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Secondary 2 Probability Worksheet

By Marius — private math tutor in Montreal. Updated 30 August 2026

A single spin or a single draw is settled by counting. Secondary 2 is where the experiments stop being single: three stages one after another, two chips taken from the same bag with the first one put back or kept out, a team picked where the order of the picking may or may not matter. The sheet works through enumerating those larger sample spaces with a table, a tree or an ordered list, then through "and" and "or", the intersection and union of sets, and the gap between the probability you calculate and the one you observe. Everything on it is done by reasoning — no counting formulas anywhere. Read it here, or print the free PDF.

Practice worksheet — free PDF

8 pages 11 questions Letter size, print-ready

Download the worksheet Read the questions here

No email, no account, no watermark. Teachers: photocopy it for your classes freely. Worked solutions and 11 harder problems come with the Secondary 2 Math bundle.

All 11 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

Q1Modes of Representation and the Enumeration of Possible Outcomes

A bubble tea counter in Rosemont prepares one drink at random for a taste test. It picks one tea — green, black, taro or mango — then one topping — tapioca, jelly or none — then one size — small or large.
1. How many different drinks are possible? Show the calculation.
2. Explain why a two-column table is awkward for this experiment, and say which mode of representation you would use instead.
3. List every possible drink that uses green tea, writing each one as a triple such as (green, jelly, large).
Q2Probability

A board game uses a spinner with 8 sectors, all the same size: 4 sectors say advance 1, 2 say advance 2, 1 says advance 3 and 1 says lose a turn.
1. Explain why the 8 sectors give 8 equally likely outcomes.
2. Find $P (advance 2)$ as a fraction in lowest terms, as a decimal and as a percentage.
3. Find the probability that the player advances at all.
4. A game lasts 200 spins. About how many times should a player expect lose a turn? Is that number guaranteed?
Q3Random Experiments Where Order Matters and Where Order Does Not Matter

A cycling club has five members: Alix, Bao, Camille, Dilan and Élodie. Use their initials A, B, C, D, E.
1. Three of them are picked to ride a relay, and the order of the picking does not matter. List every possible team, in alphabetical order, and count them.
2. Instead, three of them are picked as first rider, second rider and third rider, so the order does matter. How many possibilities? Show the calculation.
3. Divide the answer to b) by the answer to a), and explain in words what that number counts.
Q4Random Experiments With One or More Steps

A two-step game rolls a fair six-sided die numbered 1 to 6, then spins a spinner with four equal sectors numbered 1 to 4. The two numbers are added.
1. How many equally likely results does the game have? Build a table showing the total for each of them.
2. Find the probability that the total is exactly 5.
3. Find the probability that the total is 9 or more.
Q5Random Experiments With and Without Replacement

A jar holds 6 keychains: 4 blue ( $B_{1}, B_{2}, B_{3}, B_{4}$ ) and 2 orange ( $O_{1}, O_{2}$ ). Two keychains are taken.
1. The first is not put back. Treating the two keychains as an unordered pair, how many pairs are possible, and what is $P (both blue)$ ?
2. The first is put back and the results are written in order. How many results are possible now, and what is $P (both blue)$ ?
3. Which version makes “both blue” more likely? Explain why, using the contents of the jar.
Q6Sample Space

Three roommates — Ana, Ben and Coralie — draw lots for three different chores: dishes (D), recycling (R) and sweeping (S). Each person ends up with exactly one chore, and no chore is given twice.
1. Write the sample space $Ω$ , using a triple such as $(D, R, S)$ to mean “Ana gets dishes, Ben gets recycling, Coralie gets sweeping”.
2. Explain why this is a draw without replacement, and how that shows in the size of $Ω$ .
3. Find $P (Ana gets the dishes)$ and $P (Ben gets recycling and Coralie gets sweeping)$ .
Q7The Concept of ``Or'' and ``And'' in Probability

