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Cambridge IGCSE International Maths — Extended
Probabilities from Tables & Charts
🎯 Big idea: Whenever you're told "how many" outcomes fall into different categories — through a table, a grid, or a Venn diagram — probability is always just (what you want) ÷ (the total you're choosing from). The entire skill of this chapter is learning to read the picture correctly so you count the right numbers.
Summary — What This Chapter Covers
Two-way tables: compare two categories (e.g. Year group × Language) — fill in missing numbers using row/column totals, then read off probabilities.
Possibility (sample space) diagrams: grids that list every possible combined outcome of two events (like rolling two dice) — count what you want out of the total grid.
Venn diagrams for probability: circles show overlapping categories (e.g. Spanish & German) — add up the right regions, divide by the right total.
The golden rule everywhere: only count outcomes if they're equally likely. You can't just "count" the lottery.
Some questions restrict you to a smaller group (e.g. "given the child is in Class B…") — then your denominator shrinks to just that group's total, not the whole population.
1. Two-Way Tables
What is a two-way table?
A two-way table is exactly what it sounds like: a table that sorts data by two characteristics at once. Think of it like a spreadsheet where the rows are one category and the columns are another category, and each cell tells you how many things/people fall into that specific combination.
Example: a college has 55 students. They can be split by year group (Year 12 or Year 13) AND by language (Spanish or German). A two-way table lets you see both splits in one picture:
Spanish
German
Year 12
15
10
Year 13
5
25
Notice this raw table has no totals yet — that's step one you need to add yourself.
Step 1: Always add the totals
Before you can answer any probability question, add a Total row at the bottom and a Total column on the right. The bottom-right corner cell is the overall total — and it must equal both the sum of the row totals AND the sum of the column totals. This is your built-in check that you haven't made an arithmetic slip.
Spanish
German
Total
Year 12
15
10
25
Year 13
5
25
30
Total
20
35
55
Examiner Tip
Always check your row totals add up to the overall total, and your column totals add up to the overall total too. If they don't match, every probability you calculate afterwards will be wrong — so catch the mistake here, before you even start on part (a)!
Step 2: Reading off probabilities
Once your table is complete, probability questions become simple division. The key skill is figuring out what the denominator (bottom number of the fraction) should be:
If you're picking from the whole population, the denominator is the overall total (here, 55).
If you're picking from one specific category only (e.g. "a student is picked from Year 13"), the denominator is just that category's total, not the whole grand total.
Core Rule
P(event) = (frequency of what you want) ÷ (total you are selecting from)
Examples from the table above:
P(a random student studies Spanish AND is in Year 12) = 15/55
P(a random student studies Spanish) = 20/55 (using the column total — we don't care which year)
P(a student from Year 13 studies Spanish) = 5/30 (denominator is 30, the Year 13 total, because we've already restricted to Year 13 students)
Worked example: Building a table from a word problem
Scenario: At an art group, 60 children (split into Class A and Class B) choose between colouring, painting, clay modelling, and sketching.
Given info: 12 of Class A chose colouring. 13 of Class B chose clay modelling. 20 children total chose painting. 15 total chose clay modelling. 8 of the 30 children in Class A chose sketching, as did 4 in Class B.
Step 1 — Fill in what's directly given:
Colouring
Painting
Clay modelling
Sketching
Total
Class A
12
8
30
Class B
13
4
Total
20
15
60
Step 2 — Use totals to find the "easy" missing numbers (subtract what you know from the row/column total):
Class A, Clay modelling = 15 − 13 = 2
Class B total = 60 − 30 = 30
Sketching total = 8 + 4 = 12
Step 3 — Repeat until every cell is filled (working outwards from what you now know):
Class A, Painting = 30 − 12 − 2 − 8 = 8
Class B, Painting = 20 − 8 = 12
Class B, Colouring = 30 − 12 − 13 − 4 = 1
Colouring total = 12 + 1 = 13
Final table:
Colouring
Painting
Clay modelling
Sketching
Total
Class A
12
8
2
8
30
Class B
1
12
13
4
30
Total
13
20
15
12
60
Now the probability questions are pure lookups:
P(chose colouring) = 13/60
P(Class A and sketching) = 8/60 = 2/15
P(painting | picked from Class B only) = 12/30 = 2/5
Practice Question 1
A survey of 80 people records whether they own a car and whether they own a bike. 45 people own a car. 30 people own both a car and a bike. 10 people own neither. Construct a two-way table and find the probability that a randomly selected person owns a bike but not a car.
