1The idea in one paragraph

Probability measures how likely an event is on a scale from 0 (impossible) to 1 (certain). There are two ways to get the number. If you can list every possible result and they are all equally likely, count: the probability is the number of results you want divided by the number there are. If you cannot, experiment: repeat the thing many times and use the fraction of times the event happened, its relative frequency, as your estimate. The more trials, the better the estimate. The probability that something does not happen is 1 minus the probability that it does. And once you have a probability, multiplying it by the number of trials tells you how many times to expect the event: not a promise, but a long-run average.

2The vocabulary: trial, outcome, sample space, event

Every probability question uses the same six words, and the marks depend on using them precisely. Figure 1 shows four of them for one roll of a die.

• A trial is one performance of an experiment whose result is uncertain: one roll of a die, one toss of a coin, one patient given a test, one day's weather.
• An outcome is one possible result of a trial. Rolling a die has six outcomes: 1, 2, 3, 4, 5, 6.
• The sample space, written U, is the set of all possible outcomes, each listed once. For the die, U = {1, 2, 3, 4, 5, 6}, so n(U) = 6.
• An event is any collection of outcomes, a subset of U. "An even score" is the event A = {2, 4, 6}. An event can be a single outcome ("a 5"), or even every outcome ("a score below 7", which is certain).
• Outcomes are equally likely when each has the same chance of happening. A fair die makes the six scores equally likely. A loaded die does not.
• The relative frequency of an event is the number of times it happened divided by the number of trials. Section 5 is about it.

Say the event in words and in set notation. "The score is prime" is B = {2, 3, 5}. Writing it out takes five seconds and removes arguments with yourself about whether 1 counts (it does not; 1 is not prime).

Representing a sample space. The guide expects you to use whatever form fits the trial.

• A list works for one simple trial: U = {H, T} for a coin, or the 20 numbers on 20 numbered counters.
• A table or grid, the sample space diagram, works for two trials together. Put the outcomes of the first along the top, the second down the side, and fill each cell with the combined result. Figure 2 does this for two dice, with the total in each cell. There are 6 × 6 = 36 cells, and because each die is fair and the dice do not affect each other, the 36 cells are equally likely.

A two-way list such as {HH, HT, TH, TT} for two coins is a sample space too. Notice HT and TH are different outcomes: the first coin shows heads and the second tails, or the other way round. Treating "one head and one tail" as a single outcome is the error that section 3 warns against.

3Probability by counting

When all the outcomes in U are equally likely, the probability of an event A is the fraction of those outcomes that belong to A.

P(A) = n(A) ÷ n(U), valid only when every outcome in U is equally likely. Every probability satisfies 0 ≤ P(A) ≤ 1.

A bag holds 20 counters numbered 1 to 20, and one is taken at random. "At random" means every counter is equally likely to be chosen, so the formula applies.

With two dice, read the answer off Figure 2. In panel (a) the total is 7 in six cells, so P(total is 7) = 6/36 = 1/6. For "total is at least 10", count the cells showing 10, 11 or 12: 3 + 2 + 1 = 6, so the probability is again 6/36 = 1/6. Only one cell shows 12, so P(total is 12) = 1/36.

The equally-likely trap. The possible totals of two dice are 2, 3, …, 12, which is eleven values. It is tempting to say P(total is 7) = 1/11. That is wrong, because the eleven totals are not equally likely: a total of 7 can happen six ways and a total of 2 only one way. The formula counts equally likely outcomes, so count cells in the grid, not values in a list of totals. Whenever you write n(A) ÷ n(U), ask whether the things you are counting really do have the same chance.

Figure 3 places these probabilities on the probability scale. An impossible event (a total of 13) has probability 0; a certain event (a total of at least 2) has probability 1. An answer outside 0 to 1 is a signal to find your mistake, not an answer.

On Paper 1 give probabilities as exact fractions, simplified if you can: 6/36 is correct, 1/6 is neater, and either usually earns the mark. On Paper 2 decimals to 3 significant figures are accepted: 0.167.

4Complementary events

The complement of A, written A′ (read "not A"), is the event that A does not happen: every outcome in U that is not in A. A and A′ cannot both happen, and one of them must happen, so their probabilities add to 1. Figure 4 draws this: A is a region inside the rectangle U, and A′ is everything else in the rectangle.

P(A) + P(A′) = 1, so P(A′) = 1 − P(A).