A bag holds 6 tiles: $A 1$ , $A 2$ , $A 3$ , $B 1$ , $B 2$ , $B 3$ . One tile is drawn at random. Let $E$ be “the tile has the letter $A$ ” and $F$ be “the tile has an odd number”.
1. List the outcomes of $E$ and the outcomes of $F$ .
2. Find $P (E and F)$ .
3. Find $P (E or F)$ , then check your answer with $P (E) + P (F) - P (E and F)$ .
4. Give an event of this experiment that is incompatible with $E$ .
Q8The Intersection and Union of Sets

Thirty students were asked about three activities: skating ( $S$ ), music lessons ( $M$ ) and volunteering ( $V$ ). The survey found 15 skaters, 12 in music lessons and 9 volunteers; 5 students do skating and music, 4 do skating and volunteering, 3 do music and volunteering, and 2 do all three. Each of those three pair counts includes the students who do all three activities.
1. Draw a Venn diagram with three overlapping circles and write the correct number in each of its eight regions. Start from the centre.
2. Find the probability that a randomly chosen student does exactly one of the three activities.
3. Find the probability that a randomly chosen student does none of them.
Q9The Types of Probability

A thumbtack dropped on a table lands either point-up or point-down. Camille drops one 400 times and records 244 landings point-up.
1. Find the experimental probability that the thumbtack lands point-up. Give it as a decimal and as a percentage.
2. Camille's classmate says the theoretical probability must be $\frac{1}{2}$ , “since there are two ways to land”. Explain why no theoretical probability can be found for this experiment.
3. Predict about how many point-up landings there would be in 1000 drops.
Q10Types of Events

Twenty-four tiles numbered 1 to 24 are placed in a bag and one is drawn at random.
1. For each event below, give its probability and name the most precise type that applies, choosing from certain, impossible, elementary and probable: \\ $A$ : the number is a multiple of 6; $B$ : the number is greater than 24; \\ $C$ : the number is 13; $D$ : the number is less than 25.
2. Let $E$ be “the number is odd”. Are $A$ and $E$ compatible? Justify.
3. Is $E$ complementary to “the number is even”? Check both conditions.
4. Name an event that is incompatible with $D$ , and explain why every such event must be impossible.
Q11Synthesis — drawing on several sheets in this topic

A gym bin holds 6 balls: 2 blue ( $B_{1}, B_{2}$ ) and 4 white ( $W_{1}, \dots, W_{4}$ ). Two balls are taken out one after the other and the order is recorded.
1. Explain why labelling the balls, rather than just calling them “blue” and “white”, is what makes the outcomes equally likely.
2. Without replacement: how many results are in the sample space, and what is the probability of getting one ball of each colour?
3. With replacement: answer the same two questions.
4. In which version is “one of each colour” more likely? Explain why, using what is in the bin at the second draw.

The 11 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

Which stream is this for? There is no stream to choose at this level. Secondary 2 is the second year of Cycle One and every student follows the same program, so this sheet is simply Secondary 2 mathematics as the Québec Education Program's Progression of Learning sets it out — the streams begin in Secondary 4.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

Everything starts from the sample space

A probability at this level is a count divided by a count, so the whole difficulty is producing the two counts honestly. Q1, Q6 and Q11 are all the same instruction in different clothes: list every outcome, in an order that makes it obvious none is missing.

Choosing how to enumerate

The representation is not decoration — the wrong one is what loses outcomes.

1
Two stages → a two-way table
One stage down the rows, the other across the columns, and every cell is one outcome. Q4 rolls a die then spins a four-sector spinner and fills each cell with the total, which turns a probability question into reading the table.
2
Three or more stages → a tree, or an ordered list
A table has two directions and a three-stage experiment has three, which is exactly what Q1 asks you to explain. A tree branches as many times as there are stages; a systematic list works too, as long as the order is systematic.
3
Multiply the stages to know how many to expect
Four teas, three toppings, two sizes: 4 × 3 × 2 = 24 outcomes. Compute that number before listing, then count what you listed. If the two disagree, one branch is missing — and you know it immediately instead of at the end.
4
Write each outcome the same way every time
A triple like (green, jelly, large), always in the same stage order. Vary the notation and duplicates hide in the list. Writing the coin tosses in a fixed order is what makes a missing HTT impossible to overlook.

