JC2 H2 Maths Bukit Timah | Find Binomial n and p

Did you know? A binomial distribution with mean 6 and variance 4.5 must have success probability p=0.25 and 24 trials, provided those are its exact theoretical moments. You can recover both missing parameters from two ordinary numbers. But the solution is not complete until the number of trials is a positive integer and the probability lies between zero and one. Otherwise, the algebra may describe an impossible binomial experiment.

This JC2 H2 Mathematics Tutor Bukit Timah article answers the A-Level H2 Mathematics 9758 question: How can we reconstruct n and p when a binomial distribution’s mean and variance are given? We will calculate the unknown parameters, explain why the variance-to-mean ratio is the fastest route, inspect incompatible moment pairs, and distinguish exact theoretical moments from sample estimates. The guide is designed for quick answers at the top and deeper practice afterwards.

For Bukit Timah JC2 learners preparing for H2 Mathematics, knowing np and np(1−p) is not enough if a modelling question changes the direction of the calculation. The 2027 Singapore–Cambridge H2 Mathematics 9758 syllabus includes the binomial B(n,p) model and use of its mean and variance without proof. The important skill here is working backwards and assessing feasibility, not guessing a convenient trial count. Every example below is an invented mathematical exercise.

Quick answer: divide variance by mean to find p

If X~B(n,p) with positive integer n, then E(X)=np and Var(X)=np(1−p). Denote the given exact mean by m and variance by v. When m>0, divide the second equation by the first: v/m=1−p. Therefore p=1−v/m, and n=m/p. Finally verify that n is a positive integer and 0<p≤1, and substitute the values back to reproduce both moments.

Step one: write m=np and v=np(1−p).
Step two: for m>0, calculate p=1−v/m.
Step three: calculate n=m/p; do not round an exact non-integer to make it work.
Step four: check np=m and np(1−p)=v.
Step five: if m=0, treat the zero-mean case separately because division by m is undefined.
Step six: the binomial sampling assumptions still matter even when the algebra gives a valid pair.

The complete worked answer: mean 6 and variance 4.5

Suppose an invented count X~B(n,p) has exact theoretical mean m=6 and exact variance v=4.5. The ratio v/m=4.5/6=0.75, so 1−p=0.75 and p=0.25. Then n=6/0.25=24. This means the specified binomial model has 24 independent trials and one-quarter success chance per trial.

Check both equations independently. Its mean is np=24×0.25=6. Its variance is np(1−p)=24×0.25×0.75=4.5. Both are correct, n is a positive integer, and p is a valid probability. The full answer is therefore X~B(24,0.25), under the question’s stated binomial assumption.

It is vital to distinguish the average count from the possible maximum count. The mean 6 does not mean six trials. A binomial random variable ranging from zero through 24 may have expected count 6. Its realised count in a particular run could be 3, 6, 10 or another integer from zero to 24. The expectation is an average across the model’s probability distribution, not a declaration that every run has exactly six successes.

A second example: mean 3 and variance 2.4

If E(X)=3 and Var(X)=2.4, then v/m=2.4/3=0.8, giving p=0.2. Hence n=3/0.2=15. This is valid, and checking gives 15×0.2=3 for the mean and 15×0.2×0.8=2.4 for the variance. The two numbers recovered from the moments have different roles: n counts trials and must be integral, whereas p is the probability of the event counted as success.

If the problem supplies a standard deviation instead of a variance, first square it. A standard deviation √2.4 corresponds to variance 2.4; using √2.4 in v/m would be a dimensional and numerical mistake. A variance is measured in squared count units, while the standard deviation is measured in count units.

The ratio is a model-validity test

For a non-degenerate binomial with 0<p<1 and n>0, the factor 1−p lies strictly between zero and one. Therefore its variance satisfies 0<v<m. A positive variance equal to or larger than a positive mean cannot be an exact binomial moment pair. This provides a fast first diagnostic before detailed algebra. It does not say that all count distributions have variance less than mean; it says that the binomial family has this constraint.

Even when v is between zero and m, the parameters may fail the integer trial-count condition. The ratio test is necessary but not sufficient. After obtaining a valid p, you must still compute n and check that it is a positive integer. A trial count such as 12.5 is not a legitimate parameter for a standard binomial experiment.