Practice Question 2
Using the art group table above (final version), find the probability that a child selected at random from Class A only chose clay modelling.
2. Possibility (Sample Space) Diagrams
What is a sample space?
The sample space is just a fancy way of saying "the full list of everything that could possibly happen." For a coin flip, the sample space is Heads, Tails. For a single six-sided dice, it's 1, 2, 3, 4, 5, 6.
Things get trickier when two things happen at once — like rolling two dice and adding the scores. Writing out every possibility as a list would be long, messy, and easy to mess up (you could easily miss a combination or double-count one). This is where a possibility diagram (a grid) comes in — it organises every combination systematically so nothing gets missed.
Building the grid
Put one dice's outcomes along the top, the other dice's outcomes down the side, and fill each cell with the combined result (here, the sum):
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
This grid has 6 × 6 = 36 total outcomes. Every single cell is equally likely because each dice value has an equal 1-in-6 chance and the two dice don't affect each other.
Reading probabilities off the grid
Core Rule
P(event) = (number of matching cells) ÷ (total cells in the grid)
Example: how many ways can you get a total of 8? Scanning the grid: (2,6), (3,5), (4,4), (5,3), (6,2) — that's 5 cells. So P(sum = 8) = 5/36.
Common Mistake
This "just count the cells" trick only works when every outcome is equally likely. Rolling two fair dice — fine, every cell has an equal 1/36 chance. But something like "Win the lottery" vs "Lose the lottery" is NOT 50/50 just because there are two listed outcomes — winning is vastly less likely. Always check the outcomes are genuinely equally probable before you start counting cells!
More than two events: use a list instead
A grid only works for combining two sets of outcomes (it has two dimensions, after all). If you combine three or more — like flipping three coins — you have to write out the full list systematically instead:
HHH · HHT · HTH · THH · HTT · THT · TTH · TTT
That's 2 × 2 × 2 = 8 possible outcomes. A useful trick: the total number of outcomes is always (options for event 1) × (options for event 2) × (options for event 3)... — this helps you check you haven't missed one when listing.
Worked example: Conditional probability with a grid
Scenario: Two fair six-sided dice are rolled.
(a) Find P(the sum is an odd number greater than 5).
Odd numbers greater than 5 (so 5 itself doesn't count) that appear on the grid: 7, 9, 11. Circle every cell showing these values:
(b) Given that the sum is an odd number greater than 5, find the probability that one of the dice shows the number 2.
This is a conditional probability — notice the phrase "given that." This means we're no longer choosing from all 36 outcomes; we're only choosing from the 12 outcomes we found in part (a). Our new "total" is 12, not 36.
Out of those 12 outcomes, which ones include a 2? Looking through: (2,5) and (5,2) — that's 2 outcomes.
Answer
P(shows a 2 | odd sum > 5) = 2/12 = 1/6
Why the denominator changed
Whenever a question says "given that…", it's telling you to shrink your sample space down to only the outcomes satisfying that condition. You then only count the outcomes you want within that shrunk-down group, and divide by the size of that shrunk-down group — not the original 36.
Practice Question 1
Two fair six-sided dice are rolled and their scores are multiplied together (instead of added). Find the probability that the product is greater than 20.
Practice Question 2
A fair coin is flipped and a fair six-sided dice is rolled at the same time. Draw a sample space and find the probability of getting Tails and a number greater than 4.
3. Probabilities from Venn Diagrams
Reading a Venn diagram
A Venn diagram uses overlapping circles to show how categories relate. Anything inside a circle belongs to that category; anything in the overlap belongs to both categories at once; anything outside both circles (but inside the rectangle) belongs to neither.
Venn diagrams can show either:
Frequencies — a number representing how many things are in each region (most common in exams)
Individual elements — the actual items listed out in each region
Core Rule
P(event) = (sum of frequencies in the region you want) ÷ (total frequency)
Reading different regions
Consider a Venn diagram with 11 total elements, split between set A and set B, where 5 elements are in A, 2 elements are in both A and B, and 3 elements are in A but not B:
P(in A) = 5/11 — count everything inside circle A (including the overlap)
P(in both A and B) = 2/11 — count only the overlap region
P(in A but not B) = 3/11 — count only the "A-only" crescent, excluding the overlap
The Trap Examiners Love
"In A" and "in A but not B" are NOT the same thing! "In A" includes the overlap. "In A but not B" specifically excludes the overlap. Read the wording of every Venn question extremely carefully — one missing word changes which region(s) you should be adding up.
