GCSE Experiments and frequency Practice Questions

Write down what you know

There are $120$ students altogether, and the question gives the number of girls = $70$, the number of girls who walk to school = $45$ and the number of boys who do not walk to school = $30$.

Work out the number of girls who do not walk to school

The counts on the branches out of girls must add back to $70$, so subtract the ones you already know. That gives $70 - 45 = 25$.

State the answer

So the number of girls who do not walk to school is $25$.

Say what makes an estimate from an experiment a good one

A relative frequency is only an estimate of a probability. The more trials you carry out, the less the estimate jumps around and the closer it settles to the true probability.

Compare the numbers of trials

Put the numbers of spins in order: $25$ < $35$ < $80$ < $95$ < $120$. The largest is $120$.

State the answer

Gina spun the spinner $120$ times, more than anyone else, so Gina's relative frequency is the most reliable estimate.

Say what has to be compared

The number of times each spinner landed on green is not enough on its own, because the two spinners were not spun the same number of times. What must be compared is the FRACTION of spins that were successful.

Work out the relative frequency for Spinner A

Spinner A landed on green on $28$ of its $80$ spins, which is $\frac{7}{20} = 0.35$.

Work out the relative frequency for Spinner B

Spinner B landed on green on $18$ of its $40$ spins, which is $\frac{9}{20} = 0.45$.

Compare the two relative frequencies

$0.45$ is bigger than $0.35$, so Spinner B landed on green a greater FRACTION of the time.

Rule out "Spinner A has the greater relative frequency"

Spinner A's fraction is $0.35$, which is smaller, so it cannot be the greater relative frequency.

State the answer

So Spinner B has the greater relative frequency of green: $\frac{9}{20} = 0.45$ against $\frac{7}{20} = 0.35$.

Work out what a fair result would look like

If the spinner were fair, each of the $4$ outcomes would be equally likely, so the probability of red would be $\frac{1}{4} \approx 0.250$.

Work out how many times a fair spinner would be expected to land on red

In $200$ trials a fair spinner would land on red about $50$ times.

Work out the relative frequency actually observed

It actually landed on red $30$ times out of $200$, a relative frequency of $\frac{3}{20} = 0.15$.

Compare the two numbers

$0.15$ is a long way below the $0.250$ a fair spinner would give. Over $200$ trials a gap that size is far too big to put down to chance.

Rule out the opposite direction of bias

The relative frequency is less than the fair value, so the spinner favours red less than it should — it cannot be biased towards red.

Rule out "the spinner is fair"

A fair spinner would give about $50$; $30$ were observed. That is not a small difference.

Rule out "a large number of trials means it is fair"

Doing lots of trials makes the estimate more trustworthy — it does not make the object fair. Here the many trials are exactly what make the bias convincing.

Rule out "nothing can be said"

There is no magic number of trials. $200$ trials is more than enough to see a gap this size, and the results are evidence whether or not the number is round.

Write the difference as a percentage

The spinner landed on red on $15\%$ of the trials when $25.00\%$ was expected.

State the answer

So the results suggest the spinner is biased away from red.

Say what has to be compared

The number of times each spinner landed on yellow is not enough on its own, because the two spinners were not spun the same number of times. What must be compared is the FRACTION of spins that were successful.

Work out the relative frequency for Spinner A

Spinner A landed on yellow on $27$ of its $60$ spins, which is $\frac{9}{20} = 0.45$.

Work out the relative frequency for Spinner B

Spinner B landed on yellow on $30$ of its $80$ spins, which is $\frac{3}{8} = 0.375$.

Compare the two relative frequencies

$0.45$ is bigger than $0.375$, so Spinner A landed on yellow a greater FRACTION of the time.

Rule out "Spinner B has the greater relative frequency"

Spinner B's fraction is $0.375$, which is smaller, so it cannot be the greater relative frequency.

Rule out the "more successes" argument

Spinner B landed on yellow more times in total, but it was also spun a different number of times. A bigger count is not the same as a bigger fraction.

Rule out "the same relative frequency"

$\frac{9}{20}$ and $\frac{3}{8}$ are different fractions ($0.45$ against $0.375$), so the relative frequencies are not equal.

Write both as percentages to see the gap

Spinner A landed on yellow on $45\%$ of its spins and Spinner B on $37.5\%$ of its spins.

Say which estimate is more reliable

Spinner B was spun more times ($80$ against $60$), so its relative frequency is the more reliable estimate of its own probability — a separate question from which fraction is bigger.

Say what would happen with more spins

With only a few dozen spins these fractions can move about quite a lot. Hundreds of spins would pin each spinner down much more tightly.

Note the common mistake

Comparing the raw counts instead of the fractions is the classic error here. Always divide by the number of trials first.

Check the arithmetic by cross-multiplying

Cross-multiplying compares $\frac{27}{60}$ with $\frac{30}{80}$ without any division: $2160$ against $1800$ confirms that Spinner A's fraction is the bigger one.

Say what each fraction estimates

Each fraction is that spinner's own estimated probability of yellow. They are estimates for two different spinners, so there is no reason for them to agree.

Summarise the method

Turn each result into a fraction of its own number of trials, then compare the fractions — as decimals if that is easier.

State the answer

So Spinner A has the greater relative frequency of yellow: $\frac{9}{20} = 0.45$ against $\frac{3}{8} = 0.375$.