A plausible-looking but impossible exact pair

Take m=5 and v=3. The ratio gives p=1−3/5=0.4, which is a perfectly valid probability. But n=m/p=5/0.4=12.5. A binomial model cannot have half a trial, so these exact theoretical moments do not belong to any B(n,p) with positive integer n. Rounding n to 12 or 13 would alter at least one of the moments. The correct conclusion is incompatibility, not a nearest-integer guess.

If the numbers were explicitly stated to be rounded estimates from a dataset, the situation would be different: one might consider a suitable approximate model and acknowledge sampling error. This article’s impossibility conclusions concern exact stated theoretical moments. The wording of the problem determines what may reasonably be inferred.

What if the variance exceeds the mean?

Suppose an invented problem claims m=4, v=5 for a binomial variable. Then p=1−5/4=−0.25. A probability cannot be negative, so no exact binomial model fits. With m=6 and v=−1, the problem is even more basic: a variance cannot be negative for any random variable with a valid finite variance. These are not calculator errors to be patched with arbitrary rounding; they violate the mathematical definition.

This does not prevent real empirical count datasets from having sample variance greater than sample mean due to sampling variation or because the underlying process is different. The exam-style question here asserts a distribution with exact theoretical m and v. Keep those contexts distinct and make the inference only as strong as the premises support.

Boundary case one: zero mean, zero variance

For a positive integer n, m=np=0 implies p=0. Every trial then has zero probability of success, so the count is always zero and its variance is zero. But if n is unknown, many parameter pairs B(n,0) have the same moments. For example, B(5,0) and B(200,0) both have mean and variance zero. Thus the moment pair m=v=0 does not uniquely recover n.

The formula p=1−v/m must not be used here because v/m=0/0 is undefined. A careful learner first checks whether the denominator is zero and then reasons from the original equation np=m. That separate boundary treatment is a mark of genuine model understanding rather than blind substitution.

Boundary case two: zero variance, positive mean

Suppose m>0 while v=0. Then p=1−0/m=1 and n=m. Because n must be a positive integer, a positive integer mean such as m=7 gives the valid degenerate model B(7,1). Every trial succeeds, so the success count is always seven. But a stated exact m=7.5 with v=0 cannot correspond to a positive integer n under the same logic.

These boundary cases are mathematically possible or impossible for clear reasons. The objective is not to memorise a table of exceptions but to understand what happens when probability is zero or one: uncertainty about the count disappears. If the question requires all parameter pairs and the zero case allows many n, state that non-uniqueness rather than inventing a unique count.

How p changes mean and variance when n is fixed

Fixed n=20, p	Mean np	Variance np(1−p)	Interpretation
0.1	2	1.8	Successes are uncommon
0.2	4	3.2	Mean rises as p increases
0.5	10	5	Largest variance for this fixed n
0.8	16	3.2	Mean higher, variance lower
0.9	18	1.8	Success count often near the upper end

The variance does not rise forever with the mean at fixed n. It is largest around p=0.5 and becomes smaller as p approaches zero or one. This makes sense: when almost every trial fails or almost every trial succeeds, the count is comparatively predictable. The relation between mean and variance therefore carries information about p and is not just a second redundant statistic.

A model must have repeated independent trials

The mathematical inversion assumes a binomial count: a fixed number of trials, a success/failure classification for each, constant probability of success and independence between trials. If the trial probabilities change with time or the trials influence one another, the binomial moment formulas may be inappropriate. Even perfectly recovered numbers n and p do not independently prove the assumptions of an applied model.

“Success” is simply the chosen event and need not mean something desirable. A fictional testing process may count flagged items as successes. If the question defines that convention, p is the flag probability, not the pass probability. Keeping the event definition fixed prevents accidentally replacing p with 1−p mid-solution.

One-minute diagnostic before the next set of questions

Give a learner the pair m=6,v=4.5. Ask for p alone, then n, then both checks: 0.25, 24, and the original moments. Change the exact moments to m=5,v=3 and see whether the learner notices the impossible n=12.5. Finally change them to m=v=0 and ask whether n is identifiable; it is not. A student who can articulate all three situations understands more than the two formulas.

A typical diagnosis is precise. If the learner finds p correctly but rounds n without permission, the weakness is a failure to respect the integer parameter. If they use variance even when the question supplies a standard deviation, they need units and definitions. If they divide by zero in a boundary case, they need a pre-division domain check. Each error has a specific short repair.

