JC2 H2 Maths Bukit Timah | Missing Probability Tables

Did you know? A missing probability in a table is not a free number. Every probability must be between zero and one, and all probabilities in a complete discrete distribution must add to one. For example, if X takes the values 0,1,2,3 with probabilities k, 2k, 3k and 1−6k, then E(X)=3−10k. If the question states E(X)=2, the unknown is k=0.1. The probability constraints must then be checked before using that value.

This JC2 H2 Mathematics Tutor Bukit Timah article explains how to solve discrete probability tables with an unknown parameter, a supplied mean and sometimes a required variance. It gives the fastest equation order, a complete worked table and 32 varied original examples. The early explanation is for quick information; the later examples teach how to spot inconsistent or impossible probability data before calculating an attractive but invalid answer.

The 2027 SEAB H2 Mathematics syllabus 9758 includes the distribution, expectation and variance of a discrete random variable in Section 6.2. It excludes finding the cumulative distribution function of a discrete random variable, so this guide focuses on probability masses and their weighted sums. All practice distributions are invented and do not represent real student marks or official examination questions.

Quick answer: probabilities first, expectation second, variance last

For possible values x₁,…,xₙ with probabilities p₁,…,pₙ, verify that Σpᵢ=1 and 0≤pᵢ≤1. The expectation is E(X)=Σxᵢpᵢ. The second moment is E(X²)=Σxᵢ²pᵢ, and Var(X)=E(X²)−[E(X)]². When an unknown k appears in a table, use whichever supplied condition actually provides an equation for it. The total-probability condition may determine k directly, or it may hold automatically and leave the stated mean to determine k.

Step one: check that the possible values of X and their probabilities are paired in the correct columns.
Step two: sum the probability row to one; do not treat an unknown probability as unbounded.
Step three: if the sum is already one for every k, use any supplied expectation to find k.
Step four: substitute k back into every probability and confirm non-negativity.
Step five: calculate E(X²) from squared values x², not squared probabilities p².
Step six: compute variance from E(X²)−[E(X)]² and check that the result is non-negative.

Worked example: a table with one unknown k

X	0	1	2	3
P(X=x)	k	2k	3k	1−6k

Suppose the table is complete and E(X)=2. The probabilities sum to k+2k+3k+1−6k=1 for every k. The sum condition alone therefore does not tell us k. Instead, use the expected value: E(X)=0×k+1×2k+2×3k+3(1−6k)=3−10k. Setting 3−10k=2 gives k=1/10.

The probabilities are now 1/10, 1/5, 3/10 and 2/5. They are nonnegative and total one. The second moment is E(X²)=0²×(1/10)+1²×(1/5)+2²×(3/10)+3²×(2/5)=0+0.2+1.2+3.6=5. Therefore Var(X)=5−2²=1. The full table is mathematically valid.

Do not report the value 5 as the variance: 5 is E(X²), the second moment. The variance also removes the square of the mean. The distinction is explained more broadly in the separate E(X²) is not [E(X)]² guide; this article owns the missing-probability-table decision.

A different table: total probability alone can solve an unknown

Suppose X takes 0,1,2 with probabilities 0.2, k and 0.3. Here the probabilities add to 0.5+k, so the total-probability rule gives k=0.5 immediately. The mean is E(X)=0(0.2)+1(0.5)+2(0.3)=1.1. You do not need a supplied expectation to determine k in this table, because normalization supplies an independent equation.

Contrast this with the first table, in which the final probability was written as 1−6k. That expression was deliberately chosen to make the probabilities sum to one automatically. Students often write 1=1, then conclude the question is impossible. The correct next step is to inspect the other information given, such as the mean, variance or an event probability.

Missing probability versus missing parameter: which information determines it?

There are two different mathematical problems that may look similar on the page. In the first, one table entry is directly unknown, such as p, and Σp=1 determines it. In the second, a parameter k appears in several entries and a balancing expression makes the sum identically one. In that setting the expectation may determine the parameter, and normalization provides only the allowable range for k. Recognising which equation is informative is more efficient than repeating the same sum calculation.

It is also possible for the supplied conditions to contradict one another. If a mean value forces k<0 in a table whose first entry is k, the proposed model is invalid. No amount of rearranging the expectation formula can make a negative probability acceptable. The correct written conclusion is that the stated data cannot define a probability distribution with that parameter.

