB)=P(A)P(B)P(A\cap B)=P(A)\cdot P(B), equivalently P(AB)=P(A)P(A|B)=P(A).

P(AB)=P(A)P(B)A,B independentP(A\cap B)=P(A)\cdot P(B)\Leftrightarrow A,B\text{ independent}

🚨 Common Mistake

Independent ≠ Mutually Exclusive! Mutually exclusive events CANNOT both occur; independent events CAN both occur (their joint probability is just the product).

Worked Examples

Worked Example

Conditional probability from contingency table

Problem

In a class of 50: 20 boys, 30 girls. Of the boys, 12 passed; of the girls, 24 passed. Find P(passedboy)P(\text{passed}|\text{boy}) and test independence.
Activity — 8 Questions

CAPS Cognitive Level Distribution

L1 · Knowledge2 Q
L2 · Routine Procedures2 Q
L3 · Complex Procedures2 Q
L4 · Problem Solving2 Q
1
L1 · Knowledge2 marks
State the condition for events AA and BB to be independent.
2
L1 · Knowledge2 marks
P(A)=0.4P(A)=0.4, P(B)=0.5P(B)=0.5, P(AB)=0.2P(A\cap B)=0.2. Are AA and BB independent?
3
L2 · Routine Procedures2 marks
P(B)=0.3P(B)=0.3, P(AB)=0.12P(A\cap B)=0.12. Find P(AB)P(A|B).
4
L2 · Routine Procedures3 marks
A bag has 3 red and 5 blue balls. Two are drawn without replacement. Find P(both red)P(\text{both red}).
5
L3 · Complex Procedures4 marks
P(A)=0.6P(A)=0.6, P(B)=0.5P(B)=0.5, P(AB)=0.8P(A\cup B)=0.8. Find P(AB)P(A|B).
6
L3 · Complex Procedures6 marks
A disease has 2% prevalence. Test is 95% sensitive (P(+|disease)=0.95) and 90% specific (P(−|no disease)=0.90). Find P(disease|positive test).
7
L4 · Problem Solving4 marks
Three fair coins are tossed. Given at least one head occurred, find the probability of all three heads.
8
L4 · Problem Solving5 marks
Events AA and BB are independent. P(A)=pP(A)=p and P(B)=pP(B)=p. Find the range of pp for P(AB)<0.75P(A\cup B)<0.75.
Probability Grade 11 Maths CAPS Notes & Examples | MathSciBuddy