Equally likely is a condition, and it has to be earned (Q2)

The count-over-count formula is only valid when the outcomes being counted are equally likely, and Q2 opens by asking you to say why eight identical sectors are. That is not a formality. The most expensive error on this whole sheet is dividing by a number of outcomes that are not equally likely to each other.

"Red or not red" is two outcomes, not two equally likely outcomes. A bag of two red chips and one blue is not a coin toss. Draw a tree with a red branch and a non-red branch for each of two such bags, count four results, and conclude that two reds has probability one quarter: every step after the first is correct, the first is wrong, and the answer is worthless.

The fix is to label the objects. Treat the chips as R₁, R₂ and B rather than "red" and "blue", and the outcomes become genuinely interchangeable. Q11 asks for that reasoning in its own part, with balls labelled B₁, B₂, W₁ … W₄ — labelling is what makes the counting legitimate, not a habit of being tidy.

With replacement or without: the second stage is a different experiment (Q5)

Q5 takes two keychains from a jar of six, first with the first one kept out, then with it put back. Nothing else changes, and the two answers differ — which is the point of asking both.

With replacement: the jar is identical for the second draw. The same object can appear twice, and the number of ordered results is n × n.

Without replacement: the jar has one fewer object, and one fewer of whatever colour you just took. No object can repeat, and the number of ordered results is n × (n − 1).

Whether "both blue" becomes more or less likely is decided by the contents, not by a rule to memorise: removing a blue keychain leaves fewer blues in a smaller jar, and Q5 asks you to explain the comparison using those contents.

Tabulating n × n against n × (n − 1) for a few values of n is worth doing once. The gap between the two columns is small in relative terms for a large box and large for a small one, which is why a two-draw problem from a jar of six is sensitive to replacement and one from a warehouse is barely affected.

Does the order matter? (Q3)

Picking three cyclists for a relay where the order is irrelevant, and picking a first, second and third rider, are two different experiments over the same five people. Q3 asks for both, and then for the ratio between them.

That ratio is the useful idea. Each unordered team of three can be written down in 3 × 2 × 1 = 6 different orders, so the ordered count is six times the unordered one. Dividing the ordered count by the number of orderings is how you get from one to the other — and it is a piece of reasoning you can rebuild from scratch, which is exactly what this level asks for.

The ratio is not always 2. The rule unordered is always half of ordered is true for choosing two objects and false for choosing three, where the factor is 6. Testing a claim like that on a second case before accepting it is the habit worth building.

No formulas are needed, and none are used. At this level the counting is done by listing and by reasoning about how many orders each selection has. The notation for permutations and combinations arrives later; a solution that reaches for it here is answering a different question.

"And" means both at once; "or" means at least one (Q7)

Q7 draws one tile from six labelled A1, A2, A3, B1, B2, B3, with E = "has the letter A" and F = "has an odd number". List the outcomes of each event before doing anything else, then:

P(E and F) counts the outcomes in both lists P(E or F) counts the outcomes in at least one P(E or F) = P(E) + P(F) − P(E and F)

The subtraction is not a correction factor to memorise. Adding P(E) and P(F) counts every outcome that is in both lists twice, so one copy of the overlap has to come back out. Q7 asks you to compute the union by listing and then to check it with the formula, which is the fastest way to see that the two are the same statement.

Incompatible events have no outcome in common, so P(A and B) = 0 and the subtraction disappears. That is the only situation in which adding the two probabilities directly is correct.