Probability bounds are not decorative

For the first model with probabilities k,2k,3k,1−6k, we need k≥0 and 1−6k≥0, giving 0≤k≤1/6. The mean is 3−10k, so its allowable values range from 3 down to 3−10/6=4/3. If a question claimed E(X)=4, the formula would give k=−0.1, which violates non-negativity. The table could not represent that claimed mean.

You can use this allowable-mean interval as an independent check before solving: all possible values of X lie between zero and three, so E(X) cannot exceed three anyway. The stronger parameter calculation shows the particular model cannot even attain means below 4/3. Both checks come from the model and should agree.

A probability distribution is not necessarily binomial

A complete discrete probability table may describe many distributions, not just B(n,p). The four probabilities in our example are specified directly. Unless the question gives the fixed-trial, constant-success-probability and independence structure required for a binomial count, do not replace the table with an unverified binomial formula. In H2 Mathematics, the ability to work with a general discrete distribution is a separate skill from recognising a suitable binomial model.

The existing Recovering n and p from Binomial Mean and Variance article addresses the inverse binomial-parameter problem. This page instead calculates the entries, expectation and variance of a supplied finite probability table, with no assumption that its event probabilities follow a binomial pattern.

What does expectation actually mean?

The expected value is the probability-weighted average of possible outcomes. It need not be a value that the random variable can actually take. For example, X∈{0,2} with equal probabilities has E(X)=1, even though X is never one in any realised outcome. Thus a table with an expectation of 2.5 need not include 2.5 in its list of possible values.

In a fictional cost setting, E(X)=2.5 units means the mathematical average across repeated outcomes of that probability model is 2.5 units. It does not mean every outcome will be 2.5, or that a child, institution or experiment is guaranteed to realise that value. The term “expected” is mathematical, not a personal prediction or a statement of certainty.

A mean near the upper outcome constrains the parameter

Suppose the largest value in a discrete table is 5 and all probability masses are nonnegative. E(X) cannot exceed 5, because it is a weighted average of the available values. If a supplied mean is 6, the table and statement are incompatible regardless of the unknown parameter. Checking that the mean lies between the minimum and maximum values is a cheap early audit, though it does not replace checking the actual parameter constraints.

Another quick check is to compare E(X²) and [E(X)]². Because variance is nonnegative, the former must be at least the latter. A negative variance computed from a supposedly valid distribution signals a faulty probability, a missed square, a wrong expectation or some other error. A second method is to calculate Σ(x−μ)²P(X=x), which must be nonnegative term by term.

What changes when outcomes include negative numbers?

Discrete outcomes need not be positive. A fictional profit-and-loss variable might take −2,0 and 3. The expectation is still Σxp, and negative outcomes contribute negative terms. But x² is nonnegative, so E(X²) uses the squares 4,0 and 9. Students who drop the negative signs too early may distort the mean, while those who square a negative value incorrectly can distort the second moment.

For X with P(X=−2)=1/4, P(X=0)=1/4, P(X=3)=1/2, the mean is (−2)/4+0+3/2=1. The second moment is 4/4+0+9/2=11/2, giving variance 11/2−1²=9/2. Every number is exact, and the negative possible outcome does not make variance negative.

One-minute Bukit Timah diagnostic

Ask a learner to sum k+2k+3k+(1−6k). They should obtain one for every k, not an equation that identifies it. Then supply E(X)=2; solving 3−10k=2 gives k=1/10. Ask whether the four probabilities are valid: yes, each is nonnegative and they sum to one. Finally request E(X²)=5 and Var(X)=1. The sequence of questions separates model normalization, parameter identification and variance calculation.

Thirty-two original missing-parameter probability clinics

The following 32 examples use four possible outcomes, three probabilities proportional to a parameter k and a final balancing probability. A stipulated mean is given, and you must recover k from the mean. Each has been constructed from an exact rational k in its valid range, so the probability and variance checks can be verified independently. The table shapes and possible outcomes vary, rather than repeating only the numbers from the first example.

