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How do you use relative frequency to estimate probability and calculate expected outcomes?

Use relative frequency (experimental probability) to estimate probabilities from data, understand how more trials improve the estimate, and calculate expected numbers of outcomes.

A focused answer to the OCR GCSE Mathematics probability content on relative frequency and expected outcomes, covering experimental probability, the effect of more trials, fairness, and calculating expected numbers of outcomes.

Generated by Claude Opus 4.810 min answer

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  1. What this dot point is asking
  2. Relative frequency
  3. Why more trials help
  4. Judging fairness
  5. Expected outcomes
  6. Why this matters

What this dot point is asking

OCR references P4 and P5 cover relative frequency (experimental probability) and expected outcomes: estimating a probability from data, understanding why more trials give a better estimate, and calculating how many times an outcome is expected. This content bridges theoretical probability and real data, and it underpins judging fairness. It appears on every tier and is often a two-part question combining an estimate with an expected count.

Relative frequency

Relative frequency estimates probability from an experiment.

So if a drawing pin lands point-up 3535 times in 5050 drops, the relative frequency of point-up is 3550=0.7\dfrac{35}{50} = 0.7. There is no theoretical value here, because a pin is not symmetric, so the experiment is the only way to estimate the probability. Relative frequency can be written as a fraction, decimal or percentage.

Why more trials help

The estimate improves with more data.

The more trials you carry out, the closer the relative frequency tends to get to the true probability, a result sometimes called the law of large numbers. So an estimate from 10001000 spins is more trustworthy than one from 1010 spins. A graph of relative frequency against the number of trials typically swings widely at first and then settles towards a steady value. When a question asks which of several estimates is "most reliable", the answer is always the one from the largest number of trials, and the reason (more trials, closer to the true probability) earns a mark.

Judging fairness

Relative frequency reveals bias.

A fair (unbiased) object has equal theoretical probabilities, for example 16\tfrac{1}{6} for each face of a die. If the relative frequency of an outcome settles clearly away from the theoretical value after many trials, the object is probably biased. So a die landing on six with relative frequency 0.250.25 over 500500 rolls is biased towards six, because a fair die would settle near 0.1670.167. The judgement must be based on enough trials to be confident.

Expected outcomes

The expected number scales the probability up to many trials.

Why this matters

Relative frequency is how probability is estimated in the real world, from quality testing to sport, and OCR sets it in exactly such data contexts. The expected-number calculation is a direct, practical use of probability that also appears in fairness arguments and in checking model predictions against data. Choosing the largest sample for the best estimate, and justifying it, is the AO2 reasoning the board rewards.

Exam-style practice questions

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

OCR 20183 marksA biased dice is rolled 200200 times. A six comes up 5050 times. (a) Estimate the probability of rolling a six. (b) Estimate how many sixes would be expected in 360360 rolls. (Foundation, Paper 1, calculator.)
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(a) Relative frequency is the number of successes over the number of trials: 50200=0.25\dfrac{50}{200} = 0.25.

(b) Expected number == probability ×\times number of trials =0.25×360=90= 0.25 \times 360 = 90.

Markers award a mark for the relative frequency 0.250.25, a mark for the expected-number method, and a mark for 9090. Using 16\tfrac{1}{6} (the fair-dice probability) instead of the experimental 0.250.25 is the standard error, since the dice is stated to be biased.

OCR 20214 marksA spinner is spun and the relative frequency of landing on red is recorded after different numbers of spins: 0.400.40 after 1010 spins, 0.320.32 after 5050 spins, 0.300.30 after 200200 spins. (a) Which estimate is most reliable and why? (b) The spinner is spun 500500 times. Estimate the number of reds. (Higher, Paper 4, calculator.)
Show worked answer →

(a) The estimate after 200200 spins is most reliable, because relative frequency gets closer to the true probability as the number of trials increases.

(b) Use the best estimate, 0.300.30: expected reds =0.30×500=150= 0.30 \times 500 = 150.

Markers give marks for choosing the 200200-spin value, for the reason about more trials, for the expected-number method, and for 150150. Choosing the 0.400.40 value (fewest spins) is the usual error.

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