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When does the binomial distribution apply, and how do you use it?

Characteristics of a binomial distribution; the notation B(n, p); the conditions for a binomial model; the mean np; calculating binomial probabilities for n up to 10.

A focused answer to Edexcel GCSE Statistics (Higher tier) on the binomial distribution, covering its characteristics, the notation B(n, p), the conditions that make a binomial model suitable, the mean np, and calculating binomial probabilities for n up to 10.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. What the binomial distribution models
  3. The conditions for a binomial model
  4. The mean of a binomial distribution
  5. Calculating binomial probabilities
  6. Recognising a binomial situation
  7. Shape and interpretation

What this dot point is asking

Edexcel Higher tier code 3p.11 requires you to know and interpret the characteristics of a binomial distribution, to use the notation B(n,p)B(n, p), to know the conditions that make a binomial model suitable, to use the mean npnp, and to calculate binomial probabilities (questions are not set with nn larger than 1010). Probabilities may be found by any standard method (calculator functions, spreadsheets or Pascal's triangle).

What the binomial distribution models

Typical situations are the number of heads in several coin flips, the number of correct answers when guessing, or the number of faulty items in a sample. The notation XB(n,p)X \sim B(n, p) records the two parameters: nn (the number of trials) and pp (the probability of success on each trial).

The conditions for a binomial model

A binomial model is only valid when all four conditions hold:

  1. There is a fixed number of trials, nn.
  2. Each trial has two outcomes: success or failure.
  3. The probability of success pp is constant for every trial.
  4. The trials are independent of each other.

Edexcel often asks you to state the conditions or to judge whether a binomial model is appropriate in a context. If, for example, items are drawn without replacement (so pp changes), the trials are not independent and the binomial model does not strictly apply.

The mean of a binomial distribution

This formula is one Edexcel expects you to know. For 88 flips of a fair coin, the mean number of heads is 8×0.5=48 \times 0.5 = 4; for 2020 items with a 0.10.1 fault rate, the expected number of faults is 20×0.1=220 \times 0.1 = 2. The mean is the long-run average number of successes, not a guaranteed value.

Calculating binomial probabilities

To find the probability of exactly rr successes you use the binomial term

P(X=r)=(nr)pr(1p)nr,P(X = r) = \binom{n}{r} p^r (1-p)^{n-r},

where (nr)\binom{n}{r} is the number of ways to choose which rr of the nn trials are successes (read from Pascal's triangle or a calculator). Edexcel allows any standard method, and caps nn at 1010 so the calculations stay manageable. For "at least one" questions, the complement is quickest: P(X1)=1P(X=0)P(X \ge 1) = 1 - P(X = 0).

The three parts of the term each have a meaning worth understanding. The power prp^r is the probability that rr particular trials are all successes; the power (1p)nr(1-p)^{n-r} is the probability that the remaining nrn - r trials are all failures; and the coefficient (nr)\binom{n}{r} counts how many different ways those rr successes can be arranged among the nn trials. Multiplying them gives the total probability of getting exactly rr successes in any order. Seeing the term this way helps you avoid the common error of dropping the coefficient, which would only count one specific arrangement.

Recognising a binomial situation

Many exam questions do not say "binomial"; you have to recognise the situation from the four conditions. Look for a fixed number of repeated trials with a clear "success", a constant success probability, and independence. Classic contexts are coin flips, multiple choice guessing, items passing or failing a test, and shots hitting or missing a target. If the context instead involves drawing without replacement, a changing probability, or a count with no fixed maximum, the binomial model is not appropriate, and saying so (with a reason) earns marks just as readily as a calculation.

Shape and interpretation

A binomial distribution is symmetric when p=0.5p = 0.5 and skewed when pp is close to 00 or 11 (skewed towards the more likely outcome). Knowing the mean npnp tells you where the distribution is centred, so you can sense-check a probability: most of the probability sits near npnp successes, and outcomes far from npnp are unlikely. For example, in B(10,0.5)B(10, 0.5) the most likely number of successes is around 55, and getting 00 or 1010 is rare, which matches everyday intuition about flipping a fair coin ten times.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 1ST0 20214 marksA fair coin is flipped 88 times, and XX is the number of heads, so XX follows B(8,0.5)B(8, 0.5). (a) State the mean number of heads. (b) Using P(X=r)=(nr)pr(1p)nrP(X = r) = \binom{n}{r} p^r (1-p)^{n-r}, calculate the probability of exactly 22 heads.
Show worked answer →

(a) Mean =np=8×0.5=4= np = 8 \times 0.5 = 4 heads.

(b) P(X=2)=(82)(0.5)2(0.5)6=28×(0.5)8=28×1256=282560.109P(X = 2) = \binom{8}{2} (0.5)^2 (0.5)^6 = 28 \times (0.5)^8 = 28 \times \frac{1}{256} = \frac{28}{256} \approx 0.109.

Markers reward the mean np=4np = 4, the binomial term with (82)=28\binom{8}{2} = 28, and the probability 0.109\approx 0.109.

Edexcel 1ST0 20224 marksA multiple choice quiz has 55 questions, each with probability 0.20.2 of being answered correctly by guessing. (a) State two conditions that make the binomial model suitable here. (b) Find the probability of guessing exactly 11 correct answer.
Show worked answer →

(a) Any two conditions: there is a fixed number of trials (55 questions); each trial has two outcomes (correct or not); the probability of success is constant (0.20.2); the trials are independent.

(b) With n=5n = 5, p=0.2p = 0.2: P(X=1)=(51)(0.2)1(0.8)4=5×0.2×0.4096=0.4096P(X = 1) = \binom{5}{1} (0.2)^1 (0.8)^4 = 5 \times 0.2 \times 0.4096 = 0.4096.

Markers reward two valid binomial conditions, and the probability 0.410\approx 0.410 from the binomial term.

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