Thirty original feasible moment-recovery clinics

In these thirty invented exercises, m and v are exact theoretical binomial moments. Recover the two parameters and verify them. The numbers vary in size and success probability; the same algebra works each time. The important difference between exercises is the feasibility audit, the complement 1−p, and what the results mean for the model’s count. These are not real student scores or recorded research data.

1. Recover n and p from m=2.4, v=1.92

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 2.4 and the variance is 1.92. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=1.92/2.4=0.8, so p=1−v/m=0.2. The trial count is n=m/p=2.4/0.2=12. Checking both original equations gives np=2.4 and np(1−p)=1.92. Thus the parameters define the valid exact model B(12,0.2). The number of trials is integral and 0<p<1.

Transfer and interpretation. The trial count is not the mean. Even if m happens to be an integer, it describes expected successes, not total trials. Imagine the same count collected from a different experiment with changing success probabilities: even if the sample mean looked similar, you could not automatically use the binomial model. A formula is only as good as the assumptions connecting its parameters to the experiment.

2. Recover n and p from m=7.5, v=3.75

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 7.5 and the variance is 3.75. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=3.75/7.5=0.5, so p=1−v/m=0.5. The trial count is n=m/p=7.5/0.5=15. Checking both original equations gives np=7.5 and np(1−p)=3.75. Thus the parameters define the valid exact model B(15,0.5). The number of trials is integral and 0<p<1.

Transfer and interpretation. For this non-degenerate model, 0<v<m. Check the inequality before trusting the recovered p. Imagine the same count collected from a different experiment with changing success probabilities: even if the sample mean looked similar, you could not automatically use the binomial model. A formula is only as good as the assumptions connecting its parameters to the experiment.

3. Recover n and p from m=1.8, v=1.62

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 1.8 and the variance is 1.62. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=1.62/1.8=0.9, so p=1−v/m=0.1. The trial count is n=m/p=1.8/0.1=18. Checking both original equations gives np=1.8 and np(1−p)=1.62. Thus the parameters define the valid exact model B(18,0.1). The number of trials is integral and 0<p<1.

Transfer and interpretation. If you relabel successes as failures, the mean changes to n(1−p) but the variance remains np(1−p). Imagine the same count collected from a different experiment with changing success probabilities: even if the sample mean looked similar, you could not automatically use the binomial model. A formula is only as good as the assumptions connecting its parameters to the experiment.

4. Recover n and p from m=8, v=4.8

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 8 and the variance is 4.8. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=4.8/8=0.6, so p=1−v/m=0.4. The trial count is n=m/p=8/0.4=20. Checking both original equations gives np=8 and np(1−p)=4.8. Thus the parameters define the valid exact model B(20,0.4). The number of trials is integral and 0<p<1.

Transfer and interpretation. A fractional probability is legitimate. It is n, the number of trials, that must be integral. Imagine the same count collected from a different experiment with changing success probabilities: even if the sample mean looked similar, you could not automatically use the binomial model. A formula is only as good as the assumptions connecting its parameters to the experiment.

5. Recover n and p from m=19.2, v=3.84

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 19.2 and the variance is 3.84. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=3.84/19.2=0.2, so p=1−v/m=0.8. The trial count is n=m/p=19.2/0.8=24. Checking both original equations gives np=19.2 and np(1−p)=3.84. Thus the parameters define the valid exact model B(24,0.8). The number of trials is integral and 0<p<1.

Transfer and interpretation. A particular observed count can differ from the expectation; the theoretical moments describe the distribution, not a guaranteed outcome. Imagine the same count collected from a different experiment with changing success probabilities: even if the sample mean looked similar, you could not automatically use the binomial model. A formula is only as good as the assumptions connecting its parameters to the experiment.

6. Recover n and p from m=7.5, v=5.25

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 7.5 and the variance is 5.25. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=5.25/7.5=0.7, so p=1−v/m=0.3. The trial count is n=m/p=7.5/0.3=25. Checking both original equations gives np=7.5 and np(1−p)=5.25. Thus the parameters define the valid exact model B(25,0.3). The number of trials is integral and 0<p<1.

