Unit 4
Statistics and Probability
About this unit
Calculate means and averages, interpret column and line graphs, work with Venn diagrams, and apply basic probability reasoning to single and compound events.
What types of questions will you face?
- 1Find the average of one sub-group when you know the overall average and the average of the other sub-group — use the total-sum method (mean × count = total) applied to each part
- 2Evaluate three probability statements about a bag of coloured items — write each probability as a fraction, compare numerators, and apply the complement rule for 'not [colour]' claims
- 3Read a column graph or pictograph to calculate totals, differences between groups, or per-symbol values — then answer a follow-up question about the data
- 4Calculate a weighted mean when groups of different sizes have different averages — find each group total first, then combine before dividing
- 5Determine the mean, median, or mode from a small data set given in a table, then verify which of several claims about the data are correct
Skills you will build
- Converting between mean and total: total = mean × count (and back) — the essential move in every split-group or combined-group averages question
- Writing probabilities as fractions over the total and comparing by inspecting numerators rather than converting to decimals
- Applying the complement rule correctly: P(not A) = 1 − P(A) = sum of all other probabilities — essential for statement 3 in multi-claim questions
- Reading bar charts and pictographs accurately: reading bar heights, computing differences between bars, and finding the value one symbol represents
- Checking each probability or data claim independently before matching to a statement-combination option — never trust intuition on these
By the end of this unit, you will be able to
- Solve any split-group or multi-group averages problem using the total-sum method, without guessing
- Evaluate probability statements systematically and identify the correct combination of true statements
- Extract exact values from column graphs, pictographs, and data tables to answer questions accurately under time pressure
- Combine averages from groups of different sizes into a single weighted mean
- Find the mean, median, and mode of a small data set and assess whether given claims about them are true or false
Difficulty profile
Questions in this unit range from Very Easy to Very Difficult. Simple mean questions and basic graph reading are Very Easy to Easy. Multi-statement probability and split-group averages questions are Medium to Difficult. Weighted means and claim-evaluation questions on graphs can reach Very Difficult.
Exam tip: Statistics and Probability
For any averages question, your very first move is to convert every mean into a total (multiply mean × count). Never reason about averages directly — work with totals, subtract where needed, then divide at the end. For probability statements, write all fractions over the same denominator before comparing — and always check the 'not' statement using the complement: count of non-target items divided by total.
Sample Questions
Let's start with the single most important insight in every OC MR averages question: an average is just a sum in disguise. The moment you realise you can always recover the total by multiplying the average by the count, a whole family of questions becomes straightforward.
Mean and averages questions appear in almost every OC MR paper — usually one or two per test, spread across easy and difficult. The easy variants involve a single group; the harder ones split the group in two and ask you to work back and forth between both halves.
The examiner is checking whether you know that mean × count = total. Students who try to reason about averages directly — without converting to totals first — consistently pick the wrong answer on split-group questions.
You are told the average of a full group (e.g. 6 numbers averaging 18) and the average of one part of it (e.g. the first 3 average 14). You must find the average of the remaining part. The numbers are deliberately chosen so that trial-and-error is slow — the total-sum method is the only clean path.
Best approach: Step 1: Multiply the full group average by its count to get the grand total. Step 2: Multiply the sub-group average by its count to get the sub-group total. Step 3: Subtract to find the other sub-group's total. Step 4: Divide by that sub-group's count. Four arithmetic steps — no guessing needed.
Question
The average of 6 numbers is 18.
If the average of the first 3 numbers is 14, find the average of the other 3 numbers.
- A14
- B18
- C22
- D20
- E24
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25 marbles: 3 blue, 4 red, 8 yellow, 10 green. Which event is least likely? Count favourable outcomes for each: A=7, B=8, C=15, D=22, E=11. Smallest = 7 = Option A. The key skill: for 'not X' and 'neither X nor Y' events, subtract from the total.
Comparing probabilities by counting favourable outcomes is an easy-level staple in OC MR. The twist here — 'not X' and 'neither X nor Y' options — tests whether students can compute complementary event counts by subtracting from the total.
