For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.
How this chapter applies
Eclat core: single and combined events, possibility diagrams, tree diagrams, and addition or multiplication rules form the main route.
School-sensitive extension: formal conditional probability notation, reverse conditioning, and puzzle-style strategy problems may be school-sensitive depth.
2027 national comparison: K310 Topic S2 covers single and simple combined events, possibility or tree diagrams, mutually exclusive events, and independent events.
Check your school: confirm whether conditional notation is assessed or used only as enrichment.
Q: What does IP EMaths Notes (Upper Sec, Year 3-4): 14) Probability cover? A: Compute probabilities with mutually exclusive, independent, and conditional events using organised tables or tree diagrams.
The core idea is simple: Organise outcomes first, then multiply along paths and add across allowed cases.
Use it as a working check: Mutually exclusive means events cannot happen together. Independent means one event does not change the other. Conditional probability narrows the sample space before dividing.
Then go one layer deeper: Use the counter, table, and tree examples to practise choosing the right layout, writing the denominator clearly, and using complements for "at least" questions.
Probability questions reward tidy organisation. Label events clearly, state whether they are mutually exclusive or independent, and keep notation consistent across tables and tree diagrams.
Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.
When outcomes repeat, use tables or tree diagrams so that multiplication along a path and addition across paths are obvious.
Organising your work
Frequency tables: when counts are given, arrange them in a two-way table, sum rows and columns, then divide by the total to convert to probabilities.
Tree diagrams: useful for sequential events (with or without replacement). Multiply the branch probabilities, then add the relevant paths.
Complement strategy: for “at least” questions, consider the complement (e.g. “no successes”) and subtract from 1.
Decision map - choose the layout before calculating
Pick the structure from the wording before writing fractions. This prevents mixing denominators from different sample spaces.
Question clue
Best layout
First move
Common trap
Counts split by two categories, such as Biology and Chemistry.
two-way table
fill row totals, column totals, then grand total
using the grand total after the question says "given"
Events happen in order, such as two draws or two production stages.
tree diagram
write branch probabilities stage by stage
forgetting to change the denominator without replacement
Wording says "at least one", "not all", or "not none".
complement
calculate the opposite case, then subtract from 1
listing too many overlapping cases
Wording asks whether one event affects another.
independence check
compare P(A∣B) with P(A)
confusing independent with mutually exclusive
Misconception check: "given" does not mean multiply automatically. It usually means the denominator has changed to the restricted group named after "given".
Tree Denominator Checkpoint
Before multiplying along a tree path, decide whether the second branch uses the original total or the updated total.
Draw situation
What changes after the first draw
Second-branch denominator
Example wording
With replacement
The first item is put back.
Same total as before.
First red is 105, second red is still 105.
Without replacement
The first item is not put back.
Total decreases by 1.
First red is 105, then second red is 94
First draw is a different colour
One item from that colour is removed.
Total decreases by 1, but the red count may stay the same.
If the first counter is blue, red on the second draw can be 95.
Question gives a condition
Work inside the branch that satisfies the condition.
Use the branch total after the condition.
"Given first counter is red" means continue only from the red-first branch.
Misconception check: "two draws" does not automatically mean the denominators are the same. They stay the same only when the question says the item is replaced or the trials are independent.
Mutually exclusive vs independent
Mutually exclusive events cannot occur together (their intersection is zero). For example, drawing a heart and drawing a spade on a single card draw are mutually exclusive.
Independent events do not influence each other. Tossing a coin and rolling a die are independent because the outcome of one does not change the distribution of the other.
Mutually exclusive events are not independent unless one probability is zero; removing a red marble from a bag changes the probability of drawing another red marble without replacement.
Exclusive-or-independent checkpoint
Use two separate tests before choosing a formula. One test asks whether the events can happen together; the other asks whether knowing one event changes the probability of the other.
Question to ask
If the answer is yes
Formula clue
Common trap
Can both events happen in the same trial?
They are not mutually exclusive.
Use the general addition rule if finding "A or B".
Adding probabilities without subtracting the overlap.
Is the overlap impossible?
They are mutually exclusive.
Use P(A∩B)=0.
Calling them independent just because they are separate labels.
Does knowing B leave P(A) unchanged?
They are independent.
Check whether P(A∣B)=P(A).
Assuming repeated events are independent without replacement.
Does knowing B change P(A)?
They are dependent.
Update the denominator or branch probability.
Still multiplying the original probabilities.
Worked check: in one card draw, "heart" and "spade" are mutually exclusive because one card cannot be both. They are not independent, because knowing the card is a spade makes the probability of heart become 0, not the original probability of heart.
