H2 Maths Probability Formula Sheet | Conditional & Bayes
H2 Maths Probability Formula Sheet | Conditional & Bayes
Study guide/
H2 Maths probability formula sheet: addition and multiplication rules, conditional probability, independence tests, Bayes' theorem, and tree/Venn diagram methods - aligned to th...
Q: What does H2 Maths Notes (JC 1-2): 6.1) Probability cover? A: Conditional probability, independence tests, event diagrams, and common JC exam traps for the 2026 H2 syllabus.
Study cadence Tighten your language: every time you quote a probability, state the event clearly first. Keep a running list of assumptions (mutually exclusive, independent, equally likely) before applying shortcuts.
Probability is about clearly named events: Define A and B before using formulas.
Conditional probability changes the sample space: Read "given" as "restrict attention to".
Independence is a test, not a feeling: Compare the joint probability with the product of the separate probabilities.
Concrete example: If 60 students take Physics, 45 take Maths, and 35 take both, the "Physics given Maths" denominator is 45, not the whole cohort.
Status: SEAB's current H2 Mathematics (9758) syllabus PDF is labelled for 2026. Topic 6.1 is assessed in Paper 2 Section B (Probability and Statistics, 60 marks) and covers counting and conditional probability workflows.
Formulas at a glance
Every result the 9758 syllabus expects you to recall, on one screen. Bayes' theorem and the total probability rule are not in MF27, so you must derive or recall them. Worked examples for each appear in the sections below.
P(A)=i=1∑kP(A∣Bi),P(Bi) for a partition B1,…,Bk
Bayes' theorem
P(Bi∣A)=j∑P(A∣Bj),P(Bj)P(A∣Bi),P(Bi)
Core Definitions
Sample space S contains all mutually exclusive outcomes.
Event A⊆S has probability P(A)= number of favourable outcomes divided by number of total outcomes (for equally likely cases) or the frequency assigned by the model.
Independence: A and B are independent if P(A∩B)=P(A)P(B); equivalently P(A∣B)=P(A)
Conditional wording checkpoint
Before choosing a formula, translate the sentence into event notation. The word after "given" becomes the restricted sample space and therefore the denominator.
Wording in the question
Notation to write
Denominator to use
A student takes Physics given that the student takes Maths.
P(P∣M)
P(M)
An item passed Stage 1 given that it passed Stage 2.
P(S1∣S2)
P(S2)
A candidate has a portfolio given that the candidate passed the interview.
P(R∣pass)
P(pass)
At least one event happens.
P(at least one)
Use 1−P(none) when the complement is simpler.
Common trap: do not read P(A∣B) as "A then B". It means "A among the cases where B has already happened".
Independence test checkpoint
Before using P(A∩B)=P(A)P(B), prove that independence is given or verified. Independence is about whether knowing one event changes the probability of the other; it is not the same as two events being unable to happen together.
Situation in the question
First test
What it means
Common trap
The question states the events are independent
Use P(A∩B)=P(A)P(B) directly
The occurrence of one event does not change the probability of the other.
Rechecking independence and wasting time when the assumption is already given.
Assuming the events are independent because the numbers look unrelated.
You are given P(A∣B)
Compare P(A∣B) with P(A)
If the conditional probability changes, the events are dependent.
The events are mutually exclusive and both have positive probability
Use P(A∩B)=0
They cannot be independent because one event happening makes the other impossible.
Saying "separate events" means independent.
Misconception check: independent does not mean disjoint. If two positive-probability events are disjoint, knowing one happened makes the other probability zero, so dependence is forced.
Diagram Workflows
Venn diagrams
Label each region with algebraic expressions (e.g. x for the overlap, 0.4−x for the remaining part of A).
Use the total probability constraint (sum of all regions equals 1) to solve for the unknowns.
When three sets appear, build the diagram region by region, starting from the three-way overlap.
Example -- Scholarship shortlisting
Let A be "strong portfolio" and B be "strong interview". Given P(A)=0.65, P(B)=0.55, and P(A∩B)=0.42:
P(Ac∩B)=0.55−0.42=0.13.
