Define equiprobable outcomes before calculating probability
Count elementary outcomes rather than treating every summary label as equally likely.
Section 5.3 · Probability
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.3. Define the sample space and state its probability assumptions before counting. Fair independent dice give equiprobable ordered pairs; their sums are not equiprobable. Random selection means each specified individual is equally likely under the stated sampling model. Without replacement, later branch probabilities change; with replacement and independent mixing, they remain the same. A′ is a complement, A ∩ B means both, A ∪ B means either or both, and P(A|B) means A given B with P(B) > 0. Mutually exclusive and independent are different properties. The explicit general addition formula is not required by this section; explain overlap through disjoint regions. Conditional probabilities may use P(A|B) = P(A ∩ B)/P(B). Numerical models describe the stipulated experiment, not automatically a real population. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- List ordered outcomes systematically.
- Simplify fractions and count successive choices.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For two distinguishable six-sided dice, how many ordered pairs exist?
What you will learn
- Construct a complete finite sample space with stated equal-likelihood assumptions.
- Calculate event probabilities and recognise non-equiprobable summaries.
Use the elementary experiment
Two fair independent six-sided dice give 36 equally likely ordered pairs (first die, second die). The pairs (1, 6) and (6, 1) are different elementary outcomes. A total of 7 occurs in six pairs, so its probability is 6/36 = 1/6. A total of 2 occurs only at (1, 1), with probability 1/36.
The eleven possible totals from 2 to 12 are not equally likely. Dividing by 11 counts summary labels and ignores how many elementary pairs create each label. The favourable-count divided by total-count rule requires an equiprobable sample space; if outcomes have different probabilities, add their actual weights instead.
Repeated labels do not erase elementary outcomes
Three independent fair coin tosses have eight equally likely ordered H/T strings. Exactly two heads occurs at HHT, HTH and THH, so its probability is 3/8. At least two heads includes HHH as well, giving 4/8 = 1/2. Counts of heads 0, 1, 2, 3 do not have equal probabilities.
A fair spinner with four equal sectors labelled 1, 1, 2, 3 has four equally likely sector outcomes. The label 1 has probability 2/4 = 1/2, while 2 and 3 each have probability 1/4. Fairness concerns equal sector areas under the stipulated spinner model, not equality of distinct printed labels. State assumptions instead of inferring fairness from appearance alone.
Pause and explain
What is P(total 7) for two fair independent dice?
Worked example
Two fair independent dice are rolled. Find the probability of total 7 and explain why counting totals equally fails.
Show the worked solution
- There are 6 × 6 = 36 equiprobable ordered pairs.
- Total 7 has pairs (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
- Its probability is 6/36 = 1/6; totals have different numbers of contributing pairs.
Answer P(total 7) = 1/6; total labels are not equiprobable.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find P(total 7) when two fair independent six-sided dice are rolled.
Give me a hint
List all ordered pairs with the required total.
Compare my reasoning
- The sample space contains 36 equiprobable ordered pairs.
- The six favourable pairs are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
- Divide 6 by 36 and simplify to 1/6.
1/6.
Look for these in your work
- Use the ordered-pair sample space.
- State the fair independent dice assumptions.
Three independent fair coins are tossed. Find the probability of at least two heads.
Give me a hint
Include exactly two heads and the all-heads string.
Compare my reasoning
- Eight ordered H/T strings are equiprobable.
- The favourable strings are HHT, HTH, THH and HHH.
- Their probability is 4/8 = 1/2.
1/2.
Look for these in your work
- Interpret at least two correctly.
- Do not give equal weight to different head counts.
A fair spinner has four equal sectors labelled 1, 1, 2, 3. Find P(label 1) and explain why 1/3 is wrong.
Give me a hint
The equal sectors, not the distinct labels, are the elementary outcomes.
Compare my reasoning
- All four sectors have probability 1/4 in the stated model.
- Two sectors carry label 1, so its probability is 2/4 = 1/2.
- The value 1/3 incorrectly treats the three distinct labels as equiprobable.
1/2; repeated label 1 occupies two of four equal sectors.
Look for these in your work
- Retain both sectors with the repeated label.
- Identify exactly what is equally likely.
Common mistakes
- Keep the order of distinguishable dice or tosses.
- Count elementary outcomes, not equally weighted summary labels.
- State fairness and independence when the model needs them.
What can you explain now?
Are the four possible head counts for three fair coin tosses equally likely?
Compare with the explanation
No; their elementary-outcome counts are 1, 3, 3 and 1.
The eight ordered toss strings are equiprobable, not the four head-count labels.
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Make it stick
Choose the elementary outcomes and justify equal likelihood before using favourable count divided by total count.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.