Mathematics • Year 10 • Unit 4 • Lesson 20
Statistics & Probability — Mixed Challenge (Year 10 Maths Finale)
The final worksheet of Year 10 Maths. Bring every Unit 4 idea together: classifying data, five-number summary and box plots, scatter plots and correlation, addition rule, multiplication rule, tree diagrams, with/without replacement, and conditional probability. Spot a Year 10 mistake and design your own investigation.
1. Mixed problems — pick the right Unit 4 tool
Before each question, decide which technique applies. Show working. 3 marks each
1.1 A data set of 12 values: 14, 16, 18, 19, 20, 22, 24, 25, 27, 30, 32, 45. Find the median, IQR and one value you would flag as a possible outlier. Justify your outlier choice in one sentence.
1.2 A biased coin has P(head) = 0.6. The coin is flipped 3 times. Draw a tree diagram and find P(exactly 2 heads).
1.3 A bag contains 3 red, 4 blue, 5 green marbles. Two marbles are drawn without replacement. Find P(both green).
1.4 A two-way table for 200 people: 120 are female, 80 are male; 130 prefer tea, 70 prefer coffee. 90 females prefer tea. Find P(coffee | male) and decide whether sex and drink preference are independent.
1.5 A scatter plot of "temperature (°C)" vs "ice-cream sales ($)" shows points tightly clustered around an upward line. Describe the correlation in one sentence, and explain in one sentence what the slope of a line of best fit would mean in plain English.
1.6 P(A) = 0.5, P(B) = 0.4, P(A and B) = 0.2. (a) Find P(A or B). (b) Find P(A | B). (c) Use the result of (b) to decide whether A and B are independent. (d) Find P(only B, not A) using the Venn-diagram region "B but not A".
2. Find the mistake
A Year 10 student tackles this exam-style question: "A bag contains 5 red and 5 blue marbles. Two marbles are drawn without replacement. Find P(both red)." Exactly one line is wrong. 3 marks
Student's working:
Line 1: "Without replacement" → events are dependent.
Line 2: P(first red) = 5/10 = 1/2.
Line 3: After one red is removed, the bag has 4 red and 5 blue = 9 total.
Line 4: P(second red | first red) = 4/9.
Line 5: P(both red) = 1/2 + 4/9 = 17/18.
(a) Which line is wrong?
(b) Explain why, naming the rule the student should have used.
(c) Write the corrected calculation.
Stuck? "Both red" needs the multiplication rule along a tree branch — not addition. And probability cannot exceed 1.3. Open-ended challenge — design your own mini investigation
This is the last worksheet of Year 10 Maths. Pull every Unit 4 idea into one real investigation. Many valid answers. 4 marks
3.1 Design a complete Year 10 statistical investigation that uses at least three different Unit 4 techniques from this list:
- classification of data (categorical / discrete / continuous),
- five-number summary or box plot,
- scatter plot and correlation,
- two-way table or Venn diagram,
- conditional probability,
- tree diagram (with or without replacement),
- theoretical vs experimental probability.
Write up:
(i) the question your investigation answers (one sentence),
(ii) the data you would collect and how (one paragraph: what variables, who from, how many people),
(iii) the three (or more) techniques you would apply, with a sentence per technique on what you would compute,
(iv) a one-sentence prediction of what you think you would find, with a justification,
(v) one limitation (e.g. sample size, bias, missing variable) and how you would fix it next time.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Five-number summary
12 values in order. Median = average of 6th and 7th = (22+24)/2 = 23.
Lower 6: 14,16,18,19,20,22 → Q1 = (18+19)/2 = 18.5.
Upper 6: 24,25,27,30,32,45 → Q3 = (27+30)/2 = 28.5.
IQR = 28.5 − 18.5 = 10.
Possible outlier: 45. It is more than 1.5 × IQR (=15) above Q3 (28.5 + 15 = 43.5), so 45 lies beyond the upper "fence".
1.2 — Biased coin, 3 flips
P(H) = 0.6, P(T) = 0.4. Exactly 2 heads = HHT + HTH + THH.
HHT = 0.6 × 0.6 × 0.4 = 0.144 (each of the three orderings has the same value).
P(exactly 2 heads) = 3 × 0.144 = 0.432.
1.3 — Two green, no replacement
P(first green) = 5/12. P(second green | first green) = 4/11.
P(both green) = 5/12 × 4/11 = 20/132 = 5/33.
1.4 — Two-way table for tea/coffee
120 female, 80 male. 90 females tea → 30 females coffee. 130 tea total → 40 males tea. 80 male − 40 = 40 males coffee.
| Tea | Coffee | Total
F | 90 | 30 | 120
M | 40 | 40 | 80
T | 130 | 70 | 200 ✓
P(coffee | male) = 40/80 = 0.5. P(coffee) = 70/200 = 0.35. Since 0.5 ≠ 0.35, sex and drink are not independent — males in this sample are more likely to prefer coffee than the overall rate.
1.5 — Ice-cream sales scatter
Strong positive correlation. The slope of the line of best fit gives the extra dollars of ice-cream sold per 1 °C rise in temperature.
1.6 — Mixed
(a) P(A or B) = 0.5 + 0.4 − 0.2 = 0.7.
(b) P(A | B) = 0.2 / 0.4 = 0.5.
(c) P(A | B) = 0.5 = P(A), so A and B are independent.
(d) P(B and not A) = P(B) − P(A and B) = 0.4 − 0.2 = 0.2.
2 — Find the mistake
(a) Line 5.
(b) The student added when they should have multiplied. "Both red" is an AND statement, which along a tree branch uses the multiplication rule: P(both red) = P(first red) × P(second red | first red). Adding the probabilities along a branch is the Lesson 16 MCQ Q5 trap. 17/18 ≈ 0.944 is also unreasonably high for this scenario.
(c) Corrected: P(both red) = 1/2 × 4/9 = 4/18 = 2/9 ≈ 0.222.
3 — Open-ended Year 10 finale (sample solution)
(i) Question: Do Year 10 students who walk to school finish their homework on time more often than students who get a lift?
(ii) Data: Survey 80 Year 10 students. Variables: travel method (walk / lift / bus — categorical) and homework on time last week (Y/N — categorical). Also record number of after-school activities per week (discrete numerical) for an extension comparison.
(iii) Techniques (4 used):
— Classification: state each variable's type.
— Two-way table: walk vs lift × homework Y/N.
— Conditional probability: compute P(on time | walk) and P(on time | lift) and compare.
— Box plot: compare the distribution of after-school activities for walkers vs lift-takers.
(iv) Prediction: Walkers may finish homework on time slightly less often than lift-takers because walking takes longer; I expect the conditional probabilities to differ by ≈ 10 percentage points.
(v) Limitation: 80 students is small (short-run variation per Lesson 18). Next time I would survey three Year 10 cohorts (~240 students) and add a question about distance from home to school, since that confounds travel method with available time.
Marking: 1 mark — clear research question + variables; 1 — three or more techniques named with what they compute; 1 — prediction with a reason; 1 — one realistic limitation with a fix.
End of Year 10 Mathematics
Congratulations — this is the last worksheet of Year 10 Maths. Every Unit 4 tool you used here (classification, summary statistics, scatter plots, the addition / multiplication / conditional probability rules, tree diagrams) is the foundation for Senior Mathematics (Advanced, Standard or Extension). The HSC will keep using the same rules — just on harder data.