Transfer and interpretation. Recover the variance from the reconstructed parameters rather than trusting only the mean check. Imagine the same count collected from a different experiment with changing success probabilities: even if the sample mean looked similar, you could not automatically use the binomial model. A formula is only as good as the assumptions connecting its parameters to the experiment.

7. Recover n and p from m=22.5, v=5.625

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 22.5 and the variance is 5.625. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=5.625/22.5=0.25, so p=1−v/m=0.75. The trial count is n=m/p=22.5/0.75=30. Checking both original equations gives np=22.5 and np(1−p)=5.625. Thus the parameters define the valid exact model B(30,0.75). The number of trials is integral and 0<p<1.

Transfer and interpretation. The success event is the event being counted; do not silently change its definition to the complementary event. Imagine the same count collected from a different experiment with changing success probabilities: even if the sample mean looked similar, you could not automatically use the binomial model. A formula is only as good as the assumptions connecting its parameters to the experiment.

8. Recover n and p from m=8, v=6

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 8 and the variance is 6. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=6/8=0.75, so p=1−v/m=0.25. The trial count is n=m/p=8/0.25=32. Checking both original equations gives np=8 and np(1−p)=6. Thus the parameters define the valid exact model B(32,0.25). The number of trials is integral and 0<p<1.

Transfer and interpretation. If the figures were rounded empirical moments rather than exact theoretical ones, parameter recovery would require an approximation caveat. Imagine the same count collected from a different experiment with changing success probabilities: even if the sample mean looked similar, you could not automatically use the binomial model. A formula is only as good as the assumptions connecting its parameters to the experiment.

9. Recover n and p from m=21.6, v=8.64

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 21.6 and the variance is 8.64. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=8.64/21.6=0.4, so p=1−v/m=0.6. The trial count is n=m/p=21.6/0.6=36. Checking both original equations gives np=21.6 and np(1−p)=8.64. Thus the parameters define the valid exact model B(36,0.6). The number of trials is integral and 0<p<1.

Transfer and interpretation. The standard deviation is the positive square root of v, which is different from the value of v unless v is zero or one. Imagine the same count collected from a different experiment with changing success probabilities: even if the sample mean looked similar, you could not automatically use the binomial model. A formula is only as good as the assumptions connecting its parameters to the experiment.

10. Recover n and p from m=8, v=6.4

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 8 and the variance is 6.4. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=6.4/8=0.8, so p=1−v/m=0.2. The trial count is n=m/p=8/0.2=40. Checking both original equations gives np=8 and np(1−p)=6.4. Thus the parameters define the valid exact model B(40,0.2). The number of trials is integral and 0<p<1.

Transfer and interpretation. Under independence and constant success probability, the binomial moment equations determine the model parameters for positive m and feasible n. Imagine the same count collected from a different experiment with changing success probabilities: even if the sample mean looked similar, you could not automatically use the binomial model. A formula is only as good as the assumptions connecting its parameters to the experiment.

11. Recover n and p from m=22.5, v=11.25

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 22.5 and the variance is 11.25. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=11.25/22.5=0.5, so p=1−v/m=0.5. The trial count is n=m/p=22.5/0.5=45. Checking both original equations gives np=22.5 and np(1−p)=11.25. Thus the parameters define the valid exact model B(45,0.5). The number of trials is integral and 0<p<1.

12. Recover n and p from m=4.8, v=4.32

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 4.8 and the variance is 4.32. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=4.32/4.8=0.9, so p=1−v/m=0.1. The trial count is n=m/p=4.8/0.1=48. Checking both original equations gives np=4.8 and np(1−p)=4.32. Thus the parameters define the valid exact model B(48,0.1). The number of trials is integral and 0<p<1.

13. Recover n and p from m=20, v=12

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 20 and the variance is 12. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=12/20=0.6, so p=1−v/m=0.4. The trial count is n=m/p=20/0.4=50. Checking both original equations gives np=20 and np(1−p)=12. Thus the parameters define the valid exact model B(50,0.4). The number of trials is integral and 0<p<1.

14. Recover n and p from m=43.2, v=8.64

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 43.2 and the variance is 8.64. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=8.64/43.2=0.2, so p=1−v/m=0.8. The trial count is n=m/p=43.2/0.8=54. Checking both original equations gives np=43.2 and np(1−p)=8.64. Thus the parameters define the valid exact model B(54,0.8). The number of trials is integral and 0<p<1.

