GCSE · Maths · AQA · Spec 8300

Randomness, fairness and equally likely events

Nobody can predict the next roll of a dice. So how can we confidently predict about 100 sixes in 600 rolls?

Probability · Expected outcomes

How many sixes should you expect?

number of rolls: 5/10. P(six) 1/6. Rolls 300. Expected sixes 50

P(six)1/6Rolls300Expected sixes50

A fair 6-sided dice has six equally likely faces, so P(six) = 1/6 on every roll. Drag the number of rolls and watch the expected number of sixes: it is always 1/6 of the rolls. Try 300, then 600 — double the rolls, and the expected sixes double too.

Probability · Fairness

Can you just count the outcomes?

Are the outcomes of this experiment equally likely?

Still to sort

Outcomes equally likely (0)

Every outcome has the same chance, so P = outcomes you want ÷ total outcomes.

Where the line is: Being random isn't enough. Every single outcome must have the same chance.

Outcomes not equally likely (0)

Some outcomes are favoured, so counting them gives the wrong probability.

Where the line is: These are still random — you just can't find the probability by counting outcomes.

6 of 6 still to sort.

That 1/6 only works because every face of a fair dice is equally likely. Sort each experiment: are its outcomes equally likely, or not?

Probability · Expected outcomes

Your turn: fill in the two key steps

A fair spinner has 8 equal sectors: 3 blue, 3 yellow and 2 green. It is spun 240 times. How many times would you expect it to land on blue?

The spinner is fair and its 8 sectors are equal, so the 8 sectors are equally likely outcomes.
missing step

Which line is step 2?

P(blue) = 1/3, because blue is one of 3 colours.P(blue) = 3/8, because 3 of the 8 equally likely sectors are blue.P(blue) = 3/5, because 3 sectors are blue and 5 are not.

Probability · Randomness

So what does 'expected' really mean?

Jas rolls a fair 6-sided dice 60 times and counts the sixes. 1/6 × 60 = 10, so the expected number of sixes is 10.

Which is closest to what you think will happen?

Jas will get exactly 10 sixes — that's what 1/6 of 60 means.About 10, but the real number will probably be a bit more or a bit less.If Jas has rolled 20 times with no six, a six is 'due', so the next roll is more likely to be a six.It's random, so you can't say anything about how many sixes there'll be.

How sure are you?

I'm sureI think soI'm guessing

Jas will get exactly 10 sixes — that's what 1/6 of 60 means.

This is a common misconception

10 is the expected number, but expected isn't guaranteed. Each roll is random, so the count wobbles: 8, 11 or 13 sixes are all perfectly normal. Here's the surprise — exactly 10 is the single most likely count, yet it only happens about 1 time in 7.

About 10, but the real number will probably be a bit more or a bit less.

This matches the evidence

Yes. 10 is the long-run prediction. Any one set of 60 rolls is random, so the real count usually lands near 10, not on it.

If Jas has rolled 20 times with no six, a six is 'due', so the next roll is more likely to be a six.

This is a common misconception

The dice has no memory. Every roll is still 1/6 for a six, whatever happened before, so a six is never 'due'. A run with no sixes doesn't make the next roll any different.

It's random, so you can't say anything about how many sixes there'll be.

This is a common misconception

Random means you can't predict any single roll. But because the dice is fair, you know P(six) = 1/6, so 'about 10 sixes in 60 rolls' is a sensible prediction. Unpredictable one roll at a time, predictable in the long run — both are true.

The expected number is the best prediction you can make, not a promise. Real results are random and land near it.

WHAT YOU'VE LEARNED

A quick recap of today's lesson.

You can't predict the next roll — but you can predict about how many sixes in 600.

What you need to know

When outcomes are equally likely, probability = number of outcomes you want ÷ total number of equally likely outcomes.
Expected number of outcomes = probability × number of trials.
Check the outcomes really are equally likely before you count them: unequal sectors, a biased dice or the total of two dice are not.
The expected number is what you predict in the long run; actual results vary randomly around it, and a dice has no memory.

The big picture

In a fair experiment every outcome is equally likely, so you can find a probability just by counting: the outcomes you want out of all the equally likely outcomes. To predict how often something will happen over many trials, multiply that probability by the number of trials — that gives the expected number of outcomes. The expected number is a long-run prediction, not a promise: real results are random and land near it, not exactly on it. And counting only works when the outcomes really are equally likely, so always check for fairness first.

Key points

1Fair means every outcome has the same chance.

2Count first: a fair 6-sided dice gives P(six) = 1/6.

3Then multiply: 1/6 × 600 rolls = 100 expected sixes.

4Double the trials and the expected number doubles too.

5Expected means 'about', never 'exactly' — and nothing is ever 'due'.

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Worked example

Problem

A bag holds 4 red, 5 green and 1 white counter. A counter is taken at random, its colour is noted, and it is put back. This is done 150 times. How many times would you expect a red counter?

⚠ Watch out

Treating the expected number as a guarantee. 1/6 × 60 = 10 means you'd predict about 10 sixes — getting 8 or 13 doesn't mean the dice is unfair or your working is wrong.

🧠

Memory hook

Count it, multiply it, then say 'about'. Count the equally likely outcomes, multiply the probability by the trials, and remember the answer is a prediction, not a promise.

✓

Check yourself

A fair 10-sided dice is numbered 1 to 10 and rolled 200 times. How many times would you expect a number greater than 7? Would you be surprised to get 63?

Flashcards

(6)

What does it mean for outcomes to be 'equally likely'?

Each outcome has exactly the same chance — like the six faces of a fair dice or the sectors of a spinner with equal sectors.

How do you find a probability from equally likely outcomes?

Number of outcomes you want ÷ total number of equally likely outcomes. For example, 3 blue sectors out of 8 equal sectors gives 3/8.

How do you work out the expected number of outcomes?

Probability × number of trials.

Does 'expected number' mean it will definitely happen?

No. It's the long-run prediction. Real results are random and usually land near it, not exactly on it.

A fair coin has landed tails 5 times in a row. Is heads now more likely?

No. The coin has no memory — P(heads) is still 1/2 on every flip. Nothing is ever 'due'.

You double the number of trials of a fair experiment. What happens to the expected number of outcomes?

It doubles too. The probability stays the same, and expected number = probability × number of trials, so it grows in proportion to the trials.

Tap any card to flip it, or use Study as deck to go through them one at a time. In the full lesson these run as a spaced-repetition deck — you rate each card Hard, Good or Easy and the tricky ones keep coming back until they stick.

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Can you just count the outcomes?

Your turn: fill in the two key steps

So what does 'expected' really mean?

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