Expected Frequencies
A quality controller knows from experience that $2\%$ of microchips fail testing. If $5000$ chips are produced today, how many are expected to fail? This question — about expected frequency over many trials — is the bridge between probability theory and real-world planning. The rule is wonderfully simple: expected count = total trials × event probability. This lesson sharpens the calculation and the interpretation.
You already know that for a single trial with success probability $p$, the expected proportion of successes is $p$. Without using any formula — if you run the same experiment $m$ times, how many successes do you expect on average? Write your reasoning below.
Every expected-frequency question rewards two habits: identify the single-trial probability $p$ and the number of trials $m$, then multiply: expected frequency $= m \times p$. Treat each scenario as a chain of independent Bernoulli trials.
The identify-then-multiply strategy: (1) state $p$, the probability of the event on one trial, (2) state $m$, the number of independent repetitions, (3) compute $E = m \times p$ and round only at the end if a whole-number answer is required.
For $X \sim B(n, p)$ run $m$ trials: expected count $= m \times p$
Key facts
- Expected frequency: $E = m \times p$ where $m$ = number of trials and $p$ = single-trial probability
- For $X \sim B(n, p)$, $E(X) = np$ is itself an expected frequency for a single sample
- The expected count is a theoretical average — actual counts will vary trial-to-trial
Concepts
- Why expected frequencies scale linearly with the number of trials
- Why expected does not mean "exactly" — it is a long-run mean
- How to recognise that a problem fits the binomial / expected-frequency setup
Skills
- Compute expected frequencies in production, sampling, and survey contexts
- Interpret deviation from expected as natural variability around the mean
- Compare expected with observed and judge whether the gap is reasonable
If a single trial has probability $p$ of producing a particular event, and the trial is repeated $m$ times independently, then the expected frequency of the event is
$E = m \times p$.
This is exactly $E(X)$ for $X \sim B(m, p)$. The expected frequency is the long-run average count; in any particular sample of $m$ trials the observed count will fluctuate around this value.
Worked through the hook: A fair die rolled $300$ times.
- (a) Probability of rolling a six on one roll: $p = \tfrac{1}{6}$. Expected frequency: $E = 300 \times \tfrac{1}{6} = 50$ sixes.
- (b) Probability of rolling an even number on one roll: $p = \tfrac{1}{2}$. Expected frequency: $E = 300 \times \tfrac{1}{2} = 150$ even rolls.
- Will you get exactly $50$ sixes? Probably not — but typically you'll be within a few of $50$. For $B(300, \tfrac{1}{6})$ the SD is $\sqrt{300 \times \tfrac{1}{6} \times \tfrac{5}{6}} \approx 6.45$, so $50 \pm 6$ is the typical range.
Expected frequency formula: $E = m\times p$ where $m$ is the number of trials and $p$ is the probability of the event on each trial.
Pause — copy the expected frequency formula $E=mp$ and work through the hook example (die rolled 300 times, expected frequency of each face $= 300\times\frac{1}{6}=50$) into your book.
Quick check: A factory produces components, $3\%$ of which are defective. In a batch of $1500$ components, what is the expected number of defectives?
We just saw that if a single trial has probability $p$ and is repeated $m$ times, the expected frequency is $E=mp$. That raises a question: when actual observed counts differ from expected, how do you decide if the gap is "within normal variability" or genuinely suspicious? This card answers it → gaps up to about $1$–$2$ standard deviations from the expected value are typical; gaps beyond $2$ standard deviations warrant further investigation.
The expected frequency tells you what to predict; the observed frequency is what actually happens. The gap between them is natural variability. A useful rule of thumb: differences of up to about $1$–$2$ standard deviations from the expected value are common, while gaps larger than $3$ SDs are unusual enough to investigate.
Example: A coin is tossed $400$ times. Expected number of heads $= 400 \times 0.5 = 200$.
- Variance: $npq = 400 \times 0.5 \times 0.5 = 100$, so SD $= 10$.
- Observing $195$ heads (deviation $5$, that is $0.5$ SDs) — completely typical.
- Observing $230$ heads (deviation $30$, that is $3$ SDs) — unusual; the coin may not be fair.
So expected frequencies are not just calculations — they give you a baseline to compare real-world data against.
The expected frequency tells you what to predict; the observed frequency is what actually happens. The gap between them is natural variability. A useful rule of thumb: differences of up to about $1$–$2$ standard deviations from the expected value are...
Pause — copy the rule of thumb: observed frequency within $\pm 2$ SD of expected is normal variation; a larger gap suggests the model may not hold into your book.
