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Probability
Probability is used to give a numerical value to the question of how likely some event
is to occur.
Probability is always given as a fraction or decimal between zero and one.
0
cannot
happen
1
certain to
happen
Do not use ratios as probabilities. Percentages are also best avoided, especially when
probabilities need to be added or multiplied.
Leave probabilities as proper fractions where possible – this avoids the use of
repeating decimals.
Definitions
Trial:
Outcome:
Sample Space:
Event:
one roll of a die or toss of a coin etc
the result of a trial (head, tail etc)
the set of all possible outcomes
part of the sample space (rolling an even number etc)
The probability of an event occurring is written as:
P(event ) 
number of favourable outcomes
number in the sample space
This rule is valid when all outcomes are equally likely.
Example:
A fair die is rolled. What is the probability of obtaining a number larger than 4?
There are 6 equally likely outcomes.
There are 2 numbers larger than 4. (favourable outcomes for this question)
P(number larger than 4) 
2 1

6 3
(It is not usually necessary to simplify the fraction.)
Example:
The spinner shown is spun. What is the
probability of obtaining an odd number?
There are 6 possibilities and 4 odd numbers.
SC probability
1
P(odd number) 
4 2

6 3
Example:
A bag contains discs numbered from 1 to 20. One disc is drawn out at random.
Find the probability that the number is:
1. greater than 15
the numbers greater than 15 are - 16, 17, 18, 19, 20 (5 numbers)
5 1

so the probability is
20 4
2. a multiple of 6
the multiples of six are - 6, 12, 18
3
so the probability is
20
3. a multiple of 3 or 5
the multiples of 3 are - 3, 6, 9, 12, 15, 18
the multiples of 5 are - 5, 10, 15, 20
the number of multiples of 3 or 5 is 9 (do not count the 15 twice)
9
so the probability is
20
Sample space for more than one object or action
When more than one object is involved, it is important to be able to write down the
whole sample space.
Example:
What is the probability of obtaining a total of 10 when two dice are thrown?
There are six possibilities for the first die and six possibilities for the second die, so
the total number in the sample space is 6  6  36 . These possibilities are shown in
the table.
Number on
second die
1
2
3
4
5
6
1
2
3
4
5
6
7
Number on first die
2
3
4
5
3
4
5
6
4
5
6
7
5
6
7
8
6
7
8
9
7
8
9
10
8
9
10 11
6
7
8
9
10
11
12
There are 36 possibilities altogether and three ways of obtaining a total of 10.
These are (4, 6) (5, 5) and (6, 4).
SC probability
2
3
1

36 12
If three dice were thrown, the total number in the sample space would be
6  6  6  216 .
So the probability of obtaining a total of 10 is
1
, since
216
This means that the probability of throwing a total of 18 with three dice is
there is only one way of obtaining a total of 18. (6, 6 and 6)
Multiplying probabilities
When events follow each other and they are independent (the outcome of one does not
influence the outcome of another) the probabilities may be multiplied.
Example:
What is the probability that the captain of a sports team will win the toss 4 times in a
row.
The events are independent, so probabilities can be multiplied.
1 1 1 1 1
P(4wins)     
2 2 2 2 16
In general, if the question uses the word AND, the probabilities are multiplied.
If the question uses the word OR, the probabilities are added.
Tree Diagrams
When probabilities are to be calculated from an action carried out more than once, it
is often easiest to solve the problem using a tree diagram.
Each branch of a tree diagram indicates a possibility.
Each branch should be named and the probability marked.
The final probability for each outcome is found by multiplying all the probabilities
along the branches which lead to that outcome.
Example:
One fifth of customers in a shop choose tapes and the rest choose CDs.
In each case, one third choose male singers and the rest choose female singers.
What is the probability that a randomly chosen customer will choose a female singer?
There are two ways in which a customer can choose a
female singer.
1 2 2
P(tape and female) =  
5 3 15
P(CD and female) =
4 2 8
 
5 3 15
4
5
3
m
2
3
f
tapes
1
5
CD
s
SC probability
1
3
1
3
m
2
3
f
P(female) =
2 8 10 2
 

15 15 15 3
Example:
If a fair coin is tossed three times, what is the probability that there will be two heads
and one tail?
½
H
½
H
½
½
T
H
½
H
T
½
½
T H
HHT HT
H
½
½
½
½
½
H
T H
THH
½
T
T
½
T
There are three ways in which it is possible to obtain two heads and one tail. These
possibilities are marked at the bottom of the tree diagram.
The possibilities are HHT, HTH and THH.
The probability for each of these is 12  12  12  81 (multiplying the probabilities down
each branch)
So the total probability of two heads and one tail is 18  81  81  83
Example:
The probability that an adult has a flu injection is 0.7
Of those who have an injection, 0.05 will get the flu.
Of those who do not have an injection, 0.6 will catch the flu.
Draw a tree diagram to find the probability that a randomly selected person will catch
the flu.
0.7
0.3
No injection
injection
0.05
flu
0.6
0.95
flu
no flu
0.4
No flu
Multiplying probabilities down the two branches where people have caught the flu
(underlined) gives a probability of:
P(catches the flu) = 0.7  0.05  0.3 0.6  0.215
Notice that the probabilities on each set of branches must add to one.
SC probability
4
SC probability
5