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Without looking at your notes, write down the formula for probability, one thing about Venn diagrams, and one thing about tree diagrams.
All the tools we have built this unit, connected together.
Mixing up which tool to use: use a Venn diagram when you have overlapping groups/sets (e.g., students who play sport AND music). Use a tree diagram when you have sequential stages (e.g., draw a marble, then roll a die). Using the wrong tool will not give wrong answers, but the right tool makes the problem much easier.
The probability of an event for equally likely outcomes:
$$P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$$The scale: $0 \leq P \leq 1$. Impossible $= 0$. Certain $= 1$. Even chance $= 0.5$.
Complement rule: $P(A') = 1 - P(A)$.
Example: A bag has 4 red, 3 blue, 2 green marbles (9 total). $P(\text{red}) = \frac{4}{9}$. $P(\text{not red}) = 1 - \frac{4}{9} = \frac{5}{9}$.
For compound experiments, use the counting principle: $|S| = m \times n$.
Example: A coin is tossed and a 3-sector spinner (1, 2, 3) is spun. $|S| = 2 \times 3 = 6$.
Systematic list: $(H,1),\,(H,2),\,(H,3),\,(T,1),\,(T,2),\,(T,3)$.
$P(\text{Head and odd number}) = \dfrac{2}{6} = \dfrac{1}{3}$ (outcomes: $(H,1)$ and $(H,3)$).
Key steps: always start with $n(A \cap B)$, then compute A only, B only, neither.
Worked example: $n(\xi) = 50$, $n(A) = 28$, $n(B) = 22$, $n(A \cap B) = 10$.
$P(\text{neither}) = \dfrac{10}{50} = \dfrac{1}{5}$. Addition rule check: $\dfrac{28}{50} + \dfrac{22}{50} - \dfrac{10}{50} = \dfrac{40}{50} = \dfrac{4}{5} = P(A \cup B)$ ✓
Key rule: write probabilities on branches; multiply along a path; add paths for "or".
Worked example: Two spinners each numbered 1–3. Find $P(\text{product} > 4)$.
All pairs $(s_1, s_2)$: total $3 \times 3 = 9$ outcomes.
Products: $(1,1)=1,\,(1,2)=2,\,(1,3)=3,\,(2,1)=2,\,(2,2)=4,\,(2,3)=6,\,(3,1)=3,\,(3,2)=6,\,(3,3)=9$.
Products $> 4$: $(2,3),\,(3,2),\,(3,3)$, 3 outcomes.
$$P(\text{product} > 4) = \frac{3}{9} = \frac{1}{3}$$Theoretical: count equally likely outcomes. No data needed.
Experimental: $P \approx \dfrac{\text{frequency}}{\text{total trials}}$. Based on real results.
Law of large numbers: more trials $\Rightarrow$ experimental probability closer to theoretical.
Expected frequency: $\text{expected} = P \times n$ (number of trials).
Example: Fair die rolled 300 times. Expected number of sixes $= \frac{1}{6} \times 300 = 50$. An experimental result of 48 is entirely consistent with a fair die.
Master formula set for Unit 4 Probability:
A Venn diagram has $n(\xi)=50$, $n(A)=28$, $n(B)=22$, $n(A \cap B)=10$. What is $P(\text{A only})$?
Two spinners each numbered 1–3 are spun. What is $P(\text{product} > 4)$?
A spinner is spun 80 times. Outcome A appears 24 times. What is the experimental probability of outcome A?
A fair die is rolled. What is $P(\text{not rolling a 6})$?
A spinner has the letters A, B, C, D, E and a die numbered 1–4 is rolled. How many outcomes does the combined sample space have?
Q6. In a group of 50 people, 28 like tea (T), 22 like coffee (C), and 10 like both. (a) Draw and complete a Venn diagram. (b) Find $n(\text{neither})$. (c) Find $P(\text{neither T nor C})$.