Complementary is stronger than incompatible. Two events can share no outcome and still leave outcomes belonging to neither. Complementary means incompatible and covering everything, so their probabilities add to exactly 1. Q10 asks you to separate the two ideas on a bag of numbered tiles.

Venn diagrams: start from the centre (Q8)

Q8 gives three overlapping activities and eight regions to fill. Filling them in the order the numbers are given produces a mess; filling them from the middle outwards works every time.

Filling three circles

Each step subtracts what you have already placed.

1
The centre first
The number doing all three activities goes straight in, because nothing has to be subtracted from it.
2
Then each pairwise overlap, minus the centre
Q8 states explicitly that each pair count already includes the students doing all three. So the region for exactly two activities is that pair count minus the centre — and this single subtraction is what most wrong diagrams are missing.
3
Then each circle alone
The circle's total minus the three regions already written inside it.
4
Then outside
The overall total minus everything placed. A negative number here means an earlier subtraction went wrong, and it is a genuinely useful signal.

Run the same arithmetic in reverse and it detects survey figures that cannot all be true at once — a diagram whose regions will not fill without a negative number somewhere. Being able to say which numbers are consistent with which is what the diagram is actually for.

Theoretical, experimental, and the experiments that have no theory (Q9)

A theoretical probability is computed from a list of equally likely outcomes. An experimental probability is observed: the number of times something happened divided by the number of trials. Q9 drops a thumbtack 400 times, and the interesting part is part b).

"Two ways to land" does not make each way likely. A thumbtack is not symmetric, so nothing entitles you to call point-up and point-down equally likely, and there is no theoretical probability to compute at all. Repeated trials are the only source of a number here — which is precisely why experimental probability exists as a separate idea.

Predicting from an experimental probability is a proportion: multiply the observed rate by the new number of trials. The prediction is an expectation, never a guarantee — Q2 makes the same point about a spinner over 200 spins, and saying so is part of the answer.

The reverse question, whether observed counts are far enough from the expected ones to suggest a loaded die, is settled by comparing each observed count with the count you would expect and judging the size of the gaps — with real trials, small differences are normal and no single face being slightly ahead proves anything.

Naming the type of an event (Q10)

Q10 asks for the most precise name available, which is where it is easy to lose a mark by giving a true but weaker answer.

Impossible — contains no outcome, probability 0. Certain — contains every outcome, probability 1. Elementary — contains exactly one outcome. Probable — the general case, anything with a probability strictly between 0 and 1.

An elementary event is also probable, so answering "probable" for the tile numbered 13 is not false — it is just not the most precise name, and the question asked for that one.

Q11: everything at once

Two balls drawn from a bin of two blue and four white, with the order recorded, answered first without replacement and then with. It uses the labelling argument, both sample-space counts, and a comparison you have to justify from the contents of the bin rather than from a rule.

Work it in the order given. Part a) is what makes parts b) and c) legitimate, and the comparison in part d) is only convincing if the two counts before it were built the same way.

Getting the most out of it

Count the outcomes twice, by two different routes

Multiply the stages to predict how many outcomes there should be, then list them and count what you actually wrote. Almost every wrong answer on this sheet is a sample space missing one branch, and this is the check that finds it in five seconds.

Label identical objects before you count anything

B₁, B₂, B₃, B₄ rather than "the blue ones". It looks like extra writing and it is the step that makes the outcomes equally likely, which is the condition the count-over-count formula needs to be true at all.

Say which experiment you are in, in words, first

How many stages? With or without replacement? Does the order get recorded? Three short answers written at the top of your working determine the whole solution, and every one of them comes from the wording rather than from the arithmetic.

Want the solutions, or something more challenging?

The worksheet, the questions above and every explanation on this page stay free permanently. Three more PDFs exist for this topic — the worked answer key, a harder problem set, and the answer key to that. They come with the Secondary 2 Math Solutions Bundle, which is what keeps the rest of the series free.

What else exists for Probability

Three PDFs · 16 pages · all three are in the bundle below.

