IP EMaths Notes (Upper Sec, Year 3-4): 14) Probability

Study guide08 Nov 2025, 00:00 Z

Compute probabilities with mutually exclusive, independent, and conditional events using organised tables or tree diagrams.

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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.

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These notes align with SEAB GCE O-Level Mathematics (4052) content used in IP programmes (exams from 2026).

Status: SEAB O-Level Mathematics 4052 syllabus (exams from 2026) checked 2025-11-30 - scope unchanged; remains the reference for these notes.

Quick reference

$P(A)$ is the probability of event A; $P(A') = 1 - P(A)$

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Marcus Pang·Managing Director (Maths)

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 $P(A \mid B)$ with $P(A)$	confusing independent with mutually exclusive

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 $\dfrac{5}{10}$ , second red is still $\dfrac{5}{10}$ .
Without replacement	The first item is not put back.	Total decreases by 1.	First red is $\dfrac{5}{10}$ , then second red is $\dfrac{4}{9}$ .
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 $\dfrac{5}{9}$ .
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.

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 $P(A \cap B) = 0$ .	Calling them independent just because they are separate labels.
Does knowing B leave $P(A)$ unchanged?	They are independent.	Check whether $P(A \mid B) = P(A)$ .	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.

Expression	Denominator to use	Numerator to use	What the wording sounds like
$P(A \mid B)$	all cases in B	cases in both A and B	"A, given B"
$P(B \mid A)$	all cases in A	cases in both A and B	"B, given A"
$P(A \cap B)$	the whole sample space	cases in both A and B	"A and B"

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.

IP EMaths Notes (Upper Sec, Year 3-4): 14) Probability

Quick reference

Sources

Organising your work

Decision map - choose the layout before calculating

Tree Denominator Checkpoint

Mutually exclusive vs independent

Exclusive-or-independent checkpoint

Conditional probability workflow

Conditional denominator checkpoint

At-least-one checkpoint

Worked example - two draws without replacement

Worked example - conditional probability with a table

Worked example - tree diagram with conditional branches

Reverse conditional probability checkpoint

Practice Quiz

Try this

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	Chemistry	No Chemistry	Total
Biology	$38$	$22$	$60$
No Bio	$25$	$35$	$60$
Total	$63$	$57$	$120$

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".