Hard GCSE Expected outcomes Questions

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.

Notice that the outcomes are NOT equally likely

The question says the dice is not fair. So there is nothing to count: $\frac{1}{6}$ is not the probability of anything here, and the only probability you may use is the one you are given.

Read the probability off the question

The probability that the dice lands on a six is given as $0.28$.

Multiply the probability by the number of trials

Expected frequency is probability times trials: $\frac{7}{25} \times 150 = 42$.

Rule out "25"

This is what you get by ASSUMING the $6$ outcomes are equally likely: $\frac{1}{6} \times 150 = 25$. The question says they are not, so this is wrong — the true answer is $42$.

Rule out "108"

This is the expected number of times the outcome does NOT happen, $(1 - \frac{7}{25}) \times 150 = 108$. The question asked for the times it DOES happen, which is $42$.

Rule out "28"

$\frac{7}{25} \times 150 = 42$, not $28$, so this statement is false.

Rule out "84"

$\frac{7}{25} \times 150 = 42$, not $84$, so this statement is false.

Say why fairness has to be stated, not assumed

Counting outcomes to get a probability is only valid when the question SAYS the outcomes are equally likely. Here it says the opposite, in as many words.

Check the probability is between $0$ and $1$

$\frac{7}{25}$ lies strictly between $0$ and $1$, so it is a possible probability even though the device is biased.

Work out the probability of the opposite outcome

The outcome either happens or it does not, so the other probability is $\frac{18}{25}$.

Work out the expected number of times it does NOT happen

That gives $108$, and $42 + 108 = 150$, which is a good check: the two expectations must add back to the number of trials.

Write the probability as a decimal and a percentage

About $28\%$ of the trials are expected to give this outcome.

Say what "expected" does and does not promise

The dice will not give exactly $42$ every time. The expected frequency is the long-run average, not a promise about one run of trials.

Say where a biased probability comes from

A biased object has no theory to fall back on. Its probability has to be estimated from a long run of trials — which is why the question simply hands it to you.

State the answer

So the expected number of times the dice lands on a six is $42$.

Work out the expected frequency of each colour

The spinner is fair, so each of the $16$ 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 $240$ spins: red $45$, blue $75$ and green $120$.

Compare the expected frequencies

Putting them in order, the largest expected frequency is $120$, which belongs to green.

Rule out "Green, 8 times"

Green IS the right colour, but the expected frequency is $120$, not $8$.

Rule out "Blue, 75 times"

$75$ is the right expected frequency for blue, but blue is not the colour that comes up most often — green is, with $120$.

Rule out "Red, 45 times"

$45$ is the right expected frequency for red, but red is not the colour that comes up most often — green is, with $120$.

Rule out "Blue, 120 times"

Blue is expected about $75$ times, not $120$, and it is not the most likely colour either.

Check the expected frequencies add back to the number of spins

Every spin lands on exactly one colour, so the expected frequencies have to add back to the $240$ spins. They do, which is a strong check that none of them is wrong.

Note the mistake to avoid

The colour with the most sections is expected most often ONLY because the spinner is fair. If the sections were different sizes, the number of sections would tell you nothing at all.

Note that the number of sections is not the answer

A tempting wrong answer is to quote the number of sections ($8$) instead of the expected frequency ($120$). The number of sections is part of the probability, not the prediction.

Write the winning probability down

Green takes $8$ of the $16$ equally likely sections, a probability of $\frac{1}{2}$, and $\frac{1}{2} \times 240 = 120$.

Say what "expected" does and does not promise

The spinner will not land on green exactly $120$ times. It is the average over many repeats — but it is still the best single prediction, and over many spins it is very likely to be the commonest colour.

Say what more spins would do

With only a few spins any colour could come out on top by luck. The more spins there are, the more nearly the actual counts match these expected ones.

Check each probability is between $0$ and $1$

Each colour takes some but not all of the sections, so each probability lies strictly between $0$ and $1$, as every probability must.

Check the probabilities add to $1$

The colours are the only possibilities and no spin can be two colours at once, so their probabilities must add to exactly $1$.

State the answer

So the spinner is expected to land on green most often, about $120$ times in $240$ spins.

Say why the outcomes can be counted at all

The spinner is fair, so each of its $15$ sections is equally likely. Only then does the probability of a colour equal (its sections) divided by (all the sections).

Work out the expected frequency of every colour

Expected frequency is probability times trials, so work one out for each colour: red $72$, blue $48$ and green $60$ in $180$ spins. Every option can now be tested against these numbers.

Test the statement about blue

Blue takes $4$ of the $15$ sections, so its probability is $\frac{4}{15}$ and its expected frequency is $48$. That statement is TRUE.

