GCSE Expected outcomes Practice Questions

Check that the outcomes are equally likely

The question says the spinner is fair, so each of the $8$ sections is equally likely. That is what makes it safe to work the probability out by counting outcomes.

Write down the probability

For equally likely outcomes the probability is the number of favourable outcomes over the total: $\frac{3}{8}$. $3$ and $8$ have no common factor bigger than $1$, so it is already in its simplest form. The favourable outcomes are the $3$ red sections.

State the answer

So the probability that the spinner lands on red is $\frac{3}{8}$.

Work out the expected frequency

The coin is fair, so all $2$ outcomes are equally likely and the probability that the coin lands on heads is $\frac{1}{2}$. In $10$ flips the expected frequency is $\frac{1}{2} \times 10 = 5$.

Compare the expected frequency with what actually happened

It actually happened $7$ times, not $5$. These are two different things: an EXPECTED frequency is a prediction from the probability, an ACTUAL frequency is a count of what happened. They are not supposed to be equal.

State the answer

The expected number of times the coin lands on heads is $5$; the coin actually gave $7$, and that is perfectly consistent with a fair coin.

Work out the expected frequency of each colour

The spinner is fair, so each of the $10$ sections is equally likely. For each colour the probability is (its sections) $\div$ (all the sections), and the expected frequency is that probability multiplied by the $200$ spins: red $100$, blue $60$ and green $40$.

Compare the expected frequencies

Putting them in order, the largest expected frequency is $100$, which belongs to red.

Rule out "Red, 5 times"

Red IS the right colour, but the expected frequency is $100$, not $5$.

Rule out "Blue, 60 times"

$60$ is the right expected frequency for blue, but blue is not the colour that comes up most often — red is, with $100$.

Rule out "Green, 40 times"

$40$ is the right expected frequency for green, but green is not the colour that comes up most often — red is, with $100$.

State the answer

So the spinner is expected to land on red most often, about $100$ times in $200$ spins.

Work out the expected frequency

The dice is fair, so all $6$ outcomes are equally likely and the probability that the dice lands on 6 is $\frac{1}{6}$. In $60$ rolls the expected frequency is $\frac{1}{6} \times 60 = 10$.

Compare the expected frequency with what actually happened

It actually happened $14$ times, not $10$. These are two different things: an EXPECTED frequency is a prediction from the probability, an ACTUAL frequency is a count of what happened. They are not supposed to be equal.

Rule out "the expected number is 14"

$14$ is what HAPPENED. The expected number comes from the probability and the number of rolls: $\frac{1}{6} \times 60 = 10$. An expectation is never read off the results.

Rule out "the expected number is 15"

$\frac{1}{6} \times 60 = 10$, so $15$ is simply the wrong arithmetic.

Rule out "the dice must be biased"

The question SAYS the dice is fair, so the probability is $\frac{1}{6}$ by counting equally likely outcomes — it is not $\frac{7}{30}$. And a gap of $4$ between the actual and the expected is well inside the ordinary swing of chance over $60$ rolls, so the results are no evidence of bias at all.

Rule out "the next roll is less likely to give it"

The dice has no memory. Each roll is a fresh, independent, equally likely affair, so the probability that the dice lands on 6 is still $\frac{1}{6}$ — not $\frac{1}{12}$ — however many times it has already happened. Believing otherwise is the commonest mistake about randomness there is.

Work out the relative frequency that was actually observed

The outcome turned up on $\frac{7}{30}$ of the rolls against the $\frac{1}{6}$ the theory predicts. Close, but not equal — which is what random results always look like.

Say what would count as evidence of bias

A small gap over $60$ rolls proves nothing. A relative frequency still a long way from $\frac{1}{6}$ after several thousand rolls would be real evidence, because the proportion settles down as the trials mount up.

Say what the expected frequency is for

$10$ is the best single prediction you can make before the rolls happen. Afterwards, the count that actually came up is the fact, and the expectation is only the yardstick you judge it against.

State the answer

The expected number of times the dice lands on 6 is $10$; the dice actually gave $14$, and that is perfectly consistent with a fair dice.

Work out the expected frequency

The spinner is fair, so all $8$ outcomes are equally likely and the probability that the spinner lands on red is $\frac{3}{8}$. In $200$ spins the expected frequency is $\frac{3}{8} \times 200 = 75$.

Compare the expected frequency with what actually happened

It actually happened $63$ times, not $75$. These are two different things: an EXPECTED frequency is a prediction from the probability, an ACTUAL frequency is a count of what happened. They are not supposed to be equal.

Rule out "the expected number is 63"

$63$ is what HAPPENED. The expected number comes from the probability and the number of spins: $\frac{3}{8} \times 200 = 75$. An expectation is never read off the results.

Rule out "the expected number is 80"

$\frac{3}{8} \times 200 = 75$, so $80$ is simply the wrong arithmetic.

Rule out "the spinner must be biased"

The question SAYS the spinner is fair, so the probability is $\frac{3}{8}$ by counting equally likely outcomes — it is not $\frac{63}{200}$. And a gap of $12$ between the actual and the expected is well inside the ordinary swing of chance over $200$ spins, so the results are no evidence of bias at all.

Rule out "the next spin is less likely to give it"

The spinner has no memory. Each spin is a fresh, independent, equally likely affair, so the probability that the spinner lands on red is still $\frac{3}{8}$ — not $\frac{1}{4}$ — however many times it has already happened. Believing otherwise is the commonest mistake about randomness there is.

Work out the relative frequency that was actually observed

The outcome turned up on $\frac{63}{200}$ of the spins against the $\frac{3}{8}$ the theory predicts. Close, but not equal — which is what random results always look like.

Say what would count as evidence of bias

A small gap over $200$ spins proves nothing. A relative frequency still a long way from $\frac{3}{8}$ after several thousand spins would be real evidence, because the proportion settles down as the trials mount up.

Say what the expected frequency is for

$75$ is the best single prediction you can make before the spins happen. Afterwards, the count that actually came up is the fact, and the expectation is only the yardstick you judge it against.

Check the expected frequency is a sensible size

The expected frequency lies between $0$ and the $200$ spins carried out, as it must.

Work out the expected frequency of the opposite outcome

The other outcomes are expected about $125$ times, and $75 + 125 = 200$ as it must.

Note that an expectation need not be a whole number

Here $\frac{3}{8} \times 200$ happens to come out whole. In general it does not have to: an average of $12.5$ is a perfectly good expected frequency, even though no run of trials can ever produce $12.5$ successes.

Say what "fair" is doing in this question

Every number here rests on the word "fair" in the question. Without it there would be no reason to say the probability is $\frac{3}{8}$, and no expected frequency could be worked out at all.

Summarise the method

Work out the expectation from the probability, compare it with the count, and remember that a difference between them is normal — it is the size of the difference, over enough trials, that would ever suggest bias.

State the answer

The expected number of times the spinner lands on red is $75$; the spinner actually gave $63$, and that is perfectly consistent with a fair spinner.