Did you get this? True or false: in $1000$ tosses of a fair coin, you should expect to see exactly $500$ heads every time.
Worked examples · 3 in a row, reveal as you go
In a manufacturing process, $2\%$ of microchips are defective. A batch of $5000$ microchips is produced. (a) Find the expected number of defective chips. (b) If $130$ defective chips are actually found, comment on whether this is unusual.
A survey finds that $37\%$ of voters in an electorate support Party A. If $250$ voters are polled at random, what is the expected number who support Party A? Why might the actual number be a whole number near $93$?
A pharmacist knows that $15\%$ of customers ask for a generic substitute. How many customers must be served, on average, so that the expected number of generic-substitute requests reaches $30$?
Fill the gap: A spinner lands on red with probability $0.4$. If the spinner is spun $250$ times, the expected number of reds is $250 \times 0.4 = $.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: if $4\%$ of items are defective, then in $250$ items the expected number of defectives is $250 \times 4 = 1000$.
Activities · practice with the ideas
A fair die is rolled $180$ times. Find the expected number of (a) ones and (b) numbers greater than $4$.
A biased coin lands heads with probability $0.7$. If it is tossed $400$ times, find the expected number of heads and the standard deviation.
A factory produces $8000$ light bulbs per day. From experience, $1.25\%$ are defective. How many defective bulbs are expected per day?
A survey reports that $22\%$ of households own two cars. If $450$ households are surveyed, how many are expected to own two cars? Briefly justify why the actual count might differ.
In a quality test, the expected number of faulty items in a sample of $m$ is $24$. The single-item fault rate is $0.06$. Find $m$.
Odd one out: Three of these statements about expected frequencies are correct. Which one is NOT?
Earlier you predicted expected counts for $300$ die rolls and asked whether the actual count would land exactly on the expected value.
Expected sixes $= 50$, expected evens $= 150$. But in any single run of $300$ rolls, the observed sixes will scatter around $50$ with SD $\approx 6.45$, so seeing $43$ or $57$ sixes is unremarkable. "Expected" is a theoretical mean, not a guarantee. The strength of the formula $E = m \times p$ is that it gives a baseline against which real data can be compared.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. A die is rolled $600$ times. Find the expected number of times a $5$ or $6$ appears. (2 marks)
Q2. A factory produces components. The probability that a randomly chosen component is faulty is $0.025$. In a daily batch of $2000$ components, find the expected number of faulty components and the standard deviation. (3 marks)
Q3. A quality engineer expects $3\%$ of bottles to be cracked. In a sample of $500$ bottles, $25$ are found to be cracked. (a) Calculate the expected number of cracked bottles. (b) Calculate the standard deviation. (c) Decide, with reasoning, whether the observed count is unusual. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. (a) $p = 1/6$, $E = 180 \times \tfrac{1}{6} = 30$ ones. (b) Numbers greater than $4$ are $5$ and $6$, so $p = 2/6 = 1/3$; $E = 180 \times \tfrac{1}{3} = 60$.
2. $E = 400 \times 0.7 = 280$ heads. $\text{SD} = \sqrt{400 \times 0.7 \times 0.3} = \sqrt{84} \approx 9.165$.
3. $E = 8000 \times 0.0125 = 100$ defective bulbs per day.
4. $E = 450 \times 0.22 = 99$ households. The actual count varies because the survey is a random sample; SD $= \sqrt{450 \times 0.22 \times 0.78} \approx 8.79$, so typical observed counts fall roughly in $90$–$108$.
5. $24 = m \times 0.06 \Rightarrow m = 24/0.06 = 400$.
Q1 (2 marks): $p = \tfrac{2}{6} = \tfrac{1}{3}$ [1]. $E = 600 \times \tfrac{1}{3} = 200$ times [1].
Q2 (3 marks): $E = 2000 \times 0.025 = 50$ faulty components [1]. $\text{Var} = 2000 \times 0.025 \times 0.975 = 48.75$ [1]. $\text{SD} = \sqrt{48.75} \approx 6.982$ [1].
Q3 (3 marks): (a) $E = 500 \times 0.03 = 15$ cracked bottles [1]. (b) $\text{SD} = \sqrt{500 \times 0.03 \times 0.97} = \sqrt{14.55} \approx 3.814$ [1]. (c) Deviation $= 25 - 15 = 10$, which is $10/3.814 \approx 2.62$ SDs above expected — moderately unusual (between $2$ and $3$ SDs); worth investigating, but not extreme [1].
Five timed questions on expected frequencies and comparing observed counts to expected. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering expected frequency questions. Lighter alternative to the boss.
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