Calculate the expected frequency of an event from the probability of the event and the number of trials, using expected frequency = P(E) x number of trials

What this dot point is asking

NESA wants you to predict how many times an event will happen when a chance experiment is repeated many times. You already know how to find the probability of an event as a single number between $0$ and $1$. Expected frequency turns that probability into a count: if you know the chance of an event on one trial, and you know how many trials you will run, you can predict how many of those trials will give the event. The arithmetic is a single multiplication. The marks are won by choosing the right probability, multiplying by the right number of trials, and knowing that the answer is a prediction the real data will scatter around, not an exact promise. You also need to run the rule backwards: given an expected count, find how many trials produce it.

The answer

Expected frequency is the probability of an event multiplied by the number of trials. If an event $E$ has probability $P(E)$ and the experiment is repeated $n$ times, then

That is the whole rule. A fair die has $P(6) = \tfrac{1}{6}$, so in $600$ rolls you expect $\tfrac{1}{6} \times 600 = 100$ sixes. The idea is intuitive: if something happens a sixth of the time, then over $600$ goes it should happen about a sixth of $600$ times. The probability is the "rate" and the number of trials is "how many", and multiplying a rate by a count gives a total.

Two things make the topic worth a section rather than a single line. First, the answer is an expectation, not a guarantee. Roll a die $600$ times and you will rarely land on exactly $100$ sixes; you will get a number near $100$. The chart below shows this: each face of a fair die rolled $60$ times has an expected frequency of $10$, but the observed counts wobble above and below that line. Second, exam questions often hide the multiplication inside wording, or run it backwards by giving you the expected count and asking for the number of trials.

The expected-frequency rule

Every expected-frequency question reduces to filling in two slots in the same formula: the probability and the number of trials. Write the probability of the event as a fraction or decimal, write down how many trials there are, and multiply. There is no extra step. For a fair die rolled $600$ times the expected number of sixes is

and for an event with probability $0.04$ run $2500$ times the expected frequency is

The probability can come from three places, and the wording tells you which:

• Theoretical, from equally likely outcomes: a fair die gives $P(6) = \tfrac{1}{6}$, a deck gives $P(\text{heart}) = \tfrac{13}{52} = \tfrac{1}{4}$.
• Given, stated directly in the question: "the probability a globe is defective is $0.02$".
• Experimental (a relative frequency), estimated from a trial: "$24$ greens in $150$ spins" gives $P(\text{green}) \approx \tfrac{24}{150} = 0.16$.

Whichever way you get $P(E)$, the rest is the same multiplication.

Expected versus observed frequency

Expected frequency is a prediction. The observed frequency is what actually happened when you ran the experiment. They are almost never identical, and that is normal. Roll a die $60$ times and each face is expected $\tfrac{1}{6} \times 60 = 10$ times, but a real run might give $12$, $9$, $11$, $7$, $13$ and $8$ - all close to $10$, none exactly $10$, as in the chart above. The expected value is the centre that the observed counts scatter around.

Two consequences matter for the exam. First, you should usually round an expected frequency sensibly, often to a whole number, because you cannot have $49.7$ defective bottles in practice; state it as "about $50$". Second, the more trials you run, the closer the observed relative frequency tends to sit to the true probability, so a prediction made from more trials, or compared against more data, is more reliable. A small gap between expected and observed is expected; a large, persistent gap is a hint the assumed probability is wrong (a biased die, a faulty machine).

Working backwards to find the number of trials

The same rule answers a different question: given the expected count, how many trials produce it? Start from

and make $n$ the subject by dividing both sides by $P(E)$:

If a process produces faulty items with probability $0.015$ and you want the run size that is expected to contain $45$ faulty items, then

So a run of $3000$ items is expected to contain $45$ faulty ones. The danger here is doing the wrong operation: backwards problems divide the expected count by the probability, they do not multiply. A quick forward check ($0.015 \times 3000 = 45$) confirms it.

How exam questions ask about expected frequency

The wording varies, but each version points to the same multiplication (or, for the last one, its reverse):

• "How many ... would you expect?" or "Calculate the expected number of ..." is the plain rule: find $P(E)$, multiply by the number of trials $n$.
• "In $n$ trials / rolls / spins, how many times ...?" names the number of trials directly; multiply it by the probability.
• "... probability is $p$. In a batch of $n$ ..., how many ...?" hands you both numbers; the answer is $p \times n$.
• "Estimate the probability, then find the expected number ..." is a two-parter: get $P(E)$ as a relative frequency from the trial first, then multiply by the new $n$.
• "For how many trials would you expect ... [a given count]?" or "Find the run size that gives ... [count]" is the backwards version: divide the expected count by $P(E)$.

The verbs map straight to the method: "expect", "predict" and "should occur" all mean apply $P(E) \times n$; "for how many" or "what number of trials" means rearrange and divide.