1. The probability scale and its vocabulary
Measuring likelihood
- Every probability is a number from 0 to 1, where 0 means impossible and 1 means certain
- Halfway, at 0.5, the outcome is as likely to happen as not
- Answers may be given as a fraction, a decimal or a percentage, and all three are accepted, but a ratio such as 7 : 20 or words such as "7 out of 20" can lose the mark
- Where the question does not specify, a fraction is usually safest, since it stays exact
- A probability outside 0 to 1 is always an error, so check any answer above 1 or below 0

The words that carry meaning
- An experiment is repeated to produce results, and each repeat is a trial
- An outcome is one possible result; an event may be a single outcome or several of them taken together
- The sample space collects every possible outcome, and is usually set out either as a list or in a table
- Something is fair when every outcome is equally likely, and biased when they are not
- Picking at random means every item has the same chance of being chosen
2. Calculating a probability
Equally likely outcomes
- With equally likely outcomes, count how many produce the event and divide that count by how many outcomes there are altogether
- Count the total carefully, since it is the denominator of everything that follows
- A bag holds 40 counters of which 14 are red, so the probability of drawing a red one is 14 ÷ 40, which simplifies to 7/20
- Simplifying is not required unless the question asks for the simplest form: the mark schemes accept any equivalent fraction, such as 14/40
- Information may come from a table, a graph or a Venn diagram rather than from a sentence
Probabilities that must add to 1
- Every outcome of an experiment together accounts for the whole of the probability, so they add to 1
- A table with one probability missing is completed by subtracting the others from 1
- The event not happening takes up whatever is left, so for an event A, the probability of A not happening is 1 − the probability of A happening, written P(not A) = 1 − P(A) for short
- Where a train has a 0.28 chance of being late, the chance of it not being late is 1 − 0.28 = 0.72
- Events that can never occur together are mutually exclusive, and for those the chance of one or the other is simply the sum of the two
- An event and its complement are always mutually exclusive, which is why they add to 1
Exam tip
Write the subtraction from 1
When a probability table has a gap, the subtraction from 1, such as 1 − (0.3 + 0.25), earns a method mark on its own, even if a later step slips, so write it as a line first. If two gaps are linked, as in "green is twice as likely as yellow", share the 0.45 that is left, not 1: green takes 2 parts and yellow 1, so yellow is 0.45 ÷ 3 = 0.15 and green 0.3.
Worked example
Completing a probability table
A bag holds beads that are red, blue, green or yellow. A bead is taken at random. The table shows the probabilities of red, blue and green. Find the probability of yellow, and the probability of taking a green or a yellow bead.
| Colour | red | blue | green | yellow |
|---|---|---|---|---|
| Probability | 0.35 | 0.2 | 0.15 |
Solution:
- The four probabilities together must come to 1
- The three given ones total 0.35 + 0.2 + 0.15 = 0.7
- So the probability of yellow is 1 − 0.7 = 0.3
- One bead cannot be both green and yellow, so those events are mutually exclusive and their probabilities add
- P(green or yellow) = 0.15 + 0.3 = 0.45
3. Listing the outcomes
Sample space diagrams
- Where two things happen together, such as two dice or a dice and a coin, a table lays out every combination
- Put one experiment's outcomes along the top and the other's down the side, then fill in the result of each pair
- The number of cells is the total number of outcomes, and counting the ones that satisfy the event gives the numerator
- Two ordinary dice give 6 × 6 = 36 outcomes, and a systematic table makes them impossible to miscount
- A total of 9 arises from (3, 6), (4, 5), (5, 4) and (6, 3), so its probability is 4 ÷ 36 = 1/9

- Even if the dice look identical, they are two separate dice, so (3, 6) and (6, 3) are different outcomes and both are counted
- Filling the whole table before counting is faster than trying to spot the successes directly

Worked example
Reading a probability off a sample space
Two ordinary dice are rolled and their scores added. Find the probability that the total is 10 or more.
Solution:
- Each die has 6 faces, so the sample space has 6 × 6 = 36 equally likely outcomes
- List the pairs giving a total of 10 or more, taking them in order so none is missed
- Total 10: (4, 6), (5, 5), (6, 4)
- Total 11: (5, 6), (6, 5)
- Total 12: (6, 6)
- That is 6 successful outcomes out of 36
- P(total ⩾ 10) = 6/36 = 1/6
- Even if the dice look identical, they are two separate dice, so (4, 6) and (6, 4) are two different outcomes, not one
4. Relative and expected frequency
Relative frequency
- Some probabilities cannot be worked out in advance, so they are estimated from an experiment instead
- Relative frequency = number of successes ÷ total number of trials
- It is only an estimate, and the more trials are run the closer it should get to the true probability
- Comparing a relative frequency with the theoretical probability is how a spinner or dice is judged fair or biased
- A fair six-sided dice should give a five about 1/6 of the time, roughly 0.167
- Rolling 300 times and getting 48 fives gives a relative frequency of 48 ÷ 300 = 0.16, close enough to suggest the dice is fair
- A relative frequency far from the theoretical value suggests bias
Expected frequency
- Going the other way, a known probability predicts how often something should happen in a given number of trials
- Expected frequency = probability × number of trials
- The answer is a prediction rather than a guarantee, so the actual number will usually differ a little
- A non-whole answer is acceptable, since an expected frequency is an average rather than a count
- With a probability of 0.15 over 400 trials, the expected frequency is 0.15 × 400 = 60
Worked example
Relative frequency and expected frequency together
A spinner has five equal sections numbered 1 to 5. In an experiment it is spun 200 times and lands on 4 on 32 occasions. Find the relative frequency of landing on 4, say whether the spinner appears to be fair, and find the expected number of 4s in 500 spins of a fair spinner.
Solution:
- Relative frequency = 32 ÷ 200 = 0.16
- For a fair five-section spinner the theoretical probability is 1 ÷ 5 = 0.2
- 0.16 is reasonably close to 0.2, so on this evidence the spinner appears roughly fair, though more trials would give a better test
- For a fair spinner, expected frequency = probability × number of trials
- Expected number of 4s = 0.2 × 500 = 100
- Had the question asked how many 4s to expect in 500 spins of this spinner, rather than a fair one, the relative frequency would be the probability to use: 0.16 × 500 = 80