Rule out "expected to land on blue 60 times"

Blue takes $4$ of the $15$ sections, so its expected frequency is really $48$, not $60$. That statement is false.

Rule out "expected to land on red 48 times"

Red takes $6$ of the $15$ sections, so its expected frequency is really $72$, not $48$. That statement is false.

Rule out "expected to land on green 48 times"

Green takes $5$ of the $15$ sections, so its expected frequency is really $60$, not $48$. That statement is false.

Rule out "expected to land on blue 4 times"

Blue takes $4$ of the $15$ sections, so its expected frequency is really $48$, not $4$. That statement is false.

Check the expected frequencies add back to the number of spins

Each spin lands on exactly one colour, so the expected frequencies must add back to the $180$ spins — a quick check that none of them is wrong.

Check the probabilities add to $1$

The colours are the only possibilities, so their probabilities have to add to exactly $1$.

Note the mistake to avoid

Quoting the number of SECTIONS instead of the expected frequency is the classic trap in this question. The sections give the probability; only multiplying by the spins gives a prediction.

Note the other mistake to avoid

The expected frequency of one colour can never be the whole $180$ spins unless that colour takes every section. Any option quoting the total number of spins for a single colour is wrong on sight.

Say what "expected" does and does not promise

The spinner will not land on blue exactly $48$ times. The expected frequency is an average over many repeats — it is a prediction, not a promise.

Write the probability as a decimal

$\frac{4}{15}$ has no exact decimal form, so the multiplication is done with the fraction to keep the answer exact.

Say what more spins would do

Over a few spins the counts can be well away from these expectations. Over thousands of spins they settle very close to them.

State the answer

So the correct statement is that the spinner is expected to land on blue $48$ times.

Check that the outcomes are equally likely

The question says the counter is fair, so each of the $16$ counters is equally likely. That is what makes it safe to work the probability out by counting outcomes.

Count the outcomes that give "a red or blue counter is taken"

The favourable outcomes are the $5$ red and $6$ blue counters — that is $11$ of them. There are $16$ outcomes altogether.

Write down the probability

For equally likely outcomes the probability is the number of favourable outcomes over the total: $\frac{11}{16}$. $11$ and $16$ have no common factor bigger than $1$, so it is already in its simplest form. The favourable outcomes are the $5$ red and $6$ blue counters.

Write down the rule for an expected frequency

If an outcome happens on a fraction $p$ of the trials on average, then in $T$ trials it is expected about $p \times T$ times. That single rule is the whole of this topic.

Multiply the probability by the number of trials

$\frac{11}{16} \times 40 = 27.5$, so a red or blue counter is taken is expected about $27.5$ times in $40$ trials.

Check the probability is between $0$ and $1$

$11$ of the $16$ equally likely outcomes are favourable, and $11$ is neither $0$ nor all $16$, so the probability must land strictly between $0$ and $1$. Any "probability" outside that range is a mistake.

Write the probability as a decimal and as a percentage

The same probability three ways. $68.75\%$ of the trials are expected to give this outcome. Decimals and percentages are easier to compare; the fraction is exact, so the fraction is what the multiplication should use.

Say what "expected" actually means

The expected frequency is the number of times a red or blue counter is taken would happen ON AVERAGE if you did these $40$ trials over and over again. It is not a promise: a real set of trials will usually land a little above or a little below it.

Note that an expected frequency need not be a whole number

An expected frequency is an average, so it is perfectly allowed to come out as something like $12.5$. Only a COUNT of what actually happened has to be a whole number.

Note the mistake to avoid

The probability is multiplied by the NUMBER OF TRIALS. Multiplying by $16$ — the number of counters — is the commonest slip in this topic and has nothing to do with how many times the experiment was carried out.

Say why the device has to be fair

Counting outcomes only gives the probability because the question says the counter is fair, so all $16$ outcomes are equally likely. If they were not, $\frac{11}{16}$ would be worth nothing and the probability would have to be given to you.

Say what happens over more trials

A short run of trials can stray a long way from the expected frequency, in proportion. Over thousands of trials the proportion of successes settles towards $\frac{11}{16}$, which is what makes the expectation useful.

Summarise the method

Count the favourable outcomes out of the total, write that as a fraction in its simplest form, then multiply by the number of trials.

Work out the probability of the event NOT happening

The other $5$ outcomes are the ones where a red or blue counter is taken does not happen, so the probability of that is $\frac{5}{16} = \frac{5}{16}$, which is exactly $1 - \frac{11}{16}$.

State the answer

So the expected number of times a red or blue counter is taken is $27.5$. An expected frequency is an average, so it does not have to be a whole number.