Calculate relative frequencies to estimate probabilities of events, where relative frequency = frequency of the event divided by the total number of trials, recognising that as the number of trials increases the relative frequency approaches the theoretical probability

Part (a): relative frequency of red is $\tfrac{90}{250} = \tfrac{9}{25} = 0.36$.

Part (b): blue is the complement, so its estimated probability is $1 - 0.36 = 0.64$ (equivalently $\tfrac{160}{250} = 0.64$).

Part (c): apply the red relative frequency to the $20$ counters: $0.36 \times 20 = 7.2$, so about $7$ counters are red.

Markers reward the correct relative frequency in part (a), use of the complement (or the blue count over the total) in part (b), and multiplying the probability by $20$ then rounding sensibly in part (c). A bald answer with no working may be capped for showing no method.

Part (a): the relative frequency starts high ($0.30$) and decreases, then steadies, settling close to about $0.135$ as the number of tosses grows.

Part (b): the best estimate is the value after the most tosses, about $0.135$ (accept $0.13$ to $0.14$), because relative frequency approaches the theoretical probability as the number of trials increases, so the largest sample gives the most reliable estimate.

Part (c): with only $20$ tosses the sample is small, so chance variation can push the relative frequency a long way from the true probability; the value $0.30$ is more than double the long-run figure, showing how unreliable a small sample is.

Markers reward describing the trend as converging or steadying (not just "it goes down"), choosing the large-sample value with a reason tied to the number of trials, and a clear statement that small samples are unreliable.

Part (a): expected goals $= 0.68 \times 25 = 17$, so about $17$ goals.

Part (b): relative frequency for the final is $\tfrac{20}{25} = \tfrac{4}{5} = 0.8$ (that is $80\%$).

Part (c): any sensible reason that a single $25$-shot game is a small sample subject to chance variation - she had a hot game, faced an easier defence, or simply that a small number of shots can sit well above the long-run figure.

Markers reward $0.68 \times 25 = 17$ in part (a), the simplified relative frequency $0.8$ in part (b), and a reason in part (c) that links the gap to small-sample variability rather than restating the numbers.

Write the relative frequency formula. Relative frequency is the number of times the event happens divided by the total number of trials:

Substitute the numbers. Red came up $12$ times out of $50$ spins:

so the relative frequency of red is $\tfrac{6}{25}$, or $0.24$. (Check it is a sensible probability: it lies between $0$ and $1$, as every relative frequency must.)

Treat the survey as the experiment. The estimated probability is the relative frequency of a yes answer:

Simplify and convert.

so the estimated probability is $0.65$ (that is $65\%$). (Sanity check: more than half said yes, and $0.65 > 0.5$, which agrees.)

Confirm the total. The frequencies must add to the number of trials:

so the total is $60$, as stated.

Part (a) - relative frequency of a $4$. A $4$ appeared $12$ times out of $60$:

so the relative frequency of a $4$ is $\tfrac{1}{5}$, or $0.2$.

Part (b) - relative frequency of an even number. The even outcomes are $2$, $4$ and $6$, so add their frequencies:

then divide by the total:

so the relative frequency of an even number is $\tfrac{11}{20}$, or $0.55$. (Cross-check: the odd outcomes total $8 + 9 + 10 = 27$, and $\tfrac{27}{60} = 0.45$, and $0.55 + 0.45 = 1$, as the two complementary relative frequencies should.)

Part (a) - estimate the probability. The relative frequency of a successful free throw is the made shots over the total attempts:

so the estimated probability is $0.7$ (that is $70\%$).

Part (b) - expected number in $25$ attempts. Multiply the estimated probability by the number of new attempts:

so she would be expected to make about $17.5$ free throws, which we read as roughly $17$ or $18$ out of $25$. (The expected number need not be a whole number; it is a long-run average, so "about $18$" is the natural way to report it.)

Part (a) - relative frequency of a faulty globe. Divide the faulty count by the sample size:

so the relative frequency of a faulty globe is $\tfrac{3}{20}$, or $0.15$.

Part (b) - estimate the weekly faulty count. Use the relative frequency as the probability that any globe is faulty, then apply it to the week's production:

so about $3000$ globes are expected to be faulty that week. (Check the scale: $15\%$ of $20\,000$ is $3000$, which matches.)

Part (a) - the two relative frequencies. After $10$ tosses:

After $1000$ tosses:

Part (b) - the value being approached. Writing the full sequence of relative frequencies:

these settle down toward $0.5$.

Part (c) - is the coin fair? A fair coin has a theoretical probability of heads of $0.5$. The relative frequency starts well off ($0.7$ after only $10$ tosses) but steadies near $0.5$ as the number of tosses grows, which is exactly the behaviour expected of a fair coin. So the evidence is consistent with a fair coin. (Check the direction of the trend: the early value is far from $0.5$ and the later values cluster around it, which is the long-run convergence at work.)

Part (a) - the two relative frequencies. For Mia:

For Sam:

Part (b) - whose estimate to trust. Sam's, because it is based on far more trials. Relative frequency approaches the theoretical probability as the number of trials increases, so an estimate from $400$ drops is far more reliable than one from only $20$. Mia's small sample can swing a long way from the true value by chance.

Part (c) - expected point-up landings in $250$ drops. Use Sam's estimate as the probability and multiply by the number of new drops:

so about $102.5$, which to the nearest whole number is $103$ point-up landings. (Check: $0.41$ is close to $0.4$, and $0.4 \times 250 = 100$, so an answer just above $100$ is sensible.)