Express probabilities from 0 to 1 and as fractions or percentages.
Calculate probabilities for equally likely outcomes.
Use complementary events and expected frequency.
Combine independent events using multiplication and tree diagrams.
Revision notes
Probability scale
Probability ranges from 0 (impossible) to 1 (certain). For equally likely outcomes, P(event)=number of favourable outcomes/number of possible outcomes.
Complements
The probability an event does not happen is 1−P(event). Probabilities of all mutually exclusive outcomes in a complete sample space add to 1.
Combined events
For independent events, multiply probabilities along a tree branch. For mutually exclusive alternatives, add branch probabilities. A tree diagram labels each branch with conditional probability at that stage.
Expected frequency
Expected frequency=number of trials×probability. It is a long-run average prediction, not a guarantee for a small number of trials.
A worked example
A bag contains 5 red and 3 blue counters. One counter is selected and replaced, then a second is selected. Find the probability both are red.
There are 8 counters, so P(red on the first selection)=5/8.
The counter is replaced, so P(red on the second selection)=5/8 as well.
The selections are independent: (5/8)(5/8)=25/64.
25/64
Try it yourself
These are original practice questions, not official past-paper questions. Practice points count correct final answers; working is not automatically assessed.
Choose a question to begin.
Worked solution
Even outcomes are 2, 4, 6: three outcomes.
P=3/6=1/2.
Worked solution
Complement probability=1−0.2.
It equals 0.8.
Worked solution
There are 8 counters in total.
P(green)=2/8=1/4.
Worked solution
P(both)=0.4×0.5.
P(both)=0.2.
Worked solution
P(blue)=3/8.
3÷8=0.375.
Worked solution
P(heads)=1/2.
Expected heads=40×1/2=20.
Worked solution
There are 10 balls altogether.
P(not black)=4/10=0.4.
Worked solution
Each roll has P(6)=1/6.
P(two sixes)=1/6×1/6=1/36.
As a decimal to 3 significant figures, this is 0.0278.
Worked solution
P(first red)=3/10.
Then 7 blue of 9 remaining: P(second blue|first red)=7/9.
Multiply: 3/10×7/9=7/30≈0.233 to 3 significant figures.
Worked solution
Expected number=200×0.15.
=30 journeys.
Worked solution
Favourable ordered pairs are (2,5), (3,4), (4,3), (5,2).
There are 4 favourable pairs out of 5×5=25.
P=4/25=0.16.
Worked solution
They cannot occur together, so P(A or B)=P(A)+P(B).
0.35+0.20=0.55.
Coverage and sources
Selected IGCSE concepts, with a mix of introductory and more demanding questions. Check your own specification. AI drafted; independent teacher review is pending. Editorial details.