Restricted (conditional) probabilities on a Venn diagram
Just like with two-way tables, some Venn questions restrict you to a smaller group rather than the whole diagram. For example: "the probability of being in B, given that you are already in A" — this means your denominator is only the total number of elements in A, not the grand total.
Using the same numbers as above (5 in A total, 2 of which are also in B): P(in B | in A) = 2/5 — because we're only interested in the 5 elements already in A, and out of those, 2 also happen to be in B.
Worked example: Building a Venn diagram from a word problem
Scenario: In a class of 30 students, 15 study Spanish, and 3 of those Spanish students also study German. 7 students study neither Spanish nor German.
(a) Draw a Venn diagram to show this.
The trick to building these is to always start with the overlap (intersection) and work outwards, because the overlap number is usually the least ambiguous piece of information.
Overlap (Spanish AND German) = 3
Spanish-only = 15 total Spanish − 3 overlap = 12
Neither = 7 (goes outside both circles)
To make the grand total 30: German-only = 30 − 12 − 3 − 7 = 8
Spanish only
Both
German only
Neither
Total
12
3
8
7
30
(b) Find P(a random student studies Spanish but not German).
"Spanish but not German" means only the Spanish-only crescent — 12 students, out of the class total of 30.
Answer
P(Spanish but not German) = 12/30 = 2/5
Building Tip
When you're given a total for one circle (e.g. "15 study Spanish"), remember that number usually includes the overlap already. So Spanish-only = 15 − overlap, not 15 itself. This is the single most common place students lose marks on Venn diagram construction questions.
Practice Question 1
In a survey of 50 people, 28 like tea, 22 like coffee, and 10 like both. Draw a Venn diagram and find the probability that a randomly chosen person likes neither tea nor coffee.
Practice Question 2
Using the tea/coffee data above, find the probability that a person likes coffee, given that they already like tea.
What to Memorise
Basic probability rule
P(event) = (number of favourable outcomes) ÷ (total number of possible outcomes) — this underlies every method in this chapter.
Two-way table check
Every row total must sum to the grand total, and every column total must sum to the grand total too. If they don't match, find your error before answering any questions.
Restricted / conditional denominator
Phrases like "given that," "selected from Class B only," or "already in A" mean your denominator shrinks to just that sub-group's total — not the overall total.
Sample space size for combined events
Total outcomes = (outcomes of event 1) × (outcomes of event 2) × (outcomes of event 3)... e.g. two dice = 6 × 6 = 36; three coins = 2 × 2 × 2 = 8.
Equally likely condition
The "count the outcomes" method for probability ONLY works when every individual outcome has the same chance of happening. Always sanity-check this before counting cells.
Venn diagram construction order
Always fill in the intersection (overlap) first, then work outward to find each circle's "only" region, then finally the "neither" region outside both circles.
"In A" vs "In A but not B"
"In A" includes the overlap with B. "In A but not B" specifically excludes the overlap. These give different numerators — read the question wording carefully every time.
Concepts Checklist
Exam Tips & Common Mistakes
Forgetting to check totals
Always verify row totals sum to the grand total AND column totals sum to the grand total before moving on. This is the single fastest way to catch an early arithmetic error before it corrupts your entire answer.
Using the wrong denominator
Watch for wording like "from Year 13," "given that," or "in Class B only" — these all signal a smaller, restricted denominator. Using the grand total instead of the correct sub-total is one of the most common lost marks in this topic.
Double-counting the overlap in Venn diagrams
When a question gives you "15 study Spanish," that figure usually already includes anyone who studies both Spanish and German. Subtract the overlap to find the "Spanish-only" region — don't place 15 directly into the Spanish-only crescent.
Assuming all outcomes are equally likely
Counting cells/branches only works when every outcome truly has the same chance. Don't apply this shortcut to things like biased coins, weighted dice, or events like winning a prize draw where the options aren't symmetric.
Not simplifying the final fraction
Examiners often specifically ask for the answer "in simplest form." Always check whether your numerator and denominator share a common factor before writing your final answer (e.g. 12/36 simplifies to 1/3).
Drawing a possibility diagram even when not told to
Harder exam questions sometimes won't explicitly say "draw a possibility diagram" — but if two events are being combined, sketching the grid yourself (even roughly, in the margin) will almost always make counting outcomes far more reliable than trying to do it in your head.