The examiner tests whether students can translate compound events ('either/or', 'not', 'neither/nor') into counts, then rank them. Option B (yellow, 8) is the classic distractor — students who stop at 'yellow is the rarest single colour' miss that blue+red combined (7) is actually fewer.
A bag or box contains coloured objects with given counts. One is drawn at random. Students evaluate several events (including compound 'not' and 'neither/nor' events) and identify the least or most likely.
Best approach: Convert every event to a count: A=3+4=7, B=8, C=25−10=15, D=25−3=22, E=25−4−10=11. Rank: 7<8<11<15<22. Least likely = A.
Question
A box contains 25 marbles:
3 blue, 4 red, 8 yellow and 10 green.
Lemar takes one marble out of the box without looking.
Which of these events is the least likely?
- AThe marble is either blue or red.
- BThe marble is yellow.
- CThe marble is not green.
- DThe marble is not blue.
- EThe marble is neither red nor green.
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Now let's tackle one of the most common probability formats in OC MR: a question that gives you a bag of coloured items and asks you to judge whether several probability statements are true or false. The key skill is checking each statement independently and knowing the complement rule cold.
Multi-statement probability questions appear in almost every OC MR paper from mock 10 onwards, usually in the medium-to-difficult range. They are a consistent score separator — students who check each claim systematically score them reliably; students who trust intuition frequently miss statement 3 (the complement).
The examiner is testing whether you can write a probability as a fraction (count of the event / total), compare fractions with the same denominator by comparing numerators, and correctly apply the complement rule: the chance of "not red" equals the chance of all other colours combined.
A bag contains coloured balls or marbles with different counts. One is drawn at random. Three probability claims are made — typically one comparing two colours, one claiming equality, and one involving "not [colour]". You must identify which subset of claims is correct.
Best approach: Write every probability as a fraction over the total immediately — do not estimate or eyeball. Check Statement 1: compare numerators. Check Statement 2: check if the two numerators are equal. Check Statement 3: "not red" means all non-red balls — add the other counts and confirm it equals what's claimed. Work through all three before committing to an answer.
Question
A bag contains 24 balls: 10 red, 6 blue, 8 green.
One ball is drawn at random. Three students make statements:
-
Picking a red ball is more likely than picking a blue ball.
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Picking a green ball and picking a blue ball are equally likely.
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The chance of picking a ball that is not red is the same as the chance of picking a blue or green ball.
Which statements are correct?
- A1 only
- B2 only
- C3 only
- D1 and 3
- E2 and 3
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Jess has counters numbered 1–11. Three statements are made about drawing one at random. Students must test ALL three carefully: Statement 1 fails because 1–11 has 6 odd but only 5 even numbers. Statement 2 fails because >6 and <6 each have 5 numbers (6 itself is excluded from both). Only statement 3 is true (each single counter has probability 1/11).
Multi-statement probability questions are a favourite at the medium tier of OC MR. They appear with counters, bags of balls, spinners, or dice. The key skill is testing every statement methodically rather than relying on intuition. Statement 1 here is the classic trap: most students assume odd and even are always balanced, forgetting that 1–11 is an odd count of numbers.
The examiner is testing three distinct probability concepts in one question: (1) recognising that an odd-sized set of consecutive integers has more odd than even numbers; (2) understanding that 'greater than 6' and 'less than 6' exclude 6 itself, making them equal; (3) knowing that all individual counters in a fair bag have equal probability. Students who get D (1 and 3) have fallen for the even/odd trap.
A set of numbered items is described. Three (or four) probability statements are given. Students must evaluate each statement independently and identify which subset is true. One statement is always a 'trap' that looks right at first glance (here: equal odd/even chance seems intuitive).
Best approach: Test each statement separately. Statement 1: count evens (2,4,6,8,10=5) and odds (1,3,5,7,9,11=6) — not equal → FALSE. Statement 2: count >6 (7,8,9,10,11=5) and <6 (1,2,3,4,5=5) — equal, not 'more likely' → FALSE. Statement 3: each numbered counter has probability 1/11 → equally likely → TRUE. Answer: statement 3 only.
Question
Jess has a bag of eleven counters numbered 1 to 11.
Without looking, she takes one counter out of the bag at random.
Which of the following statements below is/are correct?
