GCSE · Probability

Relative Frequency and Expected Outcomes

Some probabilities can't be worked out on paper: will a drawing pin land point up, will a bus be late, is this dice loaded? Instead we experiment and count. This page shows how to turn data into an estimate, why more trials give a better one, how to judge bias, how to combine results, and how to predict what will happen.

GCSE subject content: record, describe and analyse the frequency of outcomes of probability experiments using tables and frequency trees; apply ideas of randomness, fairness and equally likely events to calculate expected outcomes of multiple future experiments; relate relative expected frequencies to theoretical probability; understand that empirical unbiased samples tend towards theoretical probability distributions, with increasing sample size

Before you start

1 · Relative frequency: probability from data

A theoretical probability comes from reasoning: a fair dice has 6 equally likely faces, so P(6) = 1/6. When outcomes aren't equally likely and you can't reason it out, you do an experiment: repeat it many times (each repeat is a trial) and count how often the event happens (its frequency). Then

relative frequency = frequency ÷ number of trials

The relative frequency is an estimate of the probability, sometimes called the experimental probability. Below is a bag of red, blue and green counters. You can't see inside. Take a counter, note its colour, put it back, and repeat. Use the relative frequencies to estimate what's in the bag, then peek.

Two things to notice. The relative frequencies always add up to 1 (every trial lands somewhere). And after a few trials the estimate can be wildly off, even though nothing is wrong with the bag. That's just randomness.

2 · More trials, better estimates

How good is an estimate? The more trials, the more reliable it is. With few trials, luck has a big effect; with many, the lucky and unlucky runs balance out and the relative frequency settles down towards the true probability. Run the experiment and watch the line. Then press New run a few times: every run starts out wild and they all close in on the same value.

Trials
Relative frequency
True probability

This is what the specification means by "empirical unbiased samples tend towards theoretical probability distributions with increasing sample size". In exam answers, the phrase to use is: "The estimate is more reliable because it is based on more trials." Note what it doesn't say: more trials make the estimate more likely to be close. They never guarantee the exact value.

3 · Theoretical or experimental? Judging bias

A dice is fair (unbiased) if every face has probability 1/6. To test it, roll it many times and compare each relative frequency with 1/6. Small differences are normal. A big difference after many trials is evidence of bias. Here's a mystery dice: it might be fair, it might be loaded. Roll it as much as you like, then give your verdict.

Agree or disagree? Each statement is the kind of reasoning exam questions ask you to judge.

4 · Expected outcomes

If you know (or have estimated) a probability, you can predict how many times something will happen in a number of trials. This is the expected frequency:

expected number = probability × number of trials

Why? If P(late) = 0.2, then "on average, 2 in every 10 journeys are late", so in 300 journeys about 0.2 × 300 = 60. The bar shows the trials with the expected part in orange. It's an average prediction, not a promise: the real number will usually be close, not exact.

5 · Combining results from several groups

Four groups in a class each test the same drawing pin, but they do different numbers of trials. What's the best estimate of P(point up)? The trap is to average the four relative frequencies: that treats a group with 20 trials as just as important as one with 150. Instead, pool the data: add up all the successes and all the trials.

best estimate = total successes ÷ total trials

In the chart, each bar's thickness shows how many trials that group did.

a) Which group's estimate is the most reliable?

b) Use all the results to give the best estimate of P(point up), as a decimal.

6 · Relative frequency tables and frequency trees

Exam questions often give a relative frequency table with a gap. The relative frequencies of all the outcomes add up to 1, so the gap is 1 minus the rest. Then use it to estimate an expected number.

Survey data in a frequency tree (built step by step on tree diagrams) works the same way: each relative frequency is a box divided by the total, and it estimates a probability for the next person. Switch the tree between counts and relative frequencies (each branch as a fraction of the box it comes from).

7 · Exam-style data questions

Put it all together. Answer each part, then press its Check button. The feedback names the slip if there is one.

Common mistakes and exam traps

Worked examples

A spinner is spun 80 times and lands on red 28 times. Estimate P(red) and the number of reds in 400 spins.
Relative frequency: 28 ÷ 80 = 28/80 = 7/20 = 0.35.
Use it as the estimate: P(red) ≈ 0.35.
Expected reds in 400 spins: 0.35 × 400 = 140.
Shortcut check: 400 is 5 × 80, so expect 5 × 28 = 140 ✓.
A spinner has four sections A, B, C and D. The relative frequencies are A 0.15, B 0.4, C x, D 0.25. Find x, and estimate how many times it lands on C in 300 spins.
They add to 1: 0.15 + 0.4 + 0.25 = 0.8, so x = 1 − 0.8 = 0.2.
Expected C: 0.2 × 300 = 60.
Three groups test a drawing pin. Group 1: 9 point up in 20. Group 2: 21 in 50. Group 3: 15 in 30. Find the best estimate of P(point up).
Pool the data: successes 9 + 21 + 15 = 45; trials 20 + 50 + 30 = 100.
Best estimate: 45 ÷ 100 = 0.45.
The trap: the group relative frequencies are 0.45, 0.42 and 0.5. Their mean is 1.37 ÷ 3 ≈ 0.457, which gives Group 1's 20 trials the same weight as Group 2's 50.
A dice is rolled 300 times and gives 75 sixes. Is it biased? Estimate how many sixes you'd get in 60 rolls.
If fair: expect 1/6 × 300 = 50 sixes.
Experimental: relative frequency of six = 75/300 = 0.25, compared with 1/6 ≈ 0.167.
Judgement: 300 is a lot of trials and 75 is far more than 50, so this is good evidence the dice is biased towards six.
Estimate: use the relative frequency, not 1/6: 0.25 × 60 = 15 sixes.
120 students sat a test. 45 used a revision app and 36 of them passed. Of the others, 39 passed. Complete a frequency tree, estimate P(a student passes), and predict how many of 400 students would pass.
Tree: app 45 → pass 36, fail 45 − 36 = 9. No app 120 − 45 = 75 → pass 39, fail 75 − 39 = 36. Check: 36 + 9 + 39 + 36 = 120 ✓.
Passes: 36 + 39 = 75 out of 120. Relative frequency = 75/120 = 5/8 = 0.625.
Prediction: 5/8 × 400 = 250 students.
Bonus: pass rate with the app 36/45 = 0.8; without 39/75 = 0.52. (The data can't prove the app caused it.)
A biased coin has P(heads) = 0.6. Sam flips it twice, and repeats this 200 times. How many times would he expect two heads?
Probability of two heads (independent flips): 0.6 × 0.6 = 0.36.
Expected number: 0.36 × 200 = 72.
Not 0.6 × 200 = 120: that's the expected number of pairs whose first flip is heads.

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