15. Recover n and p from m=18, v=12.6

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 18 and the variance is 12.6. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=12.6/18=0.7, so p=1−v/m=0.3. The trial count is n=m/p=18/0.3=60. Checking both original equations gives np=18 and np(1−p)=12.6. Thus the parameters define the valid exact model B(60,0.3). The number of trials is integral and 0<p<1.

16. Recover n and p from m=48, v=12

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 48 and the variance is 12. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=12/48=0.25, so p=1−v/m=0.75. The trial count is n=m/p=48/0.75=64. Checking both original equations gives np=48 and np(1−p)=12. Thus the parameters define the valid exact model B(64,0.75). The number of trials is integral and 0<p<1.

17. Recover n and p from m=18, v=13.5

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 18 and the variance is 13.5. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=13.5/18=0.75, so p=1−v/m=0.25. The trial count is n=m/p=18/0.25=72. Checking both original equations gives np=18 and np(1−p)=13.5. Thus the parameters define the valid exact model B(72,0.25). The number of trials is integral and 0<p<1.

18. Recover n and p from m=45, v=18

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 45 and the variance is 18. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=18/45=0.4, so p=1−v/m=0.6. The trial count is n=m/p=45/0.6=75. Checking both original equations gives np=45 and np(1−p)=18. Thus the parameters define the valid exact model B(75,0.6). The number of trials is integral and 0<p<1.

19. Recover n and p from m=16, v=12.8

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 16 and the variance is 12.8. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=12.8/16=0.8, so p=1−v/m=0.2. The trial count is n=m/p=16/0.2=80. Checking both original equations gives np=16 and np(1−p)=12.8. Thus the parameters define the valid exact model B(80,0.2). The number of trials is integral and 0<p<1.

20. Recover n and p from m=42, v=21

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 42 and the variance is 21. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=21/42=0.5, so p=1−v/m=0.5. The trial count is n=m/p=42/0.5=84. Checking both original equations gives np=42 and np(1−p)=21. Thus the parameters define the valid exact model B(84,0.5). The number of trials is integral and 0<p<1.

21. Recover n and p from m=9, v=8.1

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 9 and the variance is 8.1. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=8.1/9=0.9, so p=1−v/m=0.1. The trial count is n=m/p=9/0.1=90. Checking both original equations gives np=9 and np(1−p)=8.1. Thus the parameters define the valid exact model B(90,0.1). The number of trials is integral and 0<p<1.

22. Recover n and p from m=38.4, v=23.04

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 38.4 and the variance is 23.04. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=23.04/38.4=0.6, so p=1−v/m=0.4. The trial count is n=m/p=38.4/0.4=96. Checking both original equations gives np=38.4 and np(1−p)=23.04. Thus the parameters define the valid exact model B(96,0.4). The number of trials is integral and 0<p<1.

23. Recover n and p from m=80, v=16

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 80 and the variance is 16. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=16/80=0.2, so p=1−v/m=0.8. The trial count is n=m/p=80/0.8=100. Checking both original equations gives np=80 and np(1−p)=16. Thus the parameters define the valid exact model B(100,0.8). The number of trials is integral and 0<p<1.

24. Recover n and p from m=32.4, v=22.68

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 32.4 and the variance is 22.68. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=22.68/32.4=0.7, so p=1−v/m=0.3. The trial count is n=m/p=32.4/0.3=108. Checking both original equations gives np=32.4 and np(1−p)=22.68. Thus the parameters define the valid exact model B(108,0.3). The number of trials is integral and 0<p<1.

25. Recover n and p from m=90, v=22.5

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 90 and the variance is 22.5. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=22.5/90=0.25, so p=1−v/m=0.75. The trial count is n=m/p=90/0.75=120. Checking both original equations gives np=90 and np(1−p)=22.5. Thus the parameters define the valid exact model B(120,0.75). The number of trials is integral and 0<p<1.

26. Recover n and p from m=31.25, v=23.4375

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 31.25 and the variance is 23.4375. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=23.4375/31.25=0.75, so p=1−v/m=0.25. The trial count is n=m/p=31.25/0.25=125. Checking both original equations gives np=31.25 and np(1−p)=23.4375. Thus the parameters define the valid exact model B(125,0.25). The number of trials is integral and 0<p<1.