This is more than a tidy fact. It is often the fastest route to an answer, because the complement of an awkward event is frequently simple. The classic case is "at least one".

Two dice are rolled. Find the probability that at least one of them shows a six.

"At least one six" means one six or two sixes, which is several cases. Its complement, "no six at all", is one clean case: each die shows 1 to 5, so there are 5 × 5 = 25 cells with no six.

Check against panel (b) of Figure 2: 6 cells in the last column plus 6 in the last row, minus the (6, 6) cell counted twice, gives 11. The same answer both ways. The counting route works here because the grid is small; the complement route still works when the grid is far too big to draw, which is why it is worth learning now.

The complement is also how you check work. If a question asks for P(A) and P(A′) separately, they must add to exactly 1.

5Relative frequency: probability from experiment

Some probabilities cannot be found by counting, because the outcomes are not equally likely and there is no symmetry to lean on. What is the probability that a plastic bottle cap lands top-down when tossed? The two outcomes, top-down and top-up, are not equally likely; the shape decides. You have to toss it and see.

The relative frequency of an event is

relative frequency = number of times the event happened ÷ number of trials

and it is used as an estimate of the probability. This estimate is called an experimental probability. A probability found by counting equally likely outcomes, as in section 3, is a theoretical probability.

A student tosses a bottle cap 250 times and it lands top-down 165 times.

How good is the estimate? Figure 5 answers this with a simulation: a computer rolled a fair die 1000 times and, after each roll, recorded the fraction of rolls so far that were sixes.

Three things are worth reading off it. After 10 rolls there had been no sixes at all, so the relative frequency was 0, wildly wrong. After 100 rolls it was 0.14. After 1000 rolls it was 157/1000 = 0.157, close to the true 1/6 ≈ 0.167 but still not equal to it. So:

• A relative frequency from a few trials is unreliable. Its value depends heavily on luck.
• As the number of trials grows, the relative frequency tends to settle close to the true probability. More trials give a better estimate.
• It never becomes a guarantee. Even after many trials the relative frequency will usually differ a little from the probability.

This is how the two kinds of probability are linked. A theoretical probability predicts what the relative frequency will approach in the long run; a relative frequency from many trials estimates the probability when no theory is available. Insurers price policies this way, using the recorded frequency of past claims, and it is why a simulation, run thousands of times on a computer, can estimate a probability that nobody knows how to calculate.

Is it fair? Comparing the two tells you whether a theoretical model fits. If a spinner with five equal sectors is fair, each number has probability 1/5 = 0.2. Spun 500 times, a relative frequency of 0.236 for one number is higher than 0.2, but chance alone often produces a gap like that, so it is not proof of bias. A relative frequency of 0.4 after 500 spins would be strong evidence. At SL you are asked to comment, not to test formally: say what the model predicts, what the data show, and whether the gap is large compared with the number of trials.

6Expected number of occurrences

If an event has probability p and you carry out n independent trials, you expect it to happen about

expected number of occurrences = n × p

times. This follows from section 5 read backwards: in the long run the relative frequency is close to p, so the number of occurrences is close to n × p.

Roll two dice 180 times. P(total is 7) = 1/6, so you expect a total of 7 about 180 × 1/6 = 30 times.

Take the bottle cap. Using the estimate from section 5, in 600 more tosses you expect about 600 × 0.66 = 396 top-down landings. The answer inherits the uncertainty of the estimate: it is only as good as the 250 tosses behind it.

The answer need not be a whole number. A school bus is late on any given day with probability 0.15. Over a 22-day month, the expected number of late days is 22 × 0.15 = 3.3. You cannot have 3.3 late days in one month, and nobody claims you will. It means that over many months the average number of late days per month is 3.3. Do not round an expected number to a whole number unless the question tells you to; 3.3 is the answer.

An application. An insurer has 3500 policies of one kind, and past records suggest that each policy leads to a claim in a year with probability 0.02 (invented figures). The expected number of claims in a year is 3500 × 0.02 = 70. If an average claim costs €4000, the insurer should expect to pay about 70 × €4000 = €280,000 and must charge enough, spread over 3500 customers, to cover it: at least €80 each, before costs. Expected numbers are how insurance, staffing and stock levels get planned.

In 4.7 this idea becomes the expected value of a random variable, and in 4.8 it becomes the mean np of the binomial distribution. Same formula, same meaning: the long-run average.