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.
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) and P(B∣A)=P(B).
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.
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”:
∣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.