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Jess is equally likely to take out an even number or an odd number.
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Jess is more likely to take out a number greater than 6 than a number less than 6.
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Jess is equally likely to take out the number 2 or the number 9.
- Astatement 2 only
- Bstatement 3 only
- Cstatements 1 and 2 only
- Dstatements 1 and 3 only
- Estatements 1, 2 and 3
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Shamira has 4×$5, 2×$10, and 2×$20 notes — 8 total. Three statements are given. Statement 1 (more likely $5 than $20) and Statement 2 (equally likely $5 or not-$5) are both true because exactly half the notes are $5. Statement 3 (certain to pick >$5) is false — there's a 50% chance of picking a $5 note.
Multi-statement probability questions involving a mixed collection (notes, balls, tiles) appear consistently in OC MR at the medium tier. Statement 2 here is particularly instructive: 'equally likely to take a $5 or not' is confirmed by the exact 4/8 = 1/2 split — an elegant application of complementary probability.
The examiner tests three probability concepts simultaneously: (1) comparing probabilities of different denominations, (2) recognising when an event and its complement are equally likely (probability = 1/2), (3) understanding 'certain' means probability = 1 (not just 'likely'). Statement 3 is the trap — half the notes are $5, so it is far from certain.
A collection of items is described by type and count. Three statements are made about drawing one at random. Students must evaluate all three using fractions: P(event) = favourable / total. A key trap is confusing 'likely' with 'certain'.
Best approach: Count total: 4+2+2=8. Stmt 1: P($5)=4/8=1/2 > P($20)=2/8=1/4 → TRUE. Stmt 2: P($5)=4/8=1/2, P(not $5)=4/8=1/2 → equal → TRUE. Stmt 3: P($5)=1/2 ≠ 0 → not certain to pick >$5 → FALSE. Answer: C (statements 1 and 2 only).
Question
Shamira has these notes in her purse:
| - four $5 notes |
| - two $10 notes |
| - two $20 notes |
She takes a note from her purse without looking.
Which of these statements is/are correct?
-
She is more likely to take a $5 note than a $20 note.
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She is equally likely to take a $5 note or not to take a $5 note.
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It is certain that she takes a note worth more than $5.
- Astatement 1 only
- Bstatement 2 only
- Cstatements 1 and 2 only
- Dstatements 2 and 3 only
- Estatements 1, 2 and 3
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16 discs: 5R, 2G, 8B, 1Y. Statement 1 (certain blue) fails — P(blue)=½≠1. Statement 2 (equally likely blue or not blue) is true — P(blue)=8/16=½=P(not blue). Statement 3 (possible yellow) is true — P(yellow)=1/16>0. Answer: 2 and 3 only.
Evaluating probability language ('certain', 'equally likely', 'possible') with a statement-combination format is a medium-level OC MR staple. Students must know the precise threshold for each term: certain=1, equally likely=equal fractions, possible=>0.
The examiner tests precise probability vocabulary. Statement 1 ('certain') traps students who see 8 blue discs and think 'there are more blue than any other colour' without realising certain means probability = 1. Statement 3 ('possible') checks if students understand that even 1 disc out of 16 is a possible outcome.
A set of coloured objects is described with counts. Three statements each use a probability term ('certain', 'equally likely', 'possible', 'impossible', 'more likely', etc.). Students evaluate each statement and identify which are correct.
Best approach: Total=16. P(blue)=8/16=½. Statement 1: certain requires P=1, but P=½ → false. Statement 2: P(blue)=P(not blue)=½ → equally likely → true. Statement 3: 1 yellow disc exists → P>0 → possible → true. Answer: 2 and 3.
Question
Oliver has a box of 16 coloured discs.
- 5 discs are red.
- 2 discs are green.
- 8 discs are blue.
- 1 disc is yellow.
He picks one disc at random from the box.
Which of these statements are correct?
-
It is certain that he picks a blue disc.
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It is equally likely that he picks a blue disc or does not pick a blue disc.
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It is possible for him to pick a yellow disc.
- Anone of them
- Bstatements 1 and 2 only
- Cstatements 1 and 3 only
- Dstatements 2 and 3 only
- Estatements 1, 2 and 3
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