P(A∩Bc)=0.65−0.42=0.23.
P(Ac∩Bc)=1−0.65−0.55+0.42=0.22
P(A∣B)=0.550.42≈0.764.
Tree diagrams
Multiply along branches to obtain joint probabilities (e.g. P(A∩B)=P(A)P(B∣A)).
Sum the relevant final nodes to find the total probability of an event.
When conditional information is provided about the second stage, write it beside the branches before multiplying.
Example -- Quality control
A factory inspects items with two stages:
Stage 1: pass with probability 0.9.
Stage 2: conditional pass probability is 0.95 if Stage 1 passed, 0.6 if Stage 1 failed.
Compute the probability of an item passing both stages:
Branch values: pass then pass has probability 0.9×0.95=0.855.
If the second stage pass probability is to be found given the item passed Stage 2, apply Bayes:
Total Stage 2 pass probability = 0.9×0.95+0.1×0.6=0.915.
Probability an item passed Stage 1 given it passed Stage 2 = 0.9150.855≈0.935.
Replacement checkpoint for tree diagrams
Before writing second-branch probabilities, decide whether the first selection changes the box, bag, or group. The numerator and denominator may both change.
Sampling situation
What changes after the first draw
Second-branch move
Common trap
With replacement
The item is put back, so the composition is reset.
Reuse the original probabilities on the next branch.
Reducing the denominator even though the item was replaced.
Without replacement
One item has been removed.
Update both the favourable count and total count before writing the next probability.
Changing the numerator but leaving the old denominator.
Conditional result is already given
The branch probability is supplied by the question.
Write it directly beside the relevant branch.
Recomputing it from an unrelated overall probability.
Question asks "at least one"
Many final branches may qualify.
Use the complement "none" if it has fewer branches.
Adding only one successful route and missing another.
Worked check: a bag has 5 red and 3 blue counters. If two counters are drawn without replacement, P(red then red)=85×74, not 85×85.
Misconception check: "without replacement" is not just a label for the tree. It changes the sample space after every draw.
Bayes tables
List the prior probabilities in the Total column.
Multiply across each row with the provided conditional probabilities to obtain joint probabilities.
Sum the column totals, then apply P(A∣B)=P(B)P(A∩B) directly from the table.
Example -- Research profile screening
In a scholarship interview round, forty percent of applicants arrive with a vetted research portfolio R. Among those candidates, seventy-five percent pass the case interview. Only twenty-five percent of applicants without a portfolio pass. Build a Bayes table to estimate the probability that a randomly chosen successful candidate had a portfolio.
Pass interview
Fail interview
Total
Research portfolio
0.40×0.75=0.30
0.40×0.25=0.10
0.40
No portfolio
0.60×0.25=0.15
0.60×0.75=0.45
0.60
Total
0.45
0.55
1.00
The case-interview success rate is 0.45 overall. Using Bayes' rule,
P(R∣pass)=0.450.30=32≈0.667.
Interpret in context: about two-thirds of successful applicants had a research portfolio.
Common Question Types
Showing independence (or dependence)
Compute both P(A∩B) and P(A)P(B); equality means independent, inequality means dependent.
Alternatively, compare P(A∣B) with P(A).
Conditional complements
P(Ac∣B)=1−P(A∣B).
When the question gives the probability of "at least one" event happening, convert to the complement of "none".
Total probability rule
For a partition B1,B2,…,Bk of the sample space:
P(A)=i=1∑kP(A∣Bi)P(Bi).
In H2 questions this often appears in genetics or reliability contexts.
Three-event inclusion-exclusion
Add the individual set probabilities.
Subtract the pairwise overlaps (each intersection is counted twice in the first step).
Add back the triple overlap.
Use the complement P(none)=1−P(at least one) when a question asks for “none” or “at least one”.
Example -- Orientation workshops
During JC orientation, 180 students choose workshops in Data Science D, Economics E, and Robotics R. The participation data are
∣D∣=68,∣E∣=72,∣R∣=55
∣D∩E∣=24,∣D∩R∣=19,∣E∩R∣=17
∣D∩E∩R∣=9
Find how many students signed up for none of the workshops.
