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How do you model a random variable mathematically, and how do you use that model to find probabilities?

Discrete random variables and their probability distributions, the requirement that probabilities sum to one, the use of statistical distributions to model real situations, and an introduction to the binomial and normal models.

A focused answer to the AQA A-Level Mathematics statistical distributions content, covering discrete random variables, probability distributions, the condition that probabilities sum to one, and choosing a suitable model such as the binomial or normal distribution.

Generated by Claude Opus 4.89 min answer

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  1. What this dot point is asking
  2. Random variables
  3. The total probability condition
  4. Distributions given by a formula
  5. Choosing a model
  6. Reading and combining probabilities

What this dot point is asking

AQA wants you to understand discrete random variables, write and use a probability distribution, check that the probabilities sum to one, and select a suitable statistical distribution (such as the binomial or normal model) for a given situation, stating the assumptions you rely on. This topic provides the language and the total-probability rule that the binomial and normal distributions then build on.

Random variables

Capital letters such as XX denote the variable; lower-case xx denotes a particular value it can take. Distributions can be given as a table, a formula, or a description, and a frequent exam task is to convert between these forms.

The total probability condition

Distributions given by a formula

A distribution may be defined by a rule such as P(X=x)=cxP(X = x) = cx over a stated range of xx. Substitute each allowed value, add the results, set the sum equal to one, and solve for the constant. Then you can read off or combine individual probabilities as needed.

Choosing a model

You select a distribution by matching its assumptions to the situation, and you should always state those assumptions because every model is an approximation.

  • A fixed number of independent trials, each with two outcomes and the same probability of success, points to the binomial distribution B(n,p)B(n, p).
  • A continuous quantity that clusters symmetrically about a mean (such as heights, masses or measurement errors) points to the normal distribution N(μ,σ2)N(\mu, \sigma^2).
  • A discrete count with no fixed upper limit, occurring at a constant average rate, would in later study point to other models, but at A-level the binomial is the main discrete model.

Reading and combining probabilities

Once a distribution is known, exam questions ask for probabilities of single values or of ranges. For a range, add the probabilities of the included values, watching the inequality symbols. The strict symbols << and >> exclude the endpoint, while \le and \ge include it. The complement rule, P(Xk)=1P(Xk1)P(X \ge k) = 1 - P(X \le k - 1) for integer-valued variables, often saves work, and "at least one" is almost always fastest as one minus "none".

It is worth distinguishing a probability distribution from a frequency table of observed data. A probability distribution gives theoretical long-run proportions that must sum to one, whereas a frequency table records what actually happened in a sample and its entries sum to the sample size. Many marks are lost by treating observed frequencies as if they were probabilities without dividing by the total.

Exam-style practice questions

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

AQA 20186 marksPaper 3, Section A. The discrete random variable XX has the probability distribution given by P(X=1)=0.1P(X = 1) = 0.1, P(X=2)=0.3P(X = 2) = 0.3, P(X=3)=kP(X = 3) = k, P(X=4)=0.25P(X = 4) = 0.25. (a) Find the value of kk. (b) Find P(X3)P(X \ge 3). (c) Find P(2X<4)P(2 \le X < 4).
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For (a), the probabilities must sum to one: 0.1+0.3+k+0.25=10.1 + 0.3 + k + 0.25 = 1, so k=0.35k = 0.35. For (b), P(X3)=P(X=3)+P(X=4)=0.35+0.25=0.6P(X \ge 3) = P(X = 3) + P(X = 4) = 0.35 + 0.25 = 0.6. For (c), P(2X<4)=P(X=2)+P(X=3)=0.3+0.35=0.65P(2 \le X < 4) = P(X = 2) + P(X = 3) = 0.3 + 0.35 = 0.65, noting the strict upper inequality excludes X=4X = 4. Markers reward using the total-probability condition to find kk, and careful reading of the inequalities (the half-open interval includes 22 but not 44).

AQA 20215 marksPaper 3, Section A. A spinner has four sectors numbered 11 to 44. The discrete random variable XX is the number obtained, with P(X=x)=cxP(X = x) = cx for x=1,2,3,4x = 1, 2, 3, 4. (a) Show that c=0.1c = 0.1. (b) Find P(X>2)P(X > 2). (c) State, with a reason, whether this spinner is fair.
Show worked answer →

For (a), the probabilities are c,2c,3c,4cc, 2c, 3c, 4c and must sum to one: 10c=110c = 1, so c=0.1c = 0.1. For (b), P(X>2)=P(X=3)+P(X=4)=0.3+0.4=0.7P(X > 2) = P(X = 3) + P(X = 4) = 0.3 + 0.4 = 0.7. For (c), a fair spinner would give each outcome probability 0.250.25; here the probabilities are 0.1,0.2,0.3,0.40.1, 0.2, 0.3, 0.4, so it is not fair, as higher numbers are more likely. Markers reward setting up cx=1\sum cx = 1, evaluating the sum, and a justified statement about fairness.

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