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GCSE: AQA Unit 1 [calculator allowed]
A Concise Course in GCSE Probability
Example 1 – probability tables (mutually exclusive outcomes)
A spinner is spun, and the probability of each outcome is
detailed in the table below
|
Colour |
Probability |
|
Red |
0.2 |
|
Green |
0.5 |
|
Blue |
0.12 |
|
Yellow |
Missing |
a) To find the value of
the missing outcome we work out ![]()
This means ![]()
b) If you are asked to
find P(Red or Yellow) we work out ![]()

c) Suppose the spinner is
spun twice and you want to find the probability of TWO reds. In other words P(Red and a Red).
We work out ![]()

d) Suppose the spinner is
spun a total of 250
times and you want to find how many Green you would expect. We know P(Green)=0.5. So:
We work out ![]()
So we would expect 125 Green’s

Example 2 – Picking out People from a Group (Dependent events)
You are told there are 8 males and 4 females in a room. Two
people are picked out at random.
a) Find the probability
they are both Male
Think about this carefully.
·
When the first male is picked there are eight males out of 12.
·
Once he has been picked, there are now 7 males to choose out of
11
|
|
Male |
and |
Male |
|
|
|
P(M
and M) |
|
|
|
= |
|
b) Find the probability
one is a male and one is a female
This needs even more careful though.
Remember that after one person has been picked there are only 11 people left
·
A male can be picked (8
out of 12) followed by a female (4 out of 11.
·
The other option is that a female can be picked first (4 out of
12) followed by a male (8 out of 11)
|
Option 1 |
Male 1st |
and |
Female 2nd |
|
|
|
|
|
|
|
= |
|
|
Option 2 |
Female
1st |
And |
Male
2nd |
|
|
|
|
|
|
|
|
|
![]()

Example 3 – tree diagrams
Tree diagrams can be used to simplify complicated situations
Suppose Peter drives through two sets of traffic lights.
For the first set P(green)=0.6
For the second set P(green)=0.7
Find the probability he gets delayed by AT LEAST one of the
lights (in other words, he meets at least one red light)
Firstly, we can draw a tree diagram:
P(RR)=
0.4x0.3=0.12 P(RG)=
0.4x0.7=0.28 P(GR)=
0.6x0.3=0.18 P(GG)= 0.6x0.7=0.42 Second Set First Set
Now to answer the original question: Add up all the options
which have at least once Red Light
![]()