Inclusion-exclusion for "at least one": Result:∣D∪E∪R∣=(68+72+55)−(24+19+17)+9=144
Complement for “none”: 180−144=36.
The probability that a random student skipped all workshops is 18036=0.20, and the number of such students is 36.
Calculator and Notation Tips
Casio: use the TREE mode to visualise multi-stage events; store intermediate probabilities in variables A, B, C if you need to reuse them across branches.
Graphing calculator (GC): define events clearly and record the commands you used (e.g., list operations for complements, tree probabilities for sequential processes) so method marks are visible.
TI: use lists to capture outcomes and apply cumSum to compute running totals for "at least" or "at most" questions.
Always define events in words before introducing symbols; MOE marking schemes penalise answers that show algebra without context.
Exam Watch Points
State assumptions (mutually exclusive or independent) explicitly before cancelling terms.
For sequential processes, highlight whether sampling is with or without replacement; it changes conditional probabilities immediately.
When using Bayes, write out the denominator expansion so the marker sees the full weighting.
Reinforce independence tests, Bayes tables, and total probability applications across multi-stage contexts.
Quick Revision Checklist
Convert word problems into formal event notation consistently.
Draw and label Venn or tree diagrams before solving algebraically.
Execute conditional probability calculations without skipping intermediate steps.
Distinguish between mutually exclusive and independent events (they are not the same).
Apply total probability and Bayes reliably under time pressure.
Want weekly guided practice on Probability? Our H2 Maths tuition programme builds fluency in this topic through structured problem sets and exam-style drills.
Common exam mistakes
Confusing mutually exclusive with independent: Mutually exclusive events satisfy P(A∩B)=0, which forces P(A∣B)=0 whenever P(B)>0 - they cannot be independent (unless one has probability zero). Students often assume that "cannot happen together" implies "unrelated", which is wrong.
Forgetting to multiply along tree branches: In a two-stage tree, P(A∩B)=P(A)⋅P(B∣A). A common slip is to read off the branch probability P(B∣A)
Swapping the conditional in Bayes' theorem: P(A∣B) and P(B∣A) are not equal. When a question asks for the probability of a cause given an observed outcome (e.g. probability a batch is defective given an item failed inspection), write out the denominator P(B)=P(B∣A)P(A)+P(B∣Ac)P(Ac)
Assuming independence without verification: The factorisation P(A∩B)=P(A)⋅P(B) is only valid after confirming independence. In genetics or reliability questions, independence must be stated in the problem; do not assume it because events involve different individuals or components unless the question says so.
Arithmetic errors in the complement approach: When using P(at least one)=1−P(none), ensure P(none) accounts for all stages - a missed branch or an incorrect complement probability propagates through the entire calculation.
Frequently asked questions
Which paper and section does Probability appear in? Probability (Topic 6.1) is assessed in Paper 2 Section B - Probability and Statistics, which carries 60 marks in total. [1] You should expect one to two probability questions per paper, often combined with conditional probability or Bayes' theorem in a multi-stage context.
Are permutations and combinations (P&C) examinable in H2 Maths 9758 from 2026? No. Counting methods such as permutations and combinations were removed from the H2 Mathematics (9758) syllabus for candidates sitting from 2026 onwards. [1] Probability questions will instead define sample spaces explicitly or provide frequency tables, so you do not need to apply nPr or nCr formulas.
Is the Bayes' theorem formula provided in the exam, or must it be memorised? Bayes' theorem is not listed in the MF27 formula booklet. [1] You are expected to derive it from first principles using the conditional probability definition P(A∣B)=P(B)P(A∩B) combined with the total probability rule. Practise expanding the denominator in full so the working is visible to the marker.
Is there a formula sheet for H2 Maths probability? Yes - the "Formulas at a glance" section near the top of this page collects every result you need: the addition law, conditional probability, the multiplication rule, independence tests, the total probability rule, and Bayes' theorem. Note that MF27 does not include the conditional probability definition, the multiplication rule, or Bayes' theorem, so you must be able to derive or recall these in the exam.
Other H2 Maths formula sheets
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