1. Four outcomes, three weighted k entries

Try this original table. The possible values of X are -2, 0, 1, 4. Their corresponding probabilities are 1k, 2k, 3k and 1−6k. The mean is stated to be E(X)=13/9. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 6k and the balancing entry is 1−6k. Therefore normalization holds automatically. The expectation equation is E(X)=4−23k=13/9, so k=1/9. The actual probability row is 1/9, 2/9, 1/3, 1/3. Each probability is nonnegative, and their numerators total 1+2+3+3=9. A candidate k outside 0≤k≤1/6 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=55/9. The squared mean is (13/9)², so Var(X)=326/81. The probability of one of the two larger outcomes, X≥1, equals 2/3. Check normalization before using the supplied mean. The balancing final probability was chosen to make the total exactly one. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

2. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 0, 1, 2, 3. Their corresponding probabilities are 3k, 2k, 1k and 1−6k. The mean is stated to be E(X)=8/5. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 6k and the balancing entry is 1−6k. Therefore normalization holds automatically. The expectation equation is E(X)=3−14k=8/5, so k=1/10. The actual probability row is 3/10, 1/5, 1/10, 2/5. Each probability is nonnegative, and their numerators total 3+2+1+4=10. A candidate k outside 0≤k≤1/6 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=21/5. The squared mean is (8/5)², so Var(X)=41/25. The probability of one of the two larger outcomes, X≥2, equals 1/2. The x-values and probabilities belong in matching columns; shifting one weight to the wrong outcome changes the expected value. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

3. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 0, 2, 3, 6. Their corresponding probabilities are 3k, 1k, 2k and 1−6k. The mean is stated to be E(X)=38/11. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 6k and the balancing entry is 1−6k. Therefore normalization holds automatically. The expectation equation is E(X)=6−28k=38/11, so k=1/11. The actual probability row is 3/11, 1/11, 2/11, 5/11. Each probability is nonnegative, and their numerators total 3+1+2+5=11. A candidate k outside 0≤k≤1/6 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=202/11. The squared mean is (38/11)², so Var(X)=778/121. The probability of one of the two larger outcomes, X≥3, equals 7/11. A negative first outcome contributes negatively to the mean but positively to E(X²) after squaring. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

4. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 1, 3, 5, 8. Their corresponding probabilities are 2k, 1k, 4k and 1−7k. The mean is stated to be E(X)=73/13. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 7k and the balancing entry is 1−7k. Therefore normalization holds automatically. The expectation equation is E(X)=8−31k=73/13, so k=1/13. The actual probability row is 2/13, 1/13, 4/13, 6/13. Each probability is nonnegative, and their numerators total 2+1+4+6=13. A candidate k outside 0≤k≤1/7 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=495/13. The squared mean is (73/13)², so Var(X)=1106/169. The probability of one of the two larger outcomes, X≥5, equals 10/13. You can check the expected value independently by summing each probability times the corresponding outcome. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

5. Four outcomes, three weighted k entries

Try this original table. The possible values of X are -3, -1, 2, 4. Their corresponding probabilities are 2k, 3k, 4k and 1−9k. The mean is stated to be E(X)=27/16. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 9k and the balancing entry is 1−9k. Therefore normalization holds automatically. The expectation equation is E(X)=4−37k=27/16, so k=1/16. The actual probability row is 1/8, 3/16, 1/4, 7/16. Each probability is nonnegative, and their numerators total 2+3+4+7=16. A candidate k outside 0≤k≤1/9 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=149/16. The squared mean is (27/16)², so Var(X)=1655/256. The probability of one of the two larger outcomes, X≥2, equals 11/16. The variance is E(X²)−[E(X)]², not merely E(X²). As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

6. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 2, 4, 6, 9. Their corresponding probabilities are 2k, 2k, 3k and 1−7k. The mean is stated to be E(X)=34/5. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 7k and the balancing entry is 1−7k. Therefore normalization holds automatically. The expectation equation is E(X)=9−33k=34/5, so k=1/15. The actual probability row is 2/15, 2/15, 1/5, 8/15. Each probability is nonnegative, and their numerators total 2+2+3+8=15. A candidate k outside 0≤k≤1/7 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=796/15. The squared mean is (34/5)², so Var(X)=512/75. The probability of one of the two larger outcomes, X≥6, equals 11/15. The recovered k must make all four probabilities nonnegative; the linear equation alone is not enough. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

7. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 0, 1, 4, 7. Their corresponding probabilities are 1k, 3k, 2k and 1−6k. The mean is stated to be E(X)=74/15. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 6k and the balancing entry is 1−6k. Therefore normalization holds automatically. The expectation equation is E(X)=7−31k=74/15, so k=1/15. The actual probability row is 1/15, 1/5, 2/15, 3/5. Each probability is nonnegative, and their numerators total 1+3+2+9=15. A candidate k outside 0≤k≤1/6 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=476/15. The squared mean is (74/15)², so Var(X)=1664/225. The probability of one of the two larger outcomes, X≥4, equals 11/15. The mean may be a fraction or a number absent from the list of possible outcomes. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

8. Four outcomes, three weighted k entries

Try this original table. The possible values of X are -1, 1, 3, 5. Their corresponding probabilities are 1k, 4k, 2k and 1−7k. The mean is stated to be E(X)=12/5. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 7k and the balancing entry is 1−7k. Therefore normalization holds automatically. The expectation equation is E(X)=5−26k=12/5, so k=1/10. The actual probability row is 1/10, 2/5, 1/5, 3/10. Each probability is nonnegative, and their numerators total 1+4+2+3=10. A candidate k outside 0≤k≤1/7 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=49/5. The squared mean is (12/5)², so Var(X)=101/25. The probability of one of the two larger outcomes, X≥3, equals 1/2. A real distribution’s variance cannot be negative, so the final result should be nonnegative. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

9. Four outcomes, three weighted k entries

Try this original table. The possible values of X are -2, 0, 1, 4. Their corresponding probabilities are 2k, 1k, 4k and 1−7k. The mean is stated to be E(X)=16/11. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 7k and the balancing entry is 1−7k. Therefore normalization holds automatically. The expectation equation is E(X)=4−28k=16/11, so k=1/11. The actual probability row is 2/11, 1/11, 4/11, 4/11. Each probability is nonnegative, and their numerators total 2+1+4+4=11. A candidate k outside 0≤k≤1/7 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=76/11. The squared mean is (16/11)², so Var(X)=580/121. The probability of one of the two larger outcomes, X≥1, equals 8/11. Check normalization before using the supplied mean. The balancing final probability was chosen to make the total exactly one. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

10. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 0, 1, 2, 3. Their corresponding probabilities are 2k, 3k, 4k and 1−9k. The mean is stated to be E(X)=13/7. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 9k and the balancing entry is 1−9k. Therefore normalization holds automatically. The expectation equation is E(X)=3−16k=13/7, so k=1/14. The actual probability row is 1/7, 3/14, 2/7, 5/14. Each probability is nonnegative, and their numerators total 2+3+4+5=14. A candidate k outside 0≤k≤1/9 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=32/7. The squared mean is (13/7)², so Var(X)=55/49. The probability of one of the two larger outcomes, X≥2, equals 9/14. The x-values and probabilities belong in matching columns; shifting one weight to the wrong outcome changes the expected value. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

11. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 0, 2, 3, 6. Their corresponding probabilities are 2k, 2k, 3k and 1−7k. The mean is stated to be E(X)=49/13. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 7k and the balancing entry is 1−7k. Therefore normalization holds automatically. The expectation equation is E(X)=6−29k=49/13, so k=1/13. The actual probability row is 2/13, 2/13, 3/13, 6/13. Each probability is nonnegative, and their numerators total 2+2+3+6=13. A candidate k outside 0≤k≤1/7 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=251/13. The squared mean is (49/13)², so Var(X)=862/169. The probability of one of the two larger outcomes, X≥3, equals 9/13. A negative first outcome contributes negatively to the mean but positively to E(X²) after squaring. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

12. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 1, 3, 5, 8. Their corresponding probabilities are 1k, 3k, 2k and 1−6k. The mean is stated to be E(X)=76/13. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 6k and the balancing entry is 1−6k. Therefore normalization holds automatically. The expectation equation is E(X)=8−28k=76/13, so k=1/13. The actual probability row is 1/13, 3/13, 2/13, 7/13. Each probability is nonnegative, and their numerators total 1+3+2+7=13. A candidate k outside 0≤k≤1/6 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=526/13. The squared mean is (76/13)², so Var(X)=1062/169. The probability of one of the two larger outcomes, X≥5, equals 9/13. You can check the expected value independently by summing each probability times the corresponding outcome. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

