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.

Probability Lesson 1

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.

How likely am I to live to 100?

Probability

Am I likely to win the lottery?

Will Coronation Street ever end?

Which team is most likely to win the FA cup?

A statistics experiment will have a number of different outcomes. The set of all possible outcomes is called the sample space of the experiment.

Introduction to probability

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.

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}.

if a normal dice is thrown the sample space would be {1, 2, 3, 4, 5, 6}.

For example:

Let A be an event arising from a statistical experiment.

The probability that A occurs is denoted P(A)(where 0 ≤ P(A) ≤ 1).

Notation

The probability that A does not occur is denoted P(A )′ .

If A is impossible, then P(A) = 0.If A is impossible, then P(A) = 0.

If A is certain to happen, then P(A) = 1.If A is certain to happen, then P(A) = 1.

P(A ) = 1 – P(′ A)P(A ) = 1 – P(′ A)

When two experiments are combined, the set of possