1. Probability Lesson 1 Aims:
• To have a general understanding of probability.
• To be able to use a sample space diagram.
• To be able to know and be able to use the addition law of probability.
P(A B) = P(A) P(B) – P(A B)
• To know probabilities add up to 1.
• To know what mutually exclusive events are. P(A B) = P(A) + P(B)
• To be able to use probability notation.

2. Probability How likely am I to live to 100?
Which team is most likely to win the FA cup?
Will Coronation Street ever end?
Am I likely to win the lottery?
Uncertainty is a feature of everyday life. Probability is an area of maths that addresses how likely things are to happen.
A good understanding of probability is important in many areas of work. It is used by scientists, governments, businesses, insurance companies, betting companies and many others, to help them anticipate future events.

3. Introduction to probability
A statistics experiment will have a number of different outcomes. The set of all possible outcomes is called the sample space of the experiment.
For example:
if a normal dice is thrown the sample space would be {1, 2, 3, 4, 5, 6}.
In a general knowledge quiz with 70 questions, the sample space for the number of questions a person answers correctly is {0, 1, 2, …, 70}.
An event is a collection of some of the outcomes from an experiment. For example, getting an even number on the dice or scoring more than 40 on the quiz.

4. Notation Let A be an event arising from a statistical experiment.
The probability that A occurs is denoted P(A) (where 0 ≤ P(A) ≤ 1).
If A is certain to happen, then P(A) = 1.
If A is impossible, then P(A) = 0.
The probability that A does not occur is denoted P(A′).
P(A′) = 1 – P(A)

5. Introduction to probability
When two experiments are combined, the set of possible outcomes can be shown in a sample space diagram.
Example: A dice is thrown twice and the scores obtained are added together. Find the probability that the total score is 6.
There are 36 equally likely outcomes. 6 7 8 9 10 11 12 5 4 3 2 1
Second throw
5 of the outcomes result in a total of 6.
P(total = 6) =
First throw
This notation means “probability that the total = 6”.

6. The outcomes that satisfy event A can be represented by a circle.
Venn diagrams
Venn diagrams can be used to represent probabilities.
The outcomes that satisfy event A can be represented by a circle.
The outcomes that satisfy event B can be represented by another circle. A B
The circles can be overlapped to represent outcomes that satisfy both events.

7. Addition properties
Two events A and B are called mutually exclusive if they cannot occur at the same time.
For example, if a card is picked at random from a standard pack of 52 cards, the events “the card is a club” and “the card is a diamond” are mutually exclusive.
However the events “the card is a club” and “the card is a queen” are not mutually exclusive.
If A and B are mutually exclusive, then:
A B
In Venn diagrams representing mutually exclusive events, the circles do not overlap.
This symbol means ‘union’ or ‘OR’

8. Addition properties P(A B) = P(A) + P(B) – P(A B)
This addition rule for finding P(A B) is not true when A and B are not mutually exclusive.
The more general rule for finding P(A B) is:
P(A B) = P(A) P(B) – P(A B)
This symbol means ‘intersect’ or ‘AND’
Venn diagrams can help you to visualize probability calculations.

9. Addition properties This represents the other 3 queens.
Example: A card is picked at random from a pack of cards. Find the probability that it is either a club or a queen or both.
Card is a club = event C Card is a queen = event Q
This represents the other 3 queens.
This area represents the 12 clubs that are not queens.
This represents the queen of clubs.
So,

10. Addition properties
Example 2: If P(A′ B′) = 0.1, P(A) = 0.45 and P(B) = 0.75, find P(A B).
P(A′ B′) is the unshaded area in the Venn Diagram.
A B
0.1
We can deduce that:
P(A B) = 1 – 0.1 = 0.9
Using the formula, P(A B) = P(A) + P(B) – P(A B), we get:
0.9 = – P(A B)
0.9 = 1.2 – P(A B)
So, P(A B) = 0.3

11. Addition properties
Examination style question: There are two events, C and D. P(C) = 2P(D) = 3P(C D). Given that P(C D) = 0.52, find:
a) P(C D)
b) P(C D′).
C D x
Let P(C D) = x
Using P(C D) = P(C) + P(D) – P(C D)
0.52 = 3x + 1.5x – x
0.52 = 3.5x
Therefore x = P(C D) = 0.52 ÷ 3.5
= 0.15 (3 s.f)
So, P(C) = and P(D) =

12. Addition properties …as P(C) = 3P(C D) Question (continued): C D
b) P(C D′) corresponds to the unshaded area in this Venn diagram.
We see that:
C D
= P(C ) – P(C D)
…as P(C) = 3P(C D)
= 3x – 0.15
= 0.45 – 0.15
= 0.3
Do exercise 4A questions 1,2 and 3.
Do worksheet – this uses the notation we’ve just seen!