13. Four outcomes, three weighted k entries

Try this original table. The possible values of X are -3, -1, 2, 4. Their corresponding probabilities are 1k, 4k, 2k and 1−7k. The mean is stated to be E(X)=29/15. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 7k and the balancing entry is 1−7k. Therefore normalization holds automatically. The expectation equation is E(X)=4−31k=29/15, so k=1/15. The actual probability row is 1/15, 4/15, 2/15, 8/15. Each probability is nonnegative, and their numerators total 1+4+2+8=15. A candidate k outside 0≤k≤1/7 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=149/15. The squared mean is (29/15)², so Var(X)=1394/225. The probability of one of the two larger outcomes, X≥2, equals 2/3. The variance is E(X²)−[E(X)]², not merely E(X²). As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

14. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 2, 4, 6, 9. Their corresponding probabilities are 1k, 2k, 3k and 1−6k. The mean is stated to be E(X)=109/15. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 6k and the balancing entry is 1−6k. Therefore normalization holds automatically. The expectation equation is E(X)=9−26k=109/15, so k=1/15. The actual probability row is 1/15, 2/15, 1/5, 3/5. Each probability is nonnegative, and their numerators total 1+2+3+9=15. A candidate k outside 0≤k≤1/6 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=291/5. The squared mean is (109/15)², so Var(X)=1214/225. The probability of one of the two larger outcomes, X≥6, equals 4/5. The recovered k must make all four probabilities nonnegative; the linear equation alone is not enough. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

15. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 0, 1, 4, 7. Their corresponding probabilities are 3k, 2k, 1k and 1−6k. The mean is stated to be E(X)=3. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 6k and the balancing entry is 1−6k. Therefore normalization holds automatically. The expectation equation is E(X)=7−36k=3, so k=1/9. The actual probability row is 1/3, 2/9, 1/9, 1/3. Each probability is nonnegative, and their numerators total 3+2+1+3=9. A candidate k outside 0≤k≤1/6 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=55/3. The squared mean is (3)², so Var(X)=28/3. The probability of one of the two larger outcomes, X≥4, equals 4/9. The mean may be a fraction or a number absent from the list of possible outcomes. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

16. Four outcomes, three weighted k entries

Try this original table. The possible values of X are -1, 1, 3, 5. Their corresponding probabilities are 3k, 1k, 2k and 1−6k. The mean is stated to be E(X)=12/5. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 6k and the balancing entry is 1−6k. Therefore normalization holds automatically. The expectation equation is E(X)=5−26k=12/5, so k=1/10. The actual probability row is 3/10, 1/10, 1/5, 2/5. Each probability is nonnegative, and their numerators total 3+1+2+4=10. A candidate k outside 0≤k≤1/6 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=61/5. The squared mean is (12/5)², so Var(X)=161/25. The probability of one of the two larger outcomes, X≥3, equals 3/5. A real distribution’s variance cannot be negative, so the final result should be nonnegative. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

17. Four outcomes, three weighted k entries

Try this original table. The possible values of X are -2, 0, 1, 4. Their corresponding probabilities are 1k, 3k, 2k and 1−6k. The mean is stated to be E(X)=20/11. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 6k and the balancing entry is 1−6k. Therefore normalization holds automatically. The expectation equation is E(X)=4−24k=20/11, so k=1/11. The actual probability row is 1/11, 3/11, 2/11, 5/11. Each probability is nonnegative, and their numerators total 1+3+2+5=11. A candidate k outside 0≤k≤1/6 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=86/11. The squared mean is (20/11)², so Var(X)=546/121. The probability of one of the two larger outcomes, X≥1, equals 7/11. Check normalization before using the supplied mean. The balancing final probability was chosen to make the total exactly one. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

18. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 0, 1, 2, 3. Their corresponding probabilities are 1k, 4k, 2k and 1−7k. The mean is stated to be E(X)=2. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 7k and the balancing entry is 1−7k. Therefore normalization holds automatically. The expectation equation is E(X)=3−13k=2, so k=1/13. The actual probability row is 1/13, 4/13, 2/13, 6/13. Each probability is nonnegative, and their numerators total 1+4+2+6=13. A candidate k outside 0≤k≤1/7 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=66/13. The squared mean is (2)², so Var(X)=14/13. The probability of one of the two larger outcomes, X≥2, equals 8/13. The x-values and probabilities belong in matching columns; shifting one weight to the wrong outcome changes the expected value. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

19. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 0, 2, 3, 6. Their corresponding probabilities are 1k, 2k, 3k and 1−6k. The mean is stated to be E(X)=55/13. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 6k and the balancing entry is 1−6k. Therefore normalization holds automatically. The expectation equation is E(X)=6−23k=55/13, so k=1/13. The actual probability row is 1/13, 2/13, 3/13, 7/13. Each probability is nonnegative, and their numerators total 1+2+3+7=13. A candidate k outside 0≤k≤1/6 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=287/13. The squared mean is (55/13)², so Var(X)=706/169. The probability of one of the two larger outcomes, X≥3, equals 10/13. A negative first outcome contributes negatively to the mean but positively to E(X²) after squaring. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

20. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 1, 3, 5, 8. Their corresponding probabilities are 3k, 2k, 1k and 1−6k. The mean is stated to be E(X)=39/7. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 6k and the balancing entry is 1−6k. Therefore normalization holds automatically. The expectation equation is E(X)=8−34k=39/7, so k=1/14. The actual probability row is 3/14, 1/7, 1/14, 4/7. Each probability is nonnegative, and their numerators total 3+2+1+8=14. A candidate k outside 0≤k≤1/6 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=279/7. The squared mean is (39/7)², so Var(X)=432/49. The probability of one of the two larger outcomes, X≥5, equals 9/14. You can check the expected value independently by summing each probability times the corresponding outcome. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

21. Four outcomes, three weighted k entries

Try this original table. The possible values of X are -3, -1, 2, 4. Their corresponding probabilities are 3k, 1k, 2k and 1−6k. The mean is stated to be E(X)=2. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 6k and the balancing entry is 1−6k. Therefore normalization holds automatically. The expectation equation is E(X)=4−30k=2, so k=1/15. The actual probability row is 1/5, 1/15, 2/15, 3/5. Each probability is nonnegative, and their numerators total 3+1+2+9=15. A candidate k outside 0≤k≤1/6 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=12. The squared mean is (2)², so Var(X)=8. The probability of one of the two larger outcomes, X≥2, equals 11/15. The variance is E(X²)−[E(X)]², not merely E(X²). As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

22. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 2, 4, 6, 9. Their corresponding probabilities are 2k, 1k, 4k and 1−7k. The mean is stated to be E(X)=59/10. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 7k and the balancing entry is 1−7k. Therefore normalization holds automatically. The expectation equation is E(X)=9−31k=59/10, so k=1/10. The actual probability row is 1/5, 1/10, 2/5, 3/10. Each probability is nonnegative, and their numerators total 2+1+4+3=10. A candidate k outside 0≤k≤1/7 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=411/10. The squared mean is (59/10)², so Var(X)=629/100. The probability of one of the two larger outcomes, X≥6, equals 7/10. The recovered k must make all four probabilities nonnegative; the linear equation alone is not enough. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

23. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 0, 1, 4, 7. Their corresponding probabilities are 2k, 3k, 4k and 1−9k. The mean is stated to be E(X)=47/13. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 9k and the balancing entry is 1−9k. Therefore normalization holds automatically. The expectation equation is E(X)=7−44k=47/13, so k=1/13. The actual probability row is 2/13, 3/13, 4/13, 4/13. Each probability is nonnegative, and their numerators total 2+3+4+4=13. A candidate k outside 0≤k≤1/9 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=263/13. The squared mean is (47/13)², so Var(X)=1210/169. The probability of one of the two larger outcomes, X≥4, equals 8/13. The mean may be a fraction or a number absent from the list of possible outcomes. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

24. Four outcomes, three weighted k entries

Try this original table. The possible values of X are -1, 1, 3, 5. Their corresponding probabilities are 2k, 2k, 3k and 1−7k. The mean is stated to be E(X)=17/6. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 7k and the balancing entry is 1−7k. Therefore normalization holds automatically. The expectation equation is E(X)=5−26k=17/6, so k=1/12. The actual probability row is 1/6, 1/6, 1/4, 5/12. Each probability is nonnegative, and their numerators total 2+2+3+5=12. A candidate k outside 0≤k≤1/7 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=13. The squared mean is (17/6)², so Var(X)=179/36. The probability of one of the two larger outcomes, X≥3, equals 2/3. A real distribution’s variance cannot be negative, so the final result should be nonnegative. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