Answer key — 3 pages. All 11 questions worked step by step, including the restrictions and the justifications. Not a list of final answers.
Challenge problems — 9 pages, 11 problems. A separate sheet at exam-plus difficulty covering the same 10 concepts. Harder than anything on the free sheet.
Challenge answer key — 4 pages. Every challenge problem worked to the same standard, with the checks shown.
PDF, letter size, print-ready.

In the bundle See what's in it Not sold separately

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Every Secondary 2 Math topic — the complete Solutions Bundle

One download, one payment, the whole program. Every answer key and every challenge set for all 14 Secondary 2 Math worksheet sets — including this one.

14 sets · 42 PDFs · 181 pages$19.99

Worked solutions, not answer lists — every step written out
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Taking Secondary 1 Math as well? The Secondary 1 Math bundle covers all 15 of its sets — 45 PDFs, 182 pages — on the same terms.

Taking Secondary 3 Math as well? The Secondary 3 Math bundle covers all 11 of its sets — 33 PDFs, 154 pages — on the same terms.

Taking Secondary 4 Math as well? The Secondary 4 Math bundle covers all 17 of its sets — 51 PDFs, 154 pages — on the same terms.

Taking Secondary 5 Math as well? The Secondary 5 Math bundle covers all 21 of its sets — 63 PDFs, 228 pages — on the same terms.

Taking CEGEP Calculus I as well? The CEGEP Calculus I bundle covers all 8 of its sets — 24 PDFs, 82 pages — on the same terms.

Taking CEGEP Calculus II as well? The CEGEP Calculus II bundle covers all 8 of its sets — 24 PDFs, 90 pages — on the same terms.

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Taking AP Calculus AB as well? The AP Calculus AB bundle covers all 8 of its sets — 24 PDFs, 126 pages — on the same terms.

Common questions

Is this worksheet really free?

Yes — the questions are on this page to read, and the PDF downloads directly, no email and no account. The one paid item is optional: the complete Secondary 2 Solutions Bundle, which covers every set at this level.

What is the difference between drawing with and without replacement?

With replacement, the object is put back, so the second draw faces exactly the same contents and an object can appear twice. Without replacement, the second draw faces one fewer object — and one fewer of whatever you just took — so nothing can repeat. The two give different probabilities, and which one is larger depends on the contents rather than on a rule.

When does the order of a selection matter?

When the outcomes are distinguishable by position — a first, second and third rider are three different roles, while three team members chosen together are not. Each unordered selection of three can be written in six different orders, so the ordered count is six times the unordered one; for selections of two the factor is two.

Why is there a subtraction in P(A or B)?

Because adding P(A) and P(B) counts every outcome belonging to both events twice, so one copy of the overlap has to be removed. When the two events are incompatible there is no overlap, P(A and B) is 0, and the two probabilities can be added directly.

Do I need permutation and combination formulas for this?

No. At this level every count is produced by enumerating outcomes and reasoning about how many orders a selection has — a table, a tree or a systematic list. That is deliberate: the reasoning is what the formulas are shorthand for, and it is what makes them safe to use later.

Why can't the thumbtack question be answered theoretically?

Because a theoretical probability needs outcomes that are equally likely, and a thumbtack is not symmetric. Having two possible landings says nothing about how likely each one is. The only honest number comes from repeated trials, which is what an experimental probability is.

Which year is this sheet for?

Secondary 2, the second year of Cycle One, and it stays inside that year. Geometric probability and the counting notation belong to Secondary 3, and conditional probability comes later still, so nothing here needs any of them.

Can teachers use this in class?

Yes. Print and photocopy it for your own classes freely — I just ask that the tutorinmontreal.ca footer stays on the page.

I'm stuck on one question. Can you help?

Yes — through one-on-one tutoring, in Montreal or online. Get in touch to arrange a session, or see the current rates.

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The same topic at the other level: Secondary 3 Math · Probability.