Misconception check: mutually exclusive is about whether the overlap can happen. Independent is about whether the probability changes after new information.
Conditional probability workflow
Identify the conditioning event (e.g. “given that the first die shows an even number”).
Restrict the sample space to outcomes that satisfy that event.
Count or compute the favourable outcomes within the restricted space.
Divide by the number of outcomes in the restricted space. Use P(A∣B)=P(B)P(A∩B) as a consistency check.
Conditional denominator checkpoint
Read P(A∣B) as "probability of A, among the cases where B has already happened".
Expression
Denominator to use
Numerator to use
What the wording sounds like
P(A∣B)
all cases in B
cases in both A and B
"A, given B"
P(B∣A)
all cases in A
cases in both A and B
"B, given A"
P(A∩B)
the whole sample space
cases in both A and B
"A and B"
Misconception check: P(A∣B) and P(B∣A) use the same overlap in the numerator, but they usually have different denominators.
At-least-one checkpoint
When a question says "at least one", first test whether the complement has fewer cases. The opposite of "at least one success" is "no successes".
Wording
Easier opposite case
First calculation
Common trap
At least one red counter in two draws
No red counters
Find the probability of no red, then subtract from 1.
Adding "one red" and "two red" with repeated or missing paths.
Not all answers are correct
All answers are correct
Find the probability that every answer is correct, then subtract from 1.
Treating "not all" as "none".
At least one machine fails
No machine fails
Multiply the probabilities that each machine does not fail, then subtract from 1.
Assuming the machines are independent without checking the question.
Worked check: if the probability of no red counter in two draws is 157, then the probability of at least one red counter is 1−157=158.
Worked example - two draws without replacement
A bag contains five red, three blue and two green counters. Two counters are drawn at random without replacement. What is the probability that both counters are the same colour?
Total counters: 10.
Two reds: 105×94=9020.
Two blues: 103×92=906
Two greens: 102×91=902
Add the mutually exclusive outcomes: 9020+906+902=9028=4514
So both counters share a colour with probability 4514.
Worked example - conditional probability with a table
A survey of 120 students records whether they take Biology (B) and/or Chemistry (C). The results are summarised below.
Chemistry
No Chemistry
Total
Biology
38
22
60
No Bio
25
35
60
Total
63
57
120
A student is chosen at random. Find P(B∣C) and P(C∣B).
P(C)=12063. The intersection B∩C has 38 students, so P(B∣C)=12038÷12063=6338.
Always mention which total formed the denominator when quoting conditional probabilities.
Worked example - tree diagram with conditional branches
A factory machine produces 70% standard widgets (S) and 30% premium widgets (P). Among standard widgets, 4% are defective; among premium widgets, 9% are defective. A widget is selected at random.
Draw a tree with two stages: choose S or P, then good or defective.
Probability of a defective widget: (0.7×0.04)+(0.3×0.09)=0.028+0.027=0.055.
Probability that a defective widget came from the premium line: 0.0550.3×0.09≈0.4909.
Thus just under half of the defective widgets originated from the premium line even though premiums form only 30% of production.
Reverse conditional probability checkpoint
When the outcome is already known, the denominator is not the original group size. It is all paths that produce that known outcome.
Question wording
Numerator
Denominator
Trap to avoid
"Defective, given premium"
premium and defective path
all premium widgets
This asks for defect rate within premium.
"Premium, given defective"
premium and defective path
all defective widgets
This asks where defective widgets came from.
"Standard, given defective"
standard and defective path
all defective widgets
Use both defective paths in the denominator.
"Defective widget selected"
add all defective paths first
whole production
Do not use this as the denominator after "given defective".
Worked check: in the factory example, premium-and-defective has probability 0.3×0.09=0.027. All defective widgets have probability 0.055. Therefore P(premium∣defective)=0.0550.027≈0.491.
Misconception check: P(defective∣premium) is 0.09, but P(premium∣defective) is about 0.491. The same words in reverse use different denominators.
Practice Quiz
Run through complements, independence, and conditional probability checks with mixed-format questions.
Try this
A spinner is equally likely to show any integer from 1 to 8. Find P(X≥6∣X is even).
A box contains eight black, five white and three red marbles. Two marbles are drawn without replacement. Find P(at least one red).
In a game show, a contestant faces three doors. One door hides a prize. After the contestant chooses a door, the host opens a different door that never contains the prize. The contestant may stay or switch. Use a tree diagram to determine P(win∣switch).
