Hard GCSE Experiments and frequency Questions

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 $\frac{3}{8} = 0.375$.

Write down the total and what you are given

There are $120$ members altogether. Every count in the two-way table must add back to that.

Work out the number of members who play tennis

Each of the members who play tennis appears in exactly one row, so add down that column. That gives $22 + 14 = 36$.

Work out the number of members who play squash

Each of the members who play squash appears in exactly one row, so add down that column. That gives $18 + 26 = 44$.

Work out the number of members who play badminton

The column totals must add back to the grand total of $120$, so subtract the column totals you already know. That gives $120 - 36 - 44 = 40$.

Complete the rest of the two-way table

Filling in every branch first means each option can be checked straight from the completed two-way table.

Work out the probability the correct statement is about

There are $40$ members who play badminton out of $120$ members, so that probability is $\frac{40}{120} = \frac{1}{3}$.

Test the statement about the men who play badminton

There are $20$ men who play badminton, so that probability is really $\frac{1}{6}$, not $\frac{1}{3}$. That statement is false.

Test the statement about the women

There are $60$ women, so that probability is really $\frac{1}{2}$, not $\frac{1}{3}$. That statement is false.

Test the statement about the members who play tennis

There are $36$ members who play tennis, so that probability is really $\frac{3}{10}$, not $\frac{1}{4}$. That statement is false.

Test the statement about the women who play squash

There are $26$ women who play squash, so that probability is really $\frac{13}{60}$, not $\frac{1}{5}$. That statement is false.

Check the row totals add back to $120$

Every person is in exactly one row, so the row totals must add back to the $120$ surveyed.

Check the column totals add back to $120$

The column totals must also add back to $120$, which they do — so the completed two-way table is consistent.

Write the correct probability as a decimal

Written as a decimal the size of the probability is easier to feel.

Note the mistake to avoid

Every one of these probabilities is out of the whole group of $120$ members, because the member is chosen from all of them.

State the answer

So the correct statement is that the probability of choosing one of the members who play badminton is $\frac{1}{3}$.

Write down the total and what you are given

There are $120$ students altogether. Every count in the two-way table must add back to that.

Work out the number of boys

The row totals must add back to the grand total of $120$, so subtract the row totals you already know. That gives $120 - 60 = 60$.

Work out the number of boys who study Spanish

The counts on the branches out of boys must add back to $60$, so subtract the ones you already know. That gives $60 - 15 - 25 = 20$.

Complete the rest of the two-way table

Filling in every branch first means each option can be checked straight from the completed two-way table.

Work out the probability the correct statement is about

There are $20$ boys who study Spanish out of $120$ students, so that probability is $\frac{20}{120} = \frac{1}{6}$.

Test the statement about the girls who study French

There are $24$ girls who study French, so that probability is really $\frac{1}{5}$, not $\frac{1}{3}$. That statement is false.

Test the statement about the girls who study Spanish

There are $20$ girls who study Spanish, so that probability is really $\frac{1}{6}$, not $\frac{1}{4}$. That statement is false.

Test the statement about the boys who study German

There are $25$ boys who study German, so that probability is really $\frac{5}{24}$, not $\frac{1}{4}$. That statement is false.

Test the statement about the students who study French

There are $39$ students who study French, so that probability is really $\frac{13}{40}$, not $\frac{1}{2}$. That statement is false.

Check the row totals add back to $120$

Every person is in exactly one row, so the row totals must add back to the $120$ surveyed.

Check the column totals add back to $120$

The column totals must also add back to $120$, which they do — so the completed two-way table is consistent.

Write the correct probability as a decimal

Written as a decimal the size of the probability is easier to feel.

Note the mistake to avoid

Every one of these probabilities is out of the whole group of $120$ students, because the student is chosen from all of them.

Summarise the method

Complete the two-way table first, then every probability is just a count from it divided by $120$.

State the answer

So the correct statement is that the probability of choosing one of the boys who study Spanish is $\frac{1}{6}$.

Work out what a fair result would look like

If the spinner were fair, each of the $5$ outcomes would be equally likely, so the probability of red would be $\frac{1}{5} \approx 0.200$.

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

In $400$ trials a fair spinner would land on red about $80$ times.

Work out the relative frequency actually observed

It actually landed on red $36$ times out of $400$, a relative frequency of $\frac{9}{100} = 0.09$.

Compare the two numbers

$0.09$ is a long way below the $0.200$ a fair spinner would give. Over $400$ 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 $80$; $36$ 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. $400$ 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 $9\%$ of the trials when $20.00\%$ was expected.

Say what "biased" actually means

A biased spinner is simply one whose outcomes are not equally likely. The experiment cannot prove this, but $400$ trials give strong evidence for it.

Say what the best estimate of the probability now is

Because the spinner is not fair, theory gives no probability at all: the only estimate available is the relative frequency, $\frac{9}{100}$.

Say how the evidence could be strengthened

Repeating the experiment with $800$ trials and getting a similar relative frequency would make the case for bias stronger still.

Note the mistake to avoid

The fair value here is $\frac{1}{5}$, not $\frac{1}{2}$: "equally likely" means each of the $5$ outcomes gets the same share, not that the outcome is as likely as not.

Check the relative frequency is a valid probability

$36$ is between $0$ and $400$, so the relative frequency lies between $0$ and $1$ as any probability must.

State the answer

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

Work out what a fair result would look like

If the dice were fair, each of the $6$ outcomes would be equally likely, so the probability of a six would be $\frac{1}{6} \approx 0.167$.

Work out how many times a fair dice would be expected to land on a six

In $600$ trials a fair dice would land on a six about $100$ times.

Work out the relative frequency actually observed

It actually landed on a six $180$ times out of $600$, a relative frequency of $\frac{3}{10} = 0.3$.

Compare the two numbers

$0.3$ is a long way above the $0.167$ a fair dice would give. Over $600$ 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 greater than the fair value, so the dice favours a six more than it should — it cannot be biased away from a six.

Rule out "the dice is fair"

A fair dice would give about $100$; $180$ 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. $600$ 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 dice landed on a six on $30\%$ of the trials when $16.67\%$ was expected.

Say what "biased" actually means

A biased dice is simply one whose outcomes are not equally likely. The experiment cannot prove this, but $600$ trials give strong evidence for it.

Say what the best estimate of the probability now is

Because the dice is not fair, theory gives no probability at all: the only estimate available is the relative frequency, $\frac{3}{10}$.

Say how the evidence could be strengthened

Repeating the experiment with $1200$ trials and getting a similar relative frequency would make the case for bias stronger still.

Note the mistake to avoid

The fair value here is $\frac{1}{6}$, not $\frac{1}{2}$: "equally likely" means each of the $6$ outcomes gets the same share, not that the outcome is as likely as not.

Check the relative frequency is a valid probability

$180$ is between $0$ and $600$, so the relative frequency lies between $0$ and $1$ as any probability must.

State the answer

So the results suggest the dice is biased towards a six.