Revision notes for Edexcel IGCSE Maths Probability. Open the guide for explanations and worked examples. Written against the Edexcel IGCSE Maths (4MA1) specification, so the content matches what's examinable rather than general Maths background.
Revision notes for Edexcel IGCSE Maths Probability. Open the guide for explanations and worked examples. Written against the Edexcel IGCSE Maths (4MA1) specification, so the content matches what's examinable rather than general Maths background.
Probability is a way of measuring how likely something is to happen.
An outcome is one possible result, such as landing on red, choosing a blue pen, or rolling a 6. An event is the result we are interested in, such as “the spinner lands on green”.
Probability
A probability is a number between 0 and 1. A probability of 0 means impossible, and a probability of 1 means certain.
A biased spinner, die, or coin is one where the outcomes are not equally likely. That is fine: you just use the probabilities you are given.
The scale below is a useful mental picture for judging whether an event is unlikely, even chance, likely, impossible, or certain.

Reading a probability
A seed has probability 0.82 of growing. What does this tell you?
Check that 0.82 is between 0 and 1, so it is a valid probability.
Since 0.82 is greater than 0.5, the seed growing is more likely than not growing.
Since 0.82 is not 1, it is not guaranteed. Some seeds may still fail to grow.
Often you are told that there are only certain outcomes, such as only red, blue and white counters.
Exhaustive outcomes
Outcomes are exhaustive when the list includes every possible result. If exactly one of the outcomes happens each time, their probabilities add to 1.
Missing probability
If a table lists all possible outcomes, find a missing probability by subtracting the known probabilities from 1.
Completing a probability table
A bag contains only red, blue and white counters. The probability of red is 0.46 and the probability of blue is 0.27. Find the probability of white.
The outcomes are only red, blue and white, so their probabilities must add to 1.
Add the known probabilities:
0.46+0.27=0.730.46 + 0.27 = 0.730.46+0.27=0.73Subtract from 1 to find the missing probability:
1−0.73=0.271 - 0.73 = 0.271−0.73=0.27The probability of choosing a white counter is 0.27.
The same idea works with fractions. If one probability is 14\frac{1}{4}41 and another is 25\frac{2}{5}52, use a common denominator before subtracting from 1.
Forgetting the total
Do not make the missing probability equal to one of the numbers already in the table unless the question tells you they are the same. First find what is left out of 1.
Sometimes two missing probabilities are the same. In that case, first find the total amount left, then split it equally.
Two missing probabilities are equal
A spinner can land on red, blue, yellow or green. The probability of red is 0.18 and the probability of yellow is 0.30. The probabilities of blue and green are equal. Find them.
Add the probabilities you already know:
0.18+0.30=0.480.18 + 0.30 = 0.480.18+0.30=0.48Subtract from 1 to find the total probability left for blue and green:
1−0.48=0.521 - 0.48 = 0.521−0.48=0.52Blue and green are equal, so divide the remaining probability by 2:
0.52÷2=0.260.52 \div 2 = 0.260.52÷2=0.26The probability of blue is 0.26 and the probability of green is 0.26.
Words like twice and three times are really ratio clues.
Use parts
If one probability is twice another, think of 3 equal parts altogether: 2 parts for the bigger probability and 1 part for the smaller probability.
Using a probability relationship
A spinner can land on 1, 2, 3 or 4. The probability of landing on 2 is 0.30 and the probability of landing on 4 is 0.25. Landing on 1 is twice as likely as landing on 3. Find the probabilities of 1 and 3.
First find how much probability is already used:
0.30+0.25=0.550.30 + 0.25 = 0.550.30+0.25=0.55Subtract from 1 to find the probability left for 1 and 3:
1−0.55=0.451 - 0.55 = 0.451−0.55=0.45Since 1 is twice as likely as 3, use the ratio 1 : 3 as 2 : 1 in parts. There are 3 parts altogether.
Divide the remaining probability by 3:
0.45÷3=0.150.45 \div 3 = 0.150.45÷3=0.15Landing on 3 has probability 0.15, and landing on 1 has probability 0.30.
A ratio tells you how many “parts” each outcome has. To turn a ratio into probabilities, divide each part by the total number of parts.
Ratio of counters to probabilities
A bag contains only red, blue and white counters. The ratio red : blue : white is 4 : 3 : 5. Find the probability of each colour.
Add the ratio parts:
4+3+5=124 + 3 + 5 = 124+3+5=12Write each colour as its parts out of 12:
P(red)=412=13P(blue)=312=14P(white)=512\begin{aligned} P(\text{red}) &= \frac{4}{12} = \frac{1}{3} \\ P(\text{blue}) &= \frac{3}{12} = \frac{1}{4} \\ P(\text{white}) &= \frac{5}{12} \end{aligned}P(red)P(blue)P(white)=124=31=123=41=125Check the probabilities add to 1:
412+312+512=1212\frac{4}{12} + \frac{3}{12} + \frac{5}{12} = \frac{12}{12}124+123+125=1212When an event is repeated many times, you can estimate the number of times it happens.
