Condition on a restricted sample space
Use only outcomes consistent with the given information and keep the direction of the condition explicit.
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 joint outcomes and intersections.
- Calculate ratios of favourable and total counts.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
In P(A|B), which outcomes remain in the conditional sample space?
What you will learn
- Calculate conditional probabilities from an equiprobable sample space.
- Distinguish P(A|B) from P(B|A) and require a positive conditioning probability.
Given information changes the denominator
P(A|B) means the probability of A given that B occurred. Among equiprobable elementary outcomes, restrict the denominator to B, then count the outcomes that also satisfy A. On a fair die, given a result greater than 3, the remaining sample space is {4,5,6}. An even result occurs at 4 or 6, so its conditional probability is 2/3, not the unconditional 1/2.
The ratio is P(A|B) = P(A ∩ B)/P(B), provided P(B) > 0. The numerator uses both events while the denominator uses the given event. Conditioning on a probability-zero event cannot use this finite-probability ratio; do not divide by zero or assert an ordinary conditional value from it.
Reversing the condition usually changes the answer
Let A be even and B be greater than 2 on a fair die. Then P(A|B) = 2/4 = 1/2, while P(B|A) = 2/3. Both numerators count {4,6}, but the conditioning denominators have different sizes. Read ‘given’ carefully rather than treating the vertical bar as a symmetric intersection.
For two fair independent dice, given total at least 9, there are ten possible ordered pairs. Four have first die 6, so P(first 6 | total ≥ 9) = 4/10 = 2/5. In reverse, given first die 6, six pairs remain and four have total at least 9, giving 2/3. Count the restricted grid rather than replacing it with equally likely total labels.
Pause and explain
Given a fair die result greater than 3, what is P(even)?
Worked example
Two fair independent dice are rolled. Find P(first die 6 | total at least 9).
Show the worked solution
- There are ten pairs with totals 9, 10, 11 or 12: counts 4, 3, 2 and 1.
- The favourable pairs are (6,3), (6,4), (6,5), (6,6).
- The restricted probability is 4/10 = 2/5.
Answer 2/5.
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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.
A fair die result is known to be greater than 3. Find P(even | greater than 3).
Give me a hint
Restrict the possible results before counting even ones.
Compare my reasoning
- The given event leaves {4,5,6}.
- Two of those results, 4 and 6, are even.
- The conditional probability is 2/3.
2/3.
Look for these in your work
- Use the restricted denominator three.
- Keep both favourable results.
Two fair independent dice are rolled. Find P(total 7 | first die odd), then P(first die odd | total 7).
Give me a hint
Count the same intersection against two different conditioning sets.
Compare my reasoning
- An odd first die gives 3 × 6 = 18 pairs, three of which sum to 7.
- Thus P(total 7 | first odd) = 3/18 = 1/6.
- Total 7 gives six pairs, three with first die odd, so the reverse conditional is 1/2.
1/6 for total 7 given first odd; 1/2 for first odd given total 7.
Look for these in your work
- Identify the numerator intersection.
- Use the correct given-event count for each direction.
Two fair independent dice are rolled. Find P(first die 6 | total at least 9) and compare with P(total at least 9 | first die 6).
Give me a hint
Totals 9–12 have ten elementary pairs; a fixed first die has six.
Compare my reasoning
- There are ten pairs with total at least 9; four begin with 6.
- The first conditional is 4/10 = 2/5.
- Given first die 6, four of six second-die results give total at least 9, so the reverse is 2/3.
2/5 and 2/3 respectively.
Look for these in your work
- Keep the condition direction explicit.
- Count elementary pairs rather than equiprobable totals.
Common mistakes
- The given event supplies the denominator.
- Conditional probabilities are directional.
- Use the ratio only when the conditioning event has positive probability.
What can you explain now?
Must P(A|B) equal P(B|A)?
Compare with the explanation
No; their conditioning denominators can differ.
The intersection is shared, but the two given-event probabilities need not match.
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Make it stick
Circle the given event, restrict the denominator to it, and count the shared outcomes; check the denominator is positive.
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.