27. Recover n and p from m=76.8, v=30.72

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 76.8 and the variance is 30.72. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=30.72/76.8=0.4, so p=1−v/m=0.6. The trial count is n=m/p=76.8/0.6=128. Checking both original equations gives np=76.8 and np(1−p)=30.72. Thus the parameters define the valid exact model B(128,0.6). The number of trials is integral and 0<p<1.

28. Recover n and p from m=28.8, v=23.04

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 28.8 and the variance is 23.04. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=23.04/28.8=0.8, so p=1−v/m=0.2. The trial count is n=m/p=28.8/0.2=144. Checking both original equations gives np=28.8 and np(1−p)=23.04. Thus the parameters define the valid exact model B(144,0.2). The number of trials is integral and 0<p<1.

29. Recover n and p from m=75, v=37.5

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 75 and the variance is 37.5. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=37.5/75=0.5, so p=1−v/m=0.5. The trial count is n=m/p=75/0.5=150. Checking both original equations gives np=75 and np(1−p)=37.5. Thus the parameters define the valid exact model B(150,0.5). The number of trials is integral and 0<p<1.

30. Recover n and p from m=16, v=14.4

Try the invented moment card. X is a binomial success count with a positive integer trial count n and a constant success probability p. The exact theoretical mean is 16 and the variance is 14.4. Find n and p, then explain whether the recovered values satisfy the mathematical parameter constraints. Do the ratio step first rather than guessing the count.

Complete arithmetic. v/m=14.4/16=0.9, so p=1−v/m=0.1. The trial count is n=m/p=16/0.1=160. Checking both original equations gives np=16 and np(1−p)=14.4. Thus the parameters define the valid exact model B(160,0.1). The number of trials is integral and 0<p<1.

Eight feasibility decisions that are not all impossible

31. Is m=5, v=3 a possible binomial pair?

Question. Someone states that an unknown binomial B(n,p), with n a positive integer, has exact theoretical mean m=5 and exact variance v=3. Determine whether such a parameter pair is possible. Do not reject merely because p is a fraction or accept merely because a calculator displays two numbers.

Decision and reason. The pair is infeasible: p=0.4, but n=12.5 is not an integer. If m>0 and v≥0, start with p=1−v/m; then calculate n=m/p whenever p permits it. If m=0, reason from np=0 without division. The decisive checks are probability range, non-negative variance and integrality of the number of trials.

Follow-through. A mathematically feasible pair is only a parameter solution to the binomial model. A real dataset must still satisfy the assumptions of a fixed trial count, independent binary trials and common success probability. An infeasible exact pair cannot be fixed by choosing the nearest integer without changing the stated moments.

32. Is m=4, v=5 a possible binomial pair?

Question. Someone states that an unknown binomial B(n,p), with n a positive integer, has exact theoretical mean m=4 and exact variance v=5. Determine whether such a parameter pair is possible. Do not reject merely because p is a fraction or accept merely because a calculator displays two numbers.

Decision and reason. The pair is infeasible: p=−0.25 would be a negative probability. If m>0 and v≥0, start with p=1−v/m; then calculate n=m/p whenever p permits it. If m=0, reason from np=0 without division. The decisive checks are probability range, non-negative variance and integrality of the number of trials.

33. Is m=6, v=-1 a possible binomial pair?

Question. Someone states that an unknown binomial B(n,p), with n a positive integer, has exact theoretical mean m=6 and exact variance v=-1. Determine whether such a parameter pair is possible. Do not reject merely because p is a fraction or accept merely because a calculator displays two numbers.

Decision and reason. The pair is infeasible: A negative variance is impossible. If m>0 and v≥0, start with p=1−v/m; then calculate n=m/p whenever p permits it. If m=0, reason from np=0 without division. The decisive checks are probability range, non-negative variance and integrality of the number of trials.

34. Is m=0, v=2 a possible binomial pair?

Question. Someone states that an unknown binomial B(n,p), with n a positive integer, has exact theoretical mean m=0 and exact variance v=2. Determine whether such a parameter pair is possible. Do not reject merely because p is a fraction or accept merely because a calculator displays two numbers.

