Loav · VCE Mathematical Methods · Units 3 and 4
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The binomial distribution models a count of successes in a fixed number of suitable trials. It compresses a large probability tree into one formula, but that convenience depends on assumptions. This guide explains the assumptions, the formula, cumulative probabilities, moments and inverse problems. You should know conditional probability, independence and expected value. Your goal is to justify a model and translate the requested event before reaching for technology.
“Success” is just a label for the outcome being counted. A damaged item can be a statistical success if X counts damaged items. State what X counts, the number of trials, and the probability of the labelled success. This one sentence prevents many errors caused by switching between the probability of success and the probability of failure.
1. Check the model before calculating
A Bernoulli trial is a random trial with two outcome categories, conventionally called success and failure, with probabilities p and 1 − p. A binomial model counts successes across a fixed number of independent Bernoulli trials sharing the same p.
A binomial model requires a fixed number n of trials; two outcome categories per trial; the same success probability p on every trial; and independence between trials. The underlying experiment may have several physical outcomes, provided these are grouped into success and not-success. A die, for example, can be grouped into “six” and “not six”.
X ∼ Bi(n, p), with n a nonnegative integer and 0 ≤ p ≤ 1.
X can take 0, 1, 2, …, n.
A fixed observation period is not automatically a fixed number of trials. “Count arrivals in an hour” does not become binomial simply because arrivals can occur or not occur. Likewise, “continue until the first success” has a variable number of trials and does not fit a fixed-n success-count question.
Sampling without replacement from a small finite collection usually violates independence and constant p. If several defective items have already been removed, the remaining composition changes. A very large population relative to the sample can make the independent-trial approximation reasonable, but that is a modelling assumption. State it rather than presenting an approximation as exact.
In a sporting example, assuming constant ability and independence may be part of the specified model. Real fatigue, changing conditions or learning could violate it. A mathematically correct answer is conditional on the model, so a short sentence about limitations can be more valuable than extra decimal places.
2. Why the probability formula works
Take one particular arrangement containing r successes and n − r failures. Independence gives its probability as pr(1 − p)n−r. However, “exactly r successes” includes many arrangements. The binomial coefficient counts ways to choose which r of the n trial positions contain successes.
P(X = r) = C(n, r)pr(1 − p)n−r
C(n, r) = n!/[r!(n − r)!], for r = 0, 1, …, n.
Without the coefficient, you have calculated only one ordering. Multiplying by r or by n is not generally the correct replacement. At the extremes r = 0 and r = n, there is only one arrangement, giving P(X = 0) = (1 − p)n and P(X = n) = pn.
For interior r, the factors have separate roles: the coefficient counts paths, while the powers give the probability of each path. All paths included in the event are disjoint, so their equal probabilities can be added by multiplying by the path count.
3. Translate inequalities into integer events
Discrete endpoints matter. “At least four” means X ≥ 4, while “more than four” means X ≥ 5. “Fewer than four” means X ≤ 3. If a calculator supplies F(r) = P(X ≤ r), an upper tail is found using a complement, and an interval is found by subtracting two cumulative probabilities.
P(X ≥ r) = 1 − F(r − 1)
P(a ≤ X ≤ b) = F(b) − F(a − 1)
P(X ≥ 1) = 1 − (1 − p)n
Write the event in symbols and, if useful, list its first and last included integers. This exposes off-by-one mistakes before they reach a calculator. Do not assume that a command called “cumulative” accepts the same input format on every device. Record the probability statement as well as the numerical answer.
Worked examples
Worked example 1: exact, cumulative and interval probabilities
In an original manufacturing model, each of 12 independently tested components needs an adjustment with probability 0.30. Let X count those needing adjustment. Then X ∼ Bi(12, 0.30).
- Exactly four: P(X = 4) = C(12,4)(0.30)4(0.70)8 = 495(0.30)4(0.70)8 ≈ 0.2311.
- At least four: P(X ≥ 4) = 1 − P(X ≤ 3) ≈ 0.5075. Subtracting P(X ≤ 4) would wrongly remove the four-adjustment case.
- Between three and six inclusive: P(3 ≤ X ≤ 6) = P(X ≤ 6) − P(X ≤ 2) ≈ 0.7086.
- E(X) = np = 3.6. Var(X) = np(1 − p) = 2.52, so sd(X) = √2.52 ≈ 1.5875.
- Given that at least two components need adjustment, P(X ≥ 4 | X ≥ 2) = P(X ≥ 4)/P(X ≥ 2) = 0.5074842266/0.9149749501 ≈ 0.5546. The numerator is the intersection because X ≥ 4 is contained in X ≥ 2.
The probability of at least four is larger than the probability of exactly four, as it must be. The expected count 3.6 is sensible even though an individual batch contains a whole number of adjusted components.
4. Mean, spread and distribution shape
Each trial contributes either zero or one to the count. Across n trials, the expected number of successes is np. For independent trials, the variability of the count is np(1 − p). These are properties of the probability model, not estimates obtained from one observed count.
E(X) = np
Var(X) = np(1 − p)
sd(X) = √[np(1 − p)]
For fixed n, the variance is greatest when p = 0.5 because p(1 − p) is then largest. When p is near zero, counts are concentrated near zero and the distribution commonly has a right tail. When p is near one, the reflected picture is concentrated near n with a left tail. For p = 0.5, the distribution is symmetric about n/2. At p = 0, X = 0 with certainty; at p = 1, X = n with certainty. Both endpoint models have zero variance because their count is certain.