Q7. Two spinners are each numbered 1, 2, 3. They are both spun. (a) How many outcomes in the sample space? (b) Find $P(\text{product} > 4)$ by listing all favourable outcomes. Show all working.
Q8. A bag has 5 red (R), 3 blue (B), and 2 green (G) marbles. Two marbles are drawn with replacement. (a) Using a tree diagram, find $P(\text{both same colour})$. Show all branch probabilities and calculations.
Q9. A Year 8 class collected data on 60 students: 35 own a bike, 40 own a skateboard, and 22 own both. (a) Draw a Venn diagram for this data. (b) A different class instead recorded their results as a two-stage process: first they asked whether each student owns a bike, then whether that same student also owns a skateboard. Which representation, a Venn diagram or a tree diagram, better matches this second class's step-by-step recording method, and why? (c) Using your Venn diagram from part (a), find $P(\text{owns exactly one of the two items})$.
Q10. For each scenario, name the best technique, a tree diagram, a Venn diagram, a two-way table, or simple counting, and justify your choice in one sentence. (a) A card is drawn from a deck, replaced, then a second card is drawn, you want $P(\text{two hearts})$. (b) In a survey, some students play both rugby and soccer, some play only one, and some play neither. (c) A cafe has 3 types of bread and 4 types of filling, you want the number of possible sandwich combinations.
T only $= 28 - 10 = 18$. C only $= 22 - 10 = 12$. Both $= 10$.
$n(\text{neither}) = 50 - 18 - 10 - 12 = 10$
$P(\text{neither}) = \dfrac{10}{50} = \dfrac{1}{5}$
Sample space: $3 \times 3 = 9$ outcomes.
All products: $(1,1)=1,\,(1,2)=2,\,(1,3)=3,\,(2,1)=2,\,(2,2)=4,\,(2,3)=6,\,(3,1)=3,\,(3,2)=6,\,(3,3)=9$.
Products $> 4$: $(2,3),\,(3,2),\,(3,3)$, 3 outcomes.
$P(\text{product} > 4) = \dfrac{3}{9} = \dfrac{1}{3}$
With replacement: $P(R) = \frac{5}{10} = \frac{1}{2}$, $P(B) = \frac{3}{10}$, $P(G) = \frac{2}{10} = \frac{1}{5}$ at every draw.
$P(RR) = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$
$P(BB) = \dfrac{3}{10} \times \dfrac{3}{10} = \dfrac{9}{100}$
$P(GG) = \dfrac{1}{5} \times \dfrac{1}{5} = \dfrac{1}{25} = \dfrac{4}{100}$
$P(\text{same colour}) = \dfrac{25}{100} + \dfrac{9}{100} + \dfrac{4}{100} = \dfrac{38}{100} = \dfrac{19}{50}$
(a) Bike only $= 35 - 22 = 13$. Skateboard only $= 40 - 22 = 18$. Both $= 22$. Neither $= 60 - 13 - 22 - 18 = 7$.
(b) A tree diagram better matches the second class's method, because their data was collected as an ordered, sequential process (bike question first, then a follow-up skateboard question for that same student), which is exactly what tree diagram branches represent. A Venn diagram instead represents overlapping groups collected all at once, which suits the first class's data.
(c) $P(\text{exactly one}) = \dfrac{13 + 18}{60} = \dfrac{31}{60}$
(a) Tree diagram, because the two draws happen in sequence (one after the other) and each draw's probability is written on a branch, then multiplied along the path.
(b) Venn diagram, because the data describes overlapping groups (both, only one, neither) rather than a sequence of stages.
(c) Simple counting (the counting principle), because you are combining two independent choices made at the same time, so $|S| = 3 \times 4 = 12$, no branches or overlapping sets are needed.
Design your own probability problem that uses either a Venn diagram or a tree diagram to solve it. Your answer must involve a probability between 0.2 and 0.4. Write the full problem, draw the diagram, and show the complete solution including a check that your final probability is in the required range.
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