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How do you model a discrete random variable, and when is the binomial distribution the right model?

Discrete random variables and probability distributions, the binomial distribution, its conditions, and calculating binomial probabilities with technology.

A focused answer to the Edexcel A-Level Mathematics statistical distributions content, covering discrete random variables and probability distributions, the binomial distribution and its conditions, and calculating binomial probabilities using the formula and a calculator.

Generated by Claude Opus 4.89 min answer

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What this dot point is asking

Edexcel wants you to understand discrete random variables and their probability distributions, recognise when the binomial distribution is an appropriate model, state its conditions, and calculate binomial probabilities using the formula and a calculator, including cumulative probabilities.

The answer

Discrete random variables

A discrete random variable XX has a probability distribution that lists each value with its probability, and the probabilities must add to 11. For example, a distribution might assign P(X=1)=0.2P(X = 1) = 0.2, P(X=2)=0.5P(X = 2) = 0.5 and P(X=3)=0.3P(X = 3) = 0.3.

The binomial distribution

A handy memory aid is the four letters in "BINS": a Binomial needs Independent trials, a fixed Number of trials, and a constant Success probability. Sampling with replacement keeps pp constant and the trials independent, so it is binomial; sampling without replacement changes pp from trial to trial, so it is not. The single value P(X=r)P(X = r) uses the formula directly, whereas a phrase like "at most", "fewer than" or "at least" calls for a cumulative probability, often read straight from a calculator's cumulative binomial function.

Examples in context

Try this

Q1. State the two distribution parameters and the mean of X∼B(20,0.1)X \sim B(20, 0.1). [2 marks]

  • Cue. n=20n = 20, p=0.1p = 0.1; mean =np=2= np = 2.

Q2. For X∼B(8,0.5)X \sim B(8, 0.5), find P(X=4)P(X = 4). [3 marks]

  • Cue. (84)(0.5)8=70×1256≈0.273\binom{8}{4}(0.5)^8 = 70 \times \tfrac{1}{256} \approx 0.273.

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 20206 marksThe random variable X∼B(12,0.25)X \sim B(12, 0.25). Calculate P(X=3)P(X = 3), P(X≤2)P(X \le 2) and P(X≥4)P(X \ge 4), giving each to four decimal places.
Show worked answer →

Single value (M1): P(X=3)=(123)(0.25)3(0.75)9=220×0.015625×0.07508≈0.2581P(X = 3) = \binom{12}{3}(0.25)^3(0.75)^9 = 220 \times 0.015625 \times 0.07508 \approx 0.2581 (A1).

Cumulative from tables or calculator (M1): P(X≤2)≈0.3907P(X \le 2) \approx 0.3907 (A1).

Upper tail (M1): P(X≥4)=1−P(X≤3)P(X \ge 4) = 1 - P(X \le 3). Since P(X≤3)=P(X≤2)+P(X=3)≈0.3907+0.2581=0.6488P(X \le 3) = P(X \le 2) + P(X = 3) \approx 0.3907 + 0.2581 = 0.6488, we get P(X≥4)≈0.3512P(X \ge 4) \approx 0.3512 (A1).

Markers reward the single probability, the cumulative value, and the complement for the upper tail.

Edexcel 20234 marksA discrete random variable XX has P(X=x)=kxP(X = x) = kx for x=1,2,3,4x = 1, 2, 3, 4. Find the value of kk and calculate P(X≥3)P(X \ge 3).
Show worked answer →

The probabilities must sum to 11 (M1): k(1)+k(2)+k(3)+k(4)=10k=1k(1) + k(2) + k(3) + k(4) = 10k = 1, so k=110k = \dfrac{1}{10} (A1).

P(X≥3)=P(X=3)+P(X=4)=310+410=710=0.7P(X \ge 3) = P(X = 3) + P(X = 4) = \dfrac{3}{10} + \dfrac{4}{10} = \dfrac{7}{10} = 0.7 (M1, A1).

Markers reward setting the total probability to 11, solving for kk, and summing the required values.

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