25. Four outcomes, three weighted k entries

Try this original table. The possible values of X are -2, 0, 1, 4. Their corresponding probabilities are 3k, 2k, 1k and 1−6k. The mean is stated to be E(X)=19/12. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 6k and the balancing entry is 1−6k. Therefore normalization holds automatically. The expectation equation is E(X)=4−29k=19/12, so k=1/12. The actual probability row is 1/4, 1/6, 1/12, 1/2. Each probability is nonnegative, and their numerators total 3+2+1+6=12. A candidate k outside 0≤k≤1/6 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=109/12. The squared mean is (19/12)², so Var(X)=947/144. The probability of one of the two larger outcomes, X≥1, equals 7/12. Check normalization before using the supplied mean. The balancing final probability was chosen to make the total exactly one. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

26. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 0, 1, 2, 3. Their corresponding probabilities are 3k, 1k, 2k and 1−6k. The mean is stated to be E(X)=2. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 6k and the balancing entry is 1−6k. Therefore normalization holds automatically. The expectation equation is E(X)=3−13k=2, so k=1/13. The actual probability row is 3/13, 1/13, 2/13, 7/13. Each probability is nonnegative, and their numerators total 3+1+2+7=13. A candidate k outside 0≤k≤1/6 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=72/13. The squared mean is (2)², so Var(X)=20/13. The probability of one of the two larger outcomes, X≥2, equals 9/13. The x-values and probabilities belong in matching columns; shifting one weight to the wrong outcome changes the expected value. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

27. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 0, 2, 3, 6. Their corresponding probabilities are 2k, 1k, 4k and 1−7k. The mean is stated to be E(X)=62/15. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 7k and the balancing entry is 1−7k. Therefore normalization holds automatically. The expectation equation is E(X)=6−28k=62/15, so k=1/15. The actual probability row is 2/15, 1/15, 4/15, 8/15. Each probability is nonnegative, and their numerators total 2+1+4+8=15. A candidate k outside 0≤k≤1/7 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=328/15. The squared mean is (62/15)², so Var(X)=1076/225. The probability of one of the two larger outcomes, X≥3, equals 4/5. A negative first outcome contributes negatively to the mean but positively to E(X²) after squaring. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

28. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 1, 3, 5, 8. Their corresponding probabilities are 2k, 3k, 4k and 1−9k. The mean is stated to be E(X)=103/18. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 9k and the balancing entry is 1−9k. Therefore normalization holds automatically. The expectation equation is E(X)=8−41k=103/18, so k=1/18. The actual probability row is 1/9, 1/6, 2/9, 1/2. Each probability is nonnegative, and their numerators total 2+3+4+9=18. A candidate k outside 0≤k≤1/9 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=235/6. The squared mean is (103/18)², so Var(X)=2081/324. The probability of one of the two larger outcomes, X≥5, equals 13/18. You can check the expected value independently by summing each probability times the corresponding outcome. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

29. Four outcomes, three weighted k entries

Try this original table. The possible values of X are -3, -1, 2, 4. Their corresponding probabilities are 2k, 2k, 3k and 1−7k. The mean is stated to be E(X)=1. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 7k and the balancing entry is 1−7k. Therefore normalization holds automatically. The expectation equation is E(X)=4−30k=1, so k=1/10. The actual probability row is 1/5, 1/5, 3/10, 3/10. Each probability is nonnegative, and their numerators total 2+2+3+3=10. A candidate k outside 0≤k≤1/7 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=8. The squared mean is (1)², so Var(X)=7. The probability of one of the two larger outcomes, X≥2, equals 3/5. The variance is E(X²)−[E(X)]², not merely E(X²). As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

30. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 2, 4, 6, 9. Their corresponding probabilities are 1k, 3k, 2k and 1−6k. The mean is stated to be E(X)=31/5. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 6k and the balancing entry is 1−6k. Therefore normalization holds automatically. The expectation equation is E(X)=9−28k=31/5, so k=1/10. The actual probability row is 1/10, 3/10, 1/5, 2/5. Each probability is nonnegative, and their numerators total 1+3+2+4=10. A candidate k outside 0≤k≤1/6 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=224/5. The squared mean is (31/5)², so Var(X)=159/25. The probability of one of the two larger outcomes, X≥6, equals 3/5. The recovered k must make all four probabilities nonnegative; the linear equation alone is not enough. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

