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\( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
In Grade 8 you counted outcomes. In Grade 9 you organise them — and that is what makes compound events manageable.
This lesson covers:
\( P(E) = \dfrac{\text{favourable outcomes}}{\text{possible outcomes}} \)
This carries over unchanged from Grade 8, and it still requires the outcomes to be equally likely.
\( 0 \;\leq\; P(E) \;\leq\; 1 \)
Every probability sits between 0 and 1.
\( P(E) + P(\text{not } E) = 1 \)
Key idea: the probabilities of all the outcomes in a sample space always total 1. That fact is used constantly in Grade 9.
\( \text{rel. freq.} = \dfrac{\text{times it happened}}{\text{number of trials}} \)
Theoretical probability predicts. Relative frequency measures what actually happened.
\( 0{,}52 \;\approx\; 0{,}5 \)
The theoretical value is \(0{,}5\). The experiment gave \(0{,}52\) — close, but not identical.
\( \text{not equally likely} \Rightarrow \text{measure it} \)
Key idea: when outcomes are not equally likely — a drawing pin, a bent coin, a real survey — the formula does not apply. You must run the experiment and use relative frequency.
Two dice give \(6 \times 6 = 36\) outcomes. Every cell is one of them, and all 36 are equally likely.
\( P(\text{sum} = 7) = \dfrac{6}{36} = \dfrac{1}{6} \)
Count the shaded cells: \((1;6), (2;5), (3;4), (4;3), (5;2), (6;1)\) — that is 6 of the 36.
\( 2 \times 6 = 12 \qquad P(\text{head and } 6) = \dfrac{1}{12} \)
The same idea with two different objects: 2 coin outcomes down the side, 6 die outcomes across the top.
\( \text{every cell} = \text{one outcome} \)
Key idea: multiply to get the total number of cells, then count the cells that match the event. Drawing the table means no outcome can be missed.
A tree suits actions that happen one after another. Follow each path from left to right to read an outcome.
\( P(HH) = \dfrac{1}{4} \qquad P(\text{at least one head}) = \dfrac{3}{4} \)
\( 3 \text{ outcomes} \quad \times \)
\( \text{one path} = \text{one outcome} \)
Key idea: use a tree when the actions are sequential. Count the end-points to get the sample space, then count the paths that match.
30 learners were surveyed. 12 play soccer, 15 play netball, and 5 play both.
The shaded region is A and B — the learners who play both sports.
The shaded region is A or B — everyone in either circle, including the overlap.
The shaded region is everything outside both circles.
\( n(A \cup B) = n(A) + n(B) - n(A \cap B) \)
Adding 12 and 15 counts the 5 who do both twice. Subtract them once to correct it:
\( \text{fill the overlap first} \)
Key idea: a Venn diagram turns a wordy survey question into a picture you can read numbers off. Always start with the "both" region.
Two events are mutually exclusive when they cannot both happen. On a Venn diagram the circles do not touch.
\( P(A \text{ or } B) = P(A) + P(B) \)
When there is no overlap, there is nothing to subtract — so you simply add.
\( P(\text{even or} > 4) \neq \dfrac{3}{6} + \dfrac{2}{6} \)
Rolling a die: \(A\) = even, \(B\) = greater than 4. Are these mutually exclusive?
\( \text{ask: can both happen?} \)
Key idea: before you add two probabilities, always check whether the events can happen at the same time. If they can, subtract the overlap.
\( P(\text{not } A) = 1 - P(A) \)
The complement of an event is everything that is not that event.
\( \text{complementary} \;\neq\; \text{mutually exclusive} \)
These two ideas are close but not identical.
\( P(\text{at least one}) = 1 - P(\text{none}) \)
When a question says "at least one", the complement is usually far quicker.
\( P(A) + P(\text{not } A) = 1 \)
Key idea: when "not" or "at least one" appears in a question, subtracting from 1 is usually the shortest route to the answer.
\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \]
You now know how to: