If Heads has relative frequency 0.35, what is the expected number of Heads in 200 flips?

70. The expected frequency is the relative frequency times the number of trials: 0.35 × 200 = 70. The common wrong answer 'about 50' rounds 'estimate' to a tidy number — but 50 out of 200 would be a relative frequency of 0.25, not the 0.35 you were given. 'Expected' is a calculation, not a round-off.

How do you find the best estimate of a probability from experimental data?

Take the relative frequency from the experiment with the most trials. If a coin's largest run is 126 Heads in 500 flips, the best estimate of P(Heads) is 126 ÷ 500 = 0.252 — not 'about a quarter'. The largest sample gives the most reliable estimate, by the law of large numbers.

Aisha got 50 heads in 100 spins; Ben got 380 in 1000. Whose estimate is more reliable?

Ben's. His relative frequency is 380 ÷ 1000 = 0.38; Aisha's is 50 ÷ 100 = 0.50. Even though Aisha's relative frequency is the bigger number, Ben ran ten times as many trials, so his estimate is closer to the true probability. Reliability comes from the number of trials, not the size of the relative frequency.

GCSE Maths Foundation · AQA · Probability

Relative frequency and expected outcomes: why 'estimate' is a calculation, not a round number

Relative-frequency questions trip students who read “estimate” or “expected” as a cue to give a tidy number. Told Heads has relative frequency 0.35 over 200 flips, they answer “about 50” instead of $0.35 \times 200 = 70$ . And asked whose estimate of a probability is more reliable, they pick the bigger relative frequency rather than the experiment with the most trials.

The thirty-second fix: expected frequency = relative frequency × number of trials, and the best estimate of a probability comes from the most trials — never round “estimate” to a tidy number, and never trust the bigger relative frequency over the larger sample.

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How to spot it in your own work

You read “expected” as “about a tidy number” — writing $\text{about } 50$ instead of $0.35 \times 200 = 70$ .
You answered “about ¼” for a best estimate instead of the relative frequency from the largest run, $126 \div 500 = 0.252$ .
You found an expected count by subtracting or dividing — e.g. $125 - 80 = 45$ — instead of $250 \times 0.32 = 80$ .
You chose the experiment with the bigger relative frequency as more reliable, instead of the one with more trials (Ben’s 1000 over Aisha’s 100).

An exam question that triggers it

Here is the canonical AQA Foundation trigger (Nov24 P2 Q20a shape, calculator):

A coin is flipped many times. Heads comes up with relative frequency 0.35. In 200 flips, work out the expected number of Heads.

The misconception answer is $\text{about } 50$ — reading “expected” as an instruction to round. The examiner’s report on the related question noted many students “read the word ‘estimate’ and thought the answer was about a quarter”.

But expected frequency is relative frequency × trials: $0.35 \times 200 = 70$ .

Why students fall for this

A relative frequency is a share — the fraction of trials in which the outcome happened. To turn a share of 200 flips into a count, you multiply: $0.35 \times 200$ . But the words “estimate” and “expected” sound vague, so the instinct is to offer a rough, tidy figure rather than to calculate.

The reliability half is a different intuition: that a bigger relative frequency is a “stronger” result. In fact reliability comes from sample size. By the law of large numbers, the more trials you run, the closer the relative frequency settles to the true probability — so an estimate from 1000 trials beats one from 100, whatever the two relative frequencies happen to be.

AQA Foundation calculator papers exploit every face of this: expected frequency as relative frequency × trials (Nov24 P2 Q20a), the best estimate as the relative frequency from the largest run (Jun24 P2 Q24), and which experiment to trust (Nov24 P2 Q20b, Jun22 P3 Q20).

The fix: Expected frequency = relative frequency × trials; the best estimate comes from the most trials

Expected frequency is a multiplication. For Heads at relative frequency 0.35 over 200 flips, $0.35 \times 200 = 70$ . The round number “about 50” would mean a relative frequency of 0.25, not the 0.35 you were given.

The best estimate is the relative frequency from the largest run. 126 Heads in 500 flips gives $126 \div 500 = 0.252$ — the estimate, not “about a quarter”.

More trials means more reliable. Aisha’s $50 \div 100 = 0.50$ from 100 spins is beaten by Ben’s $380 \div 1000 = 0.38$ from 1000 spins. The larger sample is the better estimate, even though 0.50 is the bigger number.

Worked example

A spinner has relative frequency of green 0.32. Work out the expected number of greens in 250 spins.

Read the relative frequency as a share. 0.32 of every spin lands green.
Multiply by the number of trials.
$\text{expected greens} = 0.32 \times 250 = 80$
Sense-check. 80 out of 250 is $80 \div 250 = 0.32$ — back to the relative frequency, so it fits.

The trap answer $125 - 80 = 45$ subtracts two figures that have nothing to do with an expected count, and $0.35 - 0.32 = 0.03$ or $125 \div 0.32 \approx 391$ are just as wrong. Expected frequency is always relative frequency × number of trials.

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Common questions

Does “estimate” mean I should round to a tidy number?: No. “Estimate” means calculate the relative frequency. The best estimate of P(Heads) from a run of 126 Heads in 500 flips is $126 \div 500 = 0.252$ , not “about ¼”. It may land near a tidy fraction, but the answer is the actual relative frequency.
How do I work out an expected frequency?: Multiply the relative frequency by the number of trials. For Heads at relative frequency 0.35 over 200 flips, $0.35 \times 200 = 70$ . Never find it by subtracting or dividing two figures.
Which experiment gives the more reliable estimate of a probability?: The one with the most trials. Ben’s 380 Heads in 1000 spins ( $0.38$ ) is more reliable than Aisha’s 50 in 100 ( $0.50$ ), because a larger sample is closer to the true probability. The size of the relative frequency does not decide reliability.
How can I check an expected-frequency answer?: Divide your count back by the number of trials and you should recover the relative frequency. If 80 greens are expected in 250 spins, then $80 \div 250 = 0.32$ — the relative frequency you started with, so the answer is right.

Related misconceptions

Combined events: why P(A and B) multiplies, not addsThe neighbouring probability misconception — combining the chances of two events, where 'and' multiplies rather than adds.
Reading Venn diagrams: why a region is not the whole setThe other probability vertical — reading a probability off a diagram, where the denominator must be the whole set, not a region.

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