DSE Maths | Probability tutoring guide | Sample space, conditional probability, independence

2026-11-01

Tutor lesson-prep guides · Secondary Maths

Students can recite the union formula but add freely the moment they see “at least”, and treat “given that” as an ordinary probability. The fix is not another formula: it is writing the sample space first. This lesson makes that the opening move of every question.

Objectives — Students can: (1) list the sample space and count outcomes; (2) tell mutually exclusive from independent and pick addition or multiplication correctly; (3) handle “at least” by the complement; (4) read “given that” and narrow the denominator.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Fractions	simplifying and common denominators	6/36 simplifies to 1/6
Systematic listing	writing outcomes in order	a (1,1) to (6,6) table
Sets	intersection and union	A and B both happen

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-10 min	Sample space	count first, no formulas	36 outcomes for two dice
10-24 min	Addition rule	add only when exclusive	P(A or B) = P(A)+P(B)-P(A and B)
24-38 min	Multiplication	multiply only when independent	0.6×0.5=0.3
38-48 min	Conditional	narrow the denominator	given a face card, denominator is 12
48-55 min	“At least”	use the complement	three coins, at least one head = 1-1/8
55-60 min	Exit check	three questions, marked at once	independent work

03 Teaching script (read it out)

Scene 1: Opening move: write the sample space out

Two dice give 36 ordered outcomes; draw the 6×6 table instead of imagining it. The pairs summing to 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), so 6 outcomes, and P(sum 7) = 6/36 = 1/6.
The usual error is merging (1,6) with (6,1), which shrinks the total to 21 and ruins everything after it. Ask the student: why are two identical numbers still two outcomes? (The first and second die are different objects.) Student task: tabulate every outcome summing to 8 and count them.

Scene 2: Addition and multiplication: exclusive or independent first

Mutually exclusive means both cannot happen, so P(A or B) = P(A) + P(B). If they can happen together, subtract the intersection, as when drawing a heart or a king. Independent means one does not change the other, so multiply: 0.6 x 0.5 = 0.3.
The deciding phrase is “with replacement” or “no effect”. Drawing without replacement is not independent, so the second denominator changes. Student task: compute the chance that neither happens, (1-0.6)x(1-0.5) = 0.2. Reminder: mixing up exclusive and independent is the biggest mark loser in this chapter.

Scene 3: Conditional probability and “at least”: two fixed moves

“Given a face card, find the chance it is a king”: the denominator becomes the 12 face cards and the numerator is 4 kings, so 4/12 = 1/3, not 4/52. “At least one head in 3 coins”: the complement is no heads at all, 1/8, so the answer is 1-1/8 = 7/8.
Write P(A|B) = P(A and B) / P(B), but teach the narrow-the-denominator move first and the formula last. Ask the student: why use the complement for “at least”? (Counting one, two or three heads invites omissions.) Student task: two dice, chance of at least one 6 (11/36).

04 Question types and worked steps

Type	Example	Steps	Answer
Single event	sum of two dice is 7	count the sample space	6/36=1/6
Union	heart or king	subtract the intersection	13/52+4/52-1/52
Independent product	A 0.6, B 0.5	multiply directly	0.3
Conditional	given a face card, a king	narrow the denominator	4/12=1/3

Sanity check (teach this): Every probability lies between 0 and 1; an answer of 1.2 is wrong on sight. Count the space properly: 36 for two dice, 8 for three coins, 52 for a deck. A conditional probability is usually larger than the unconditional one because the denominator shrank: 1/3 beats 4/52. Multiplying independent events must give a smaller number: 0.3 is below 0.6 and 0.5. “At least” and “at most” are complements, so the two answers must add to 1.

05 Independent practice (3 questions)

Question	Answer	What to watch
Chance of a sum of 7	1/6	6/36 simplified
At least one head in three coins	7/8	1 – 1/8
King given a face card	1/3	denominator narrowed to 12

06 Common mistakes and what to say

Mistake	Student writes	Say this
Treating (1,6) and (6,1) as one	total becomes 21	“Two dice are two objects, reversed counts as another outcome.”
Confusing exclusive with independent	adding independent events	“If they can happen together, you cannot just add.”
Conditional keeps 52	answers 4/52	“We already know it is a face card, so the bottom is 12.”
Adding every case for “at least”	misses two heads	“Take 1 minus none at all: faster and nothing is missed.”

07 Homework (paper, section and question numbers)

Secondary maths chapter index — questions 4 and 6 are the same type
Prep guide home — other chapters in this subject
Primary maths papers — fraction warm-up

08 Exit check (last 5 minutes)

Question	Answer	Decision
Chance of a sum of 8	5/36	all three correct means move to statistics
Denominator of P(A\|B)	the sample space of B	one wrong means reteach
Method for “at least one”	1 minus none at all	two wrong means back to act one

09 Tutor FAQ

How do exclusive and independent differ?

Exclusive asks whether both can happen at once; independent asks whether one affects the other. Neither implies the other.

Why narrow the denominator for conditional probability?

Because “given B” restricts the sample space to B, so every chance is recomputed inside B.

Must “at least” always use the complement?

Usually, especially with three or more trials; with only two or three cases, listing works but must be complete.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.