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How do you measure and combine the chances of events?

The probability scale, theoretical and experimental probability, relative frequency, expected frequency, and the addition and multiplication rules.

A focused answer to AQA GCSE Statistics on probability basics, covering the probability scale, theoretical and experimental probability, relative frequency, expected frequency, and the addition and multiplication rules for events.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The probability scale
  3. Theoretical and experimental probability
  4. Expected frequency
  5. The addition and multiplication rules

What this dot point is asking

AQA wants you to use the probability scale, distinguish theoretical and experimental probability, use relative frequency and expected frequency, and apply the addition and multiplication rules to combine events. These ideas are the foundation for tree diagrams, Venn diagrams and probability distributions later in the module.

The probability scale

Because the total is always 11, the probability of an event not happening is 11 minus the probability that it does: P(not A)=1P(A)P(\text{not } A) = 1 - P(A). This complement rule is the quickest route to many marks, for example finding "at least one" by subtracting the probability of "none".

Theoretical and experimental probability

Theoretical probability applies when outcomes are equally likely (fair coins, dice, spinners). Experimental probability is used when they may not be, or when the mechanism is unknown, for example a possibly biased spinner or the chance of a bus arriving late. The law of large numbers says that as the number of trials grows, the relative frequency settles toward the true probability, which is why examiners reward the comment that "more trials give a more reliable estimate".

Expected frequency

Expected frequency converts a probability into a predicted count for a real number of trials. It is a long-run average, so the actual count varies around it: rolling that die 6060 times will rarely give exactly 1010 sixes, but repeated experiments will average close to 1010.

The addition and multiplication rules

"Mutually exclusive" means the events cannot both happen at once (rolling a 11 or a 22 on one die). "Independent" means one event does not change the probability of the other (a coin and a die). When events can overlap, the general addition rule subtracts the overlap: P(A or B)=P(A)+P(B)P(A and B)P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B), which prevents double-counting outcomes in both events. Keeping these two ideas separate is essential: mutually exclusive is about whether events can co-occur (used for "or"), while independence is about whether one affects the other (used for "and"). A pair of events can be mutually exclusive but not independent, so do not assume one from the other.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20184 marksA spinner has four colours. A student spins it 200200 times and records red 4646 times, blue 5454 times, green 5252 times and yellow 4848 times. (a) Calculate the relative frequency of green. (b) The maker claims the spinner is fair. Use the results to comment on this claim.
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(a) Relative frequency of green =52200=0.26= \frac{52}{200} = 0.26.

(b) A fair four-colour spinner has theoretical probability 14=0.25\frac{1}{4} = 0.25 for each colour. All four relative frequencies (0.23,0.27,0.26,0.240.23, 0.27, 0.26, 0.24) are very close to 0.250.25, so the results support the claim that the spinner is fair.

Markers reward the relative frequency calculation, comparison with 0.250.25, and a contextual conclusion (the closeness, plus that more spins would give stronger evidence).

AQA 20203 marksA fair six-sided die is rolled 9090 times. (a) Calculate the expected number of times an even number is rolled. (b) State why the actual number of even rolls may differ from this.
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(a) P(even)=36=12P(\text{even}) = \frac{3}{6} = \frac{1}{2}. Expected frequency =12×90=45= \frac{1}{2} \times 90 = 45.

(b) Expected frequency is a long-run prediction, not a guarantee; random variation means the actual number will usually be near 4545 but not exactly 4545.

Markers reward P(even)=12P(\text{even}) = \frac{1}{2}, the product 12×90=45\frac{1}{2} \times 90 = 45, and a clear statement that it is an expectation subject to chance variation.

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