For a larger n at fixed 0 < p < 1, the expected count increases and the standard deviation of the count increases. Yet the proportion X/n becomes less variable. More trials can produce a more stable proportion while producing a more variable raw count. The sample-proportions guide develops this distinction.
Worked example 2: find the minimum number of trials
A device independently detects a rare signal on each trial with probability 0.12. How many trials are needed to make the probability of at least one detection at least 0.95?
- For n trials, P(at least one) = 1 − 0.88n. Therefore 1 − 0.88n ≥ 0.95.
- Rearrange to 0.88n ≤ 0.05 and take logarithms: n ln(0.88) ≤ ln(0.05).
- Because ln(0.88) is negative, dividing reverses the inequality: n ≥ ln(0.05)/ln(0.88).
- The least integer satisfying this condition is n = 24. Check n = 23 gives about 0.9471, while n = 24 gives about 0.9535.
The neighbouring checks prove minimality and guard against rounding the trial count down. The answer relies on each extra trial retaining probability 0.12 and remaining independent.
5. Inverse questions need constraints
Sometimes probabilities or moments are supplied and n or p must be found. Write an equation using the model, then impose 0 ≤ p ≤ 1 and integer n. A calculator may return algebraic solutions that are not valid parameters. If solving an inequality for n, round in the direction required by the condition and test the neighbouring integer.
Worked example 3: recover the success probability
Five independent trials have a common unknown success probability p. The probability of no successes is 0.16807. Find p, the expected count and the probability of five successes.
- Let X ∼ Bi(5,p). The zero-success equation is (1 − p)5 = 0.16807.
- Since 0.16807 = 0.75 and 1 − p must be nonnegative, 1 − p = 0.7. Therefore p = 0.3.
- E(X) = 5(0.3) = 1.5 and Var(X) = 5(0.3)(0.7) = 1.05.
- P(X = 5) = 0.35 = 0.00243.
The probability of no successes is not 1 − p; that is only the probability of failure on one trial. The exponent encodes all five independent failures.
6. Technology and communication checks
For an exact binomial calculation, keep n and p unchanged across parts unless the experiment changes. Use a probability-mass calculation for “exactly”, and cumulative calculations for ranges. A normal approximation is a separate method with its own conditions; label it when used and do not silently replace an exact binomial calculation.
Keep several guard digits and round the final probability to the requested accuracy. Scientific notation can be sensible for a very small probability; writing zero erases information. Check that every result is in [0,1], that a wider event has at least as much probability as a narrower event, and that complementary events add to one.
Practice and reasoned answers
1. X ∼ Bi(8,0.2). Find P(X = 2) and P(X ≥ 2).
P(X = 2) = C(8,2)(0.2)2(0.8)6 = 0.29360128. P(X ≥ 2) = 1 − P(X ≤ 1) = 0.49668352. To four decimal places these are 0.2936 and 0.4967. The tail includes the exact-two event and additional counts.
2. X ∼ Bi(10,0.4). Find the probability of fewer than four successes.
Fewer than four means X ≤ 3. Sum P(X = r) for r = 0,1,2,3, or use the cumulative function, giving 0.3822806016 ≈ 0.3823. Using X ≤ 4 would include an unwanted endpoint.
3. X ∼ Bi(20,0.65). Find the mean, variance and P(X ≥ 15).
The mean is 20(0.65) = 13; variance is 20(0.65)(0.35) = 4.55. P(X ≥ 15) = 1 − P(X ≤ 14) ≈ 0.2454. The standard deviation, if required, is √4.55 rather than 4.55.
4. Four cards are drawn without replacement from a small deck containing five blue and five orange cards. Is the blue count exactly binomial?
No. After a blue card is drawn, the next blue probability becomes 4/9 rather than 5/10. The trials are dependent and the conditional success probability changes. A fixed count and two colour categories alone are insufficient to justify a binomial model.
5. A binomial count has mean 6 and variance 3.6. Find n and p.
Divide variance by mean: 1 − p = 3.6/6 = 0.6, so p = 0.4. Then n = 6/0.4 = 15. Check that n is an integer and the recovered variance is 15(0.4)(0.6) = 3.6.
6. Six independent fair trials each have success probability 0.5. Find P(X = 3), and explain why it is not 0.5.
P(X = 3) = C(6,3)/26 = 20/64 = 0.3125. Symmetry makes complementary counts equally likely; it does not assign half the total probability to the centre. The other counts still carry positive probability.
7. Independent trials have p = 0.25. Find the least n making at least one success more likely than not.
Require 1 − 0.75n > 0.5. At n = 2 the probability is 0.4375; at n = 3 it is 0.578125. The probability increases with n, so the minimum is three trials. The wording specifies a strict inequality.
Learning check
State the four binomial conditions without looking. Explain the coefficient in the mass function, convert three verbal inequalities to integer endpoints, and solve one minimum-n problem with a neighbouring check. If these steps are secure, move to continuous densities and then the normal model, where probability is represented by area.
Curriculum and study route
Loav Mathematical Methods Units 3 and 4. These module divisions are a teaching sequence. Use the current VCAA Mathematics study design, Mathematical Methods Units 3 and 4, Area of Study 4, and the VCAA Mathematical Methods FAQ for course requirements. Scope checked 15 October 2026. All worked scenarios, questions and explanations in this guide are original.
