What is the difference between relative frequency and probability?

Probability is what should happen in the long run, worked out from the outcomes: 1/6 for a 6 on a fair die. Relative frequency is what did happen: count divided by trials. With more trials the relative frequency gets closer to the probability, so it is used to estimate probabilities you can't work out, like a drawing pin landing point up.

If I roll a die 60 times, why don't I get exactly ten 6s?

Ten is the expected number, the long-run average, not a promise. The count of 6s in 60 rolls has a standard deviation of √(60 × 1/6 × 5/6) ≈ 2.9, so anything from about 4 to 16 is normal. The relative frequency still settles down as you keep rolling.

How do I do a tree diagram without replacement?

On the first branches write the chances of each colour. For the second set, take one of that colour out of the bag first: with 3 red and 2 blue, after a red there are 2 red and 2 blue left, so P(red then red) = 3/5 × 2/4 = 3/10. Multiply along a path for 'and', add separate paths for 'or'.

Why is 7 the most common total with two dice?

There are 36 equally likely pairs. Six of them make 7 (1 and 6, 2 and 5, 3 and 4, and each the other way round), more than any other total. Only one pair makes 2 or 12, which is why those are rare.

A bag has 3 red and 2 blue counters. Two are taken without replacement. What is P(at least one red)?

P(both blue) = 2/5 × 1/4 = 1/10, so P(at least one red) = 1 − 1/10 = 9/10. P(both red) = 3/5 × 2/4 = 3/10 and P(exactly one red) = 3/5 × 2/4 + 2/5 × 3/4 = 3/5. In Solve for, Two draws, no replacement gives all three as exact fractions.

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Probability tells you what happens in the long run. Run thousands of trials and watch the relative frequency settle on the theoretical probability, never landing exactly on it.

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Relative frequency as the trials pile up

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What's happening

When all outcomes are equally likely, the theoretical probability of an event is the number of outcomes in it divided by the total number of outcomes: a 6 on a fair die is 1/6. Two dice give 36 equally likely pairs, and six of them add up to 7, so P(7) = 6/36 = 1/6 while P(2) is only 1/36. An experiment gives a relative frequency instead: the count divided by the number of trials. With a few trials it jumps about. With many it settles towards the probability, which is the law of large numbers, and the expected count after n trials is n × P. The settling is slow: the typical gap shrinks like 1/√n, so 100 times as many trials only makes it 10 times smaller. For two events one after another, a tree diagram multiplies along the branches. Draw two counters from a bag without putting the first back and the second probabilities change, because the bag has changed.

P(A) = favourable outcomes / total outcomesrelative frequency = count / trialsexpected count = n × P(A)P(A and B) = P(A) × P(B given A)

GCSE Maths (AQA, Edexcel, OCR): relative frequency, expected outcomes, sample spaces for two dice, tree diagrams with and without replacement. A-Level Statistics: the binomial standard error.

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Predict first: with two dice, which total is most likely, and how many 7s should you expect in 600 throws? Type your count into the box and check it, then press Start and see how close a real run gets.

FAQ

What is the difference between relative frequency and probability?: Probability is what should happen in the long run, worked out from the outcomes: 1/6 for a 6 on a fair die. Relative frequency is what did happen: count divided by trials. With more trials the relative frequency gets closer to the probability, so it is used to estimate probabilities you can't work out, like a drawing pin landing point up.
If I roll a die 60 times, why don't I get exactly ten 6s?: Ten is the expected number, the long-run average, not a promise. The count of 6s in 60 rolls has a standard deviation of √(60 × 1/6 × 5/6) ≈ 2.9, so anything from about 4 to 16 is normal. The relative frequency still settles down as you keep rolling.
How do I do a tree diagram without replacement?: On the first branches write the chances of each colour. For the second set, take one of that colour out of the bag first: with 3 red and 2 blue, after a red there are 2 red and 2 blue left, so P(red then red) = 3/5 × 2/4 = 3/10. Multiply along a path for 'and', add separate paths for 'or'.
Why is 7 the most common total with two dice?: There are 36 equally likely pairs. Six of them make 7 (1 and 6, 2 and 5, 3 and 4, and each the other way round), more than any other total. Only one pair makes 2 or 12, which is why those are rare.
A bag has 3 red and 2 blue counters. Two are taken without replacement. What is P(at least one red)?: P(both blue) = 2/5 × 1/4 = 1/10, so P(at least one red) = 1 − 1/10 = 9/10. P(both red) = 3/5 × 2/4 = 3/10 and P(exactly one red) = 3/5 × 2/4 + 2/5 × 3/4 = 3/5. In Solve for, Two draws, no replacement gives all three as exact fractions.