This mini-lesson covers the whole Probability strand of Edexcel GCSE Maths (1MA1): the probability scale, sample space, mutually exclusive and independent events, tree and Venn diagrams, relative frequency, and the Higher-tier topic of conditional probability.
Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Press Start when you're ready.
The probability of an event is a measure of how likely it is, written as a fraction, decimal or percentage. Every probability must lie between 0 (impossible) and 1 (certain):
For equally likely outcomes, use the basic rule:
A fair six-sided die is rolled. Find P(rolling a 4).
There is 1 favourable outcome (the 4) out of 6 possible outcomes.
P(4) = 1/6
Common slip: a probability can never be more than 1 or less than 0. An answer of 1.4 or −0.2 means you've made a mistake.
The sample space is the list of all possible outcomes of an experiment. A sample space diagram is a grid that lays them out for two events at once.
For two events, the total number of outcomes is (outcomes of first) × (outcomes of second). Flipping a fair coin twice gives 2 × 2 = 4 outcomes:
Careful: HT and TH are different outcomes. There are 4 outcomes, not 3 — don't merge "one head, one tail" into a single case when listing the sample space.
When outcomes are equally likely, count the favourable ones and divide by the total. Probabilities are often left as fractions, but you may be asked for a decimal or percentage.
A bag holds 3 red and 7 blue counters. One is taken at random. Find P(red).
Favourable = 3 red. Total = 3 + 7 = 10 counters.
P(red) = 3/10 = 0.3
Always use the total: the denominator is the whole collection (10 counters here), not just the counters of the other colour.
Two events are mutually exclusive if they cannot happen at the same time (like rolling a 2 and rolling a 5 on one roll). For mutually exclusive events you add the probabilities:
Because exhaustive outcomes cover everything, all their probabilities must add up to 1. This gives the "not" rule:
The probability it rains tomorrow is P(rain) = 0.35. Find P(no rain).
P(no rain) = 1 − 0.35 = 0.65
Don't add for everything: you may only add probabilities when the events are mutually exclusive. If both can happen at once, adding double-counts the overlap.
Two events are independent if one happening does not change the probability of the other (like two separate coin flips). For independent events you multiply:
P(A) = 0.5 and P(B) = 0.2, and A and B are independent. Find P(A and B).
P(A and B) = 0.5 × 0.2 = 0.1
And vs or: "and" (both happen, independent) means multiply; "or" (either happens, mutually exclusive) means add. Mixing these up is the classic exam error.
A tree diagram shows a sequence of events. Multiply along the branches for a combined outcome, and add the separate final outcomes you want. Each set of branches from a point must sum to 1.
Multiply along, add down: multiply probabilities along a path, then add the probabilities of the different paths that satisfy the question.
All events use one roll of a fair six-sided die. Tap an event on the left, then its matching probability on the right.
A Venn diagram sorts a set of items into overlapping groups. The overlap is "A and B" (both), the whole of both circles is "A or B", and anything outside is "neither".
Read the regions carefully: the "12" means French only, not the whole French circle. The whole French circle is 12 + 5 = 17.
When outcomes are not equally likely (a biased spinner, a drawing pin), you estimate the probability from experiments. This is the relative frequency:
A spinner landed on red 12 times in 40 spins. Estimate P(red).
relative frequency = 12 ÷ 40 = 0.3
Expected number: to predict how many times an event happens in N trials, work out P(event) × N. Here 0.3 × 100 spins ≈ 30 reds.
A probability must be between 0 and 1 inclusive. Tap a value, then tap the box it belongs in.
A conditional probability is the probability of an event given that another has already happened. It matters most when objects are taken out and not replaced — the totals change on the second pick.
A bag has 3 red and 7 blue counters. Two are taken without replacement. Find P(red then red).
First red: 3/10. Now 2 red and 7 blue remain (9 total).
Second red given first red: 2/9.
P(red, red) = 3/10 × 2/9 = 6/90 = 1/15
Without replacement changes both numbers: after taking one red, both the reds and the total drop by one (3/10 then 2/9), so the second probability is not the same as the first.
Probability scale: every probability is between 0 (impossible) and 1 (certain).
Basic rule: P(event) = favourable ÷ total, for equally likely outcomes.
Sample space: list every outcome; two coins give 4 outcomes (HH, HT, TH, TT).
Mutually exclusive: P(A or B) = P(A) + P(B); and P(not A) = 1 − P(A).
Independent: P(A and B) = P(A) × P(B) — "and" means multiply.
Tree diagrams: multiply along branches, add the paths you want.
Venn diagrams: overlap = "and", both circles = "or".
Relative frequency: successes ÷ trials estimates probability from data.
Higher only: conditional probability, especially "without replacement".
You've covered every part of the Edexcel 1MA1 Probability strand — Foundation content plus the Higher-tier conditional probability. Press Finish to see your score.
You've worked through Probability for Edexcel GCSE Maths (1MA1). 🎉
Your stars: 0 / 0
Next: test yourself in the Evaluate stage Confidence Quiz, then lock it in with Verify.