Decision and reason. The pair is infeasible: A positive-trial binomial with mean zero must also have variance zero. If m>0 and v≥0, start with p=1−v/m; then calculate n=m/p whenever p permits it. If m=0, reason from np=0 without division. The decisive checks are probability range, non-negative variance and integrality of the number of trials.

35. Is m=2, v=2 a possible binomial pair?

Question. Someone states that an unknown binomial B(n,p), with n a positive integer, has exact theoretical mean m=2 and exact variance v=2. Determine whether such a parameter pair is possible. Do not reject merely because p is a fraction or accept merely because a calculator displays two numbers.

Decision and reason. The pair is infeasible: The ratio would demand p=0 while the mean is positive. If m>0 and v≥0, start with p=1−v/m; then calculate n=m/p whenever p permits it. If m=0, reason from np=0 without division. The decisive checks are probability range, non-negative variance and integrality of the number of trials.

36. Is m=5.5, v=0 a possible binomial pair?

Question. Someone states that an unknown binomial B(n,p), with n a positive integer, has exact theoretical mean m=5.5 and exact variance v=0. Determine whether such a parameter pair is possible. Do not reject merely because p is a fraction or accept merely because a calculator displays two numbers.

Decision and reason. The pair is infeasible: The zero variance forces p=1 and n=5.5, not a valid integer count. If m>0 and v≥0, start with p=1−v/m; then calculate n=m/p whenever p permits it. If m=0, reason from np=0 without division. The decisive checks are probability range, non-negative variance and integrality of the number of trials.

37. Is m=3, v=2 a possible binomial pair?

Question. Someone states that an unknown binomial B(n,p), with n a positive integer, has exact theoretical mean m=3 and exact variance v=2. Determine whether such a parameter pair is possible. Do not reject merely because p is a fraction or accept merely because a calculator displays two numbers.

Decision and reason. The pair is feasible: p=1/3 and n=9 reproduce the exact given moments. If m>0 and v≥0, start with p=1−v/m; then calculate n=m/p whenever p permits it. If m=0, reason from np=0 without division. The decisive checks are probability range, non-negative variance and integrality of the number of trials.

38. Is m=9, v=6 a possible binomial pair?

Question. Someone states that an unknown binomial B(n,p), with n a positive integer, has exact theoretical mean m=9 and exact variance v=6. Determine whether such a parameter pair is possible. Do not reject merely because p is a fraction or accept merely because a calculator displays two numbers.

Decision and reason. The pair is feasible: p=1/3 and n=27 reproduce the exact given moments. If m>0 and v≥0, start with p=1−v/m; then calculate n=m/p whenever p permits it. If m=0, reason from np=0 without division. The decisive checks are probability range, non-negative variance and integrality of the number of trials.

Four boundary reminders before calculation

39. Zero mean and zero variance

Before dividing. The theoretical moments are m=0, v=0. Ask whether the mean is zero, whether the variance is non-negative and whether the trial count would have to be an integer. Do not automatically use a ratio whose denominator may vanish.

Correct conclusion. For positive integer n, p=0. The moment pair does not identify n uniquely. Explaining which constraint governs the answer is more useful than reporting an error message from a calculator. If any model parameter is not uniquely identified, say so instead of inventing a value.

40. Positive integer mean, zero variance

Before dividing. The theoretical moments are m=7, v=0. Ask whether the mean is zero, whether the variance is non-negative and whether the trial count would have to be an integer. Do not automatically use a ratio whose denominator may vanish.

Correct conclusion. p=1 and n=7. The count is constant at seven. Explaining which constraint governs the answer is more useful than reporting an error message from a calculator. If any model parameter is not uniquely identified, say so instead of inventing a value.

41. Equal positive mean and variance

Before dividing. The theoretical moments are m=3, v=3. Ask whether the mean is zero, whether the variance is non-negative and whether the trial count would have to be an integer. Do not automatically use a ratio whose denominator may vanish.

Correct conclusion. No finite positive-trial binomial has these exact moments; p would be zero despite positive mean. Explaining which constraint governs the answer is more useful than reporting an error message from a calculator. If any model parameter is not uniquely identified, say so instead of inventing a value.

42. Variance above a positive mean

Before dividing. The theoretical moments are m=8, v=8.5. Ask whether the mean is zero, whether the variance is non-negative and whether the trial count would have to be an integer. Do not automatically use a ratio whose denominator may vanish.

