Normalise probability tables and check parameter bounds
Use the total probability equation, then verify every entry rather than accepting a sum of one alone.
Section 5.4 · Discrete random variables
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.4. A probability table needs non-negative probabilities summing to one. E(X) is a probability-weighted mean; Var(X) = E(X²) − [E(X)]², with squared units. A binomial model B(n,p) counts successes in a fixed non-negative integer number n of independent two-outcome trials with a constant success probability p. A geometric model Geo(p) counts the trial of the first success, starting at 1, with independent trials and a constant 0 < p ≤ 1. Formula proofs are not required. The expectation of the geometric distribution is required; its variance is not required by section 5.4. Normal approximations belong to section 5.5 and are not used here. Exact model probabilities do not establish real-world independence or constant rates without evidence. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Solve linear equations.
- Check non-negative fractions and inequalities.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What must the probabilities in a complete distribution sum to?
What you will learn
- Find an unknown probability or normalising constant.
- Determine parameter ranges that keep every probability valid.
The complete table totals one
For values 0, 1, 2 with probabilities k, 2k, 3k, normalisation gives 6k = 1, so k = 1/6. The entries become 1/6, 1/3, 1/2. This is a probability-weighted distribution, not equal column weighting. An unlisted possible outcome would invalidate this total equation.
For a complete table with probabilities 0.15, a, 0.35, the missing value is a = 1 − 0.15 − 0.35 = 0.50. Each entry must be between zero and one. A zero probability is allowed in a candidate table; a negative probability is not, even if other entries make the sum one.
A normalised family can still contain invalid entries
Probabilities k, 2k, 1 − 3k automatically sum to one for every k. Validity additionally requires k ≥ 0, 2k ≥ 0 and 1 − 3k ≥ 0, giving 0 ≤ k ≤ 1/3. In that range all three entries also lie at most one.
At k = 0.20 the table is 0.20, 0.40, 0.40 and is valid. At k = 0.40 it is 0.40, 0.80, −0.20 and is invalid. Normalisation alone does not prove validity. Preserve exact fractions while deriving bounds; rounding near an endpoint can hide a small negative entry.
Pause and explain
A complete table has probabilities k, 2k, 3k. What is k?
Worked example
A complete table has probabilities k, 2k, 3k. Find k and check the entries.
Show the worked solution
- Add all entries: k + 2k + 3k = 6k.
- Set 6k = 1 to obtain k = 1/6.
- The entries 1/6, 1/3, 1/2 are non-negative and sum to one.
Answer k = 1/6.
Try a topic test
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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 complete distribution has probabilities k, 2k, 3k. Find k and the three entries.
Give me a hint
Set their sum to one, then substitute back.
Compare my reasoning
- The total is 6k = 1.
- Thus k = 1/6 and the entries are 1/6, 1/3, 1/2.
- They are all non-negative and sum to one.
k = 1/6; entries 1/6, 1/3, 1/2.
Look for these in your work
- Use all columns in the total.
- Verify the resulting probabilities.
A complete table has probabilities 0.15, a, 0.35. Find a and check validity.
Give me a hint
The missing probability is the complement of the known total.
Compare my reasoning
- The known total is 0.50.
- a = 1 − 0.50 = 0.50.
- The entries 0.15, 0.50, 0.35 are non-negative and sum to one.
a = 0.50; the table is valid.
Look for these in your work
- Use a probability total rather than a value total.
- Check the missing entry is within bounds.
Probabilities are k, 2k, 1 − 3k. Find the valid k range and evaluate k = 0.20 and k = 0.40.
Give me a hint
The sum is already one; non-negativity supplies the missing restrictions.
Compare my reasoning
- k ≥ 0 and 1 − 3k ≥ 0 give 0 ≤ k ≤ 1/3.
- k = 0.20 gives 0.20, 0.40, 0.40 and is valid.
- k = 0.40 gives 0.40, 0.80, −0.20 and is invalid.
0 ≤ k ≤ 1/3; 0.20 is valid, 0.40 is invalid.
Look for these in your work
- Check individual entries as well as their sum.
- Include the admissible boundary values.
Common mistakes
- The table must include every possible value.
- A sum of one does not excuse a negative entry.
- Check bounds before rounding parameters.
What can you explain now?
Is the table 0.4, 0.8, −0.2 valid because it sums to one?
Compare with the explanation
No; one entry is negative.
Normalisation and non-negativity are separate requirements.
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Make it stick
Normalise the complete table, then test each entry and derive any parameter bounds.
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.