Expected frequency
The expected frequency is an estimate of how many times an event will happen. Use probability multiplied by the number of trials.
The key formula is:
expected frequency=probability×number of trials\text{expected frequency} = \text{probability} \times \text{number of trials}expected frequency=probability×number of trialsEstimating from a probability
A biased die has probability 0.31 of landing on 6. The die is rolled 200 times. Estimate the number of times it lands on 6.
Identify the probability and the number of trials.
probability=0.31,trials=200\text{probability} = 0.31, \quad \text{trials} = 200probability=0.31,trials=200Multiply:
0.31×200=620.31 \times 200 = 620.31×200=62An estimate for the number of sixes is 62.
Estimate, not guarantee
An expected frequency is not a promise. If you roll the die 200 times, you might not get exactly 62 sixes, but 62 is the best estimate using the given probability.
Sometimes you know a probability and the actual number of objects in that category. You can use this to find the total number.
The basic idea is:
P(event)=number in eventtotal numberP(\text{event}) = \frac{\text{number in event}}{\text{total number}}P(event)=total numbernumber in eventFinding the total number of counters
A bag contains only red, blue and white counters. The probability of choosing a red counter is 0.2. The probabilities of blue and white are equal. There are 14 red counters. Find the total number of counters.
Let the total number of counters be TTT.
Use the probability of red:
0.2=14T0.2 = \frac{14}{T}0.2=T14Rearrange by dividing 14 by 0.2:
T=14÷0.2=70T = 14 \div 0.2 = 70T=14÷0.2=70The total number of counters is 70.
If needed, you can also find blue and white: 70 - 14 = 56 counters remain, so blue and white would have 28 each.
If two outcomes cannot happen at the same time, such as rolling a 2 or rolling a 4 on one die roll, then “or” means add their probabilities.
P(A or B)=P(A)+P(B)P(A \text{ or } B) = P(A) + P(B)P(A or B)=P(A)+P(B)Finding a missing probability, then estimating
A biased die can land on 1, 2, 3, 4, 5 or 6. The probabilities of 1, 2, 3, 5 and 6 are 0.10, 0.24, 0.16, 0.12 and 0.18. The die is rolled 150 times. Estimate the number of times it lands on 2 or 4.
First find the probability of rolling a 4 by subtracting the known probabilities from 1:
1−(0.10+0.24+0.16+0.12+0.18)=0.201 - (0.10 + 0.24 + 0.16 + 0.12 + 0.18) = 0.201−(0.10+0.24+0.16+0.12+0.18)=0.20Add the probabilities of rolling 2 or 4:
0.24+0.20=0.440.24 + 0.20 = 0.440.24+0.20=0.44Multiply by the number of rolls:
0.44×150=660.44 \times 150 = 660.44×150=66The estimate is 66 times.
Harder questions often combine several ideas: completing probabilities, using a relationship like “twice as likely”, and then using a known number of objects.
Finding the number in one category
A bag contains only red, blue, green and yellow counters. The probability of green is 0.20 and the probability of yellow is 0.35. Red is twice as likely as blue. There are 16 green counters. Find the number of red counters.
Find the probability left for red and blue:
1−0.20−0.35=0.451 - 0.20 - 0.35 = 0.451−0.20−0.35=0.45Red is twice as likely as blue, so use the ratio red : blue = 2 : 1. There are 3 parts altogether.
Find one part:
0.45÷3=0.150.45 \div 3 = 0.150.45÷3=0.15Red is 2 parts, so the probability of red is 0.30.
Use the green counters to find the total. If 0.20 of the bag is 16 counters, then:
16÷0.20=8016 \div 0.20 = 8016÷0.20=80Find the number of red counters:
0.30×80=240.30 \times 80 = 240.30×80=24There are 24 red counters.
In the exam
Check whether the outcomes are the only possible outcomes. If they are, their probabilities add to 1.
For “same probability”, split the remainder equally. For “twice” or “three times”, use ratio parts.
For estimates, multiply probability by number of trials. For totals, use probability as a fraction of the whole.
Check yourself
Can you explain why the probabilities in a complete table must add to 1?
If two missing probabilities are equal, what do you do after finding the amount left over?
How would you estimate the number of successes from 80 trials if the probability of success is 0.35?
Test yourself with 5 quick questions on this guide. Answer them all correctly to complete it.
Test yourself on this topic, or move on to the next guide.
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