31. Four outcomes, three weighted k entries

Try this original table. The possible values of X are 0, 1, 4, 7. Their corresponding probabilities are 1k, 4k, 2k and 1−7k. The mean is stated to be E(X)=47/12. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 7k and the balancing entry is 1−7k. Therefore normalization holds automatically. The expectation equation is E(X)=7−37k=47/12, so k=1/12. The actual probability row is 1/12, 1/3, 1/6, 5/12. Each probability is nonnegative, and their numerators total 1+4+2+5=12. A candidate k outside 0≤k≤1/7 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=281/12. The squared mean is (47/12)², so Var(X)=1163/144. The probability of one of the two larger outcomes, X≥4, equals 7/12. The mean may be a fraction or a number absent from the list of possible outcomes. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

32. Four outcomes, three weighted k entries

Try this original table. The possible values of X are -1, 1, 3, 5. Their corresponding probabilities are 1k, 2k, 3k and 1−6k. The mean is stated to be E(X)=10/3. Find the parameter k, check the four probabilities and calculate E(X²), Var(X) and the probability that X is one of the two largest outcomes. Treat the numbers as exact theoretical quantities.

Find k and verify the distribution. The weights total 6k and the balancing entry is 1−6k. Therefore normalization holds automatically. The expectation equation is E(X)=5−20k=10/3, so k=1/12. The actual probability row is 1/12, 1/6, 1/4, 1/2. Each probability is nonnegative, and their numerators total 1+2+3+6=12. A candidate k outside 0≤k≤1/6 would have violated the probability bounds.

Second moment, variance and a probability check. Squaring the four possible outcomes and weighting them gives E(X²)=15. The squared mean is (10/3)², so Var(X)=35/9. The probability of one of the two larger outcomes, X≥3, equals 3/4. A real distribution’s variance cannot be negative, so the final result should be nonnegative. As a transfer exercise, change only the supplied mean and solve the equation again: determine whether the new k still produces a valid probability distribution, rather than assuming every mean is feasible.

Ten fast self-checks with answers

For probabilities k,2k,3k,1−6k at X=0,1,2,3, E(X)=3−10k.
If E(X)=2, k=1/10.
The resulting probability row is 0.1,0.2,0.3,0.4.
For that row, E(X²)=5 and Var(X)=1.
For probabilities 0.2,k,0.3, normalization gives k=0.5.
The corresponding values X=0,1,2 have E(X)=1.1.
A valid probability mass cannot be negative or greater than one.
The variance is always nonnegative for a genuine distribution with finite second moment.
A mean can lie between two possible outcomes rather than matching one.
The 2027 H2 syllabus excludes calculating the discrete cumulative distribution function.

A five-day revision and three-learner method

Day one: review the nonnegative probability and total-one conditions using small three- and four-outcome tables. Day two: calculate expectations with negative and positive outcomes and use a supplied mean to recover an unknown parameter. Day three: check whether the parameter produces a valid distribution, including deliberately impossible means. Day four: calculate E(X²) and variance in two ways, including Σ(x−μ)²p. Day five: solve mixed questions without being told which equation identifies the unknown and explain why the other equations are only checks.

In a small group of up to three learners, one student can normalise the table, one derive the expectation equation and one independently audit the variance and validity of k. Rotate the roles and change the outcome row while keeping the probability pattern. The teaching goal is independent model checking, not memorising a single weighted-average expression.

Frequently asked questions

Why do all probabilities sum to one?

A complete probability distribution lists all possible mutually exclusive values of its random variable. One of them occurs, so their probabilities add to one.

Can a mean be absent from the outcome list?

Yes. An expectation is a weighted average of possible values, not necessarily an outcome that can occur.

Why can normalization fail to determine k?

A balancing entry such as 1−6k may have been chosen so the sum is identically one. Another supplied condition, such as E(X), must then determine k.

What if the mean gives a negative probability?

The proposed exact model is invalid for that claimed mean. An algebraic solution is not acceptable unless every probability lies in [0,1].

Is E(X²) the same as [E(X)]²?

No. The difference E(X²)−[E(X)]² is the variance.

Can outcomes be negative?

Yes. A discrete random variable can have negative possible values, provided the associated probabilities are valid.

Is every four-outcome table binomial?

No. A generic probability table does not require a binomial trial structure unless the model conditions are supplied.

Is this in the 2027 H2 Mathematics syllabus?

Yes. Section 6.2 includes discrete probability distributions, expectations and variances, while excluding the cumulative distribution function.

JC2 H2 Mathematics Tutor Bukit Timah | Discrete Probability Tables: Missing p and Expected Value