Correct conclusion. The formula forces a negative p, so the exact pair is impossible. Explaining which constraint governs the answer is more useful than reporting an error message from a calculator. If any model parameter is not uniquely identified, say so instead of inventing a value.

A probability question after the parameters are recovered

Once you have recovered B(24,0.25) from the mean and variance, you can also calculate probabilities within the model. For exactly two successes, P(X=2)=C(24,2)(0.25)²(0.75)²². For at least one success, the complement yields P(X≥1)=1−(0.75)²⁴. These are follow-on applications, not the same calculation as finding n and p.

The separate Binomial “At Least One” guide owns how to translate probability inequalities. This article owns reconstruction of the binomial model from its moments. Keeping those two intents separate helps readers find the exact explanation they need without repeating a whole probability textbook.

A five-day practice routine for JC2

Day one: solve ten positive-mean moment pairs and check p before calculating n. Day two: practise the integer trial-count requirement, including cases that produce non-integer n. Day three: handle m=0 and v=0 separately and explain why n can be unidentified. Day four: solve questions that supply standard deviation, converting it to variance before using the formulas. Day five: mix feasible and infeasible exact pairs without headings that reveal the answer.

In an eduKateSG-style three-student tutorial, one learner can derive p, another can calculate n and a third can test feasibility and reconstruct the moments. Then the roles rotate. The aim is not to produce more identical arithmetic, but to make the assumptions and checks visible. A confident pupil should be able to say why a model is mathematically impossible before reaching for a different calculator operation.

Eight quick self-checks

1. m=6, v=4.5 gives p=0.25 and n=24.
2. m=3, v=2.4 gives p=0.2 and n=15.
3. m=4, v=5 is incompatible with exact binomial moments.
4. m=5, v=3 implies n=12.5 and is incompatible with an integer trial count.
5. m=0, v=0 determines p=0 for positive n but does not determine n.
6. m=8, v=0 gives p=1 and n=8.
7. B(20,0.5) has mean 10 and variance 5.
8. B(20,0.9) has mean 18 and variance 1.8.

Frequently asked questions

Why should I divide variance by mean?

For positive mean m=np, the ratio v/m cancels np and leaves 1−p, directly revealing the success probability.

Can n be a decimal?

Not for the standard binomial trial count. It must be an integer; a calculated exact n=12.5 means the supplied exact moments are incompatible.

Can a binomial variance exceed its mean?

No, not for exact theoretical moments of a non-degenerate binomial. Its variance is its mean times 1−p, which is less than one.

What if m=0?

With positive n, p=0 and v=0. The number of trials cannot be uniquely recovered from the two zero moments.

What if v=0 and m is positive?

Then p=1 and n=m. The mean must be a positive integer if the exact binomial model is valid.

Can I treat sample variance as exact theoretical variance?

Not automatically. Sample variance is calculated from observed data and is subject to sampling variability; this inverse problem uses exact moments specified by the question.

Why check independence and constant p?

They are conditions for modelling a count as binomial. Finding feasible numbers n and p is not proof that a real process satisfies those conditions.

Is normal approximation to binomial part of 2027 H2 Mathematics?

No. The 2027 syllabus explicitly excludes normal approximation to the binomial distribution. This guide uses the binomial formulas directly.

Learn the neighbouring topics and verify the source

Complex Conjugation or Multiplying by i?
Regression Transforms Back to Original Units
Critical Region or p-Value?
Recovering n and p from Binomial Mean and Variance — this guide

For broader related concepts, use Bukit Timah H2 Binomial Probability, JC H2 Discrete Random Variables and Expectation and the JC2 Bukit Timah tuition hub. The immutable eduKateSG small-group mathematics tutorial example in Clementi illustrates a teaching format, not the mathematical curriculum.

Official syllabus and editorial note

Official reference: SEAB Singapore–Cambridge H2 Mathematics 9758, 2027 examination syllabus, Section 6.2 Discrete Random Variables. All counts, numerical moments and hypothetical events are original instructional problems rather than real learner records, actual experimental data or official examination questions. The examples make no promise of particular grades, class openings or future examination content.

JC2 H2 Mathematics Tutor Bukit Timah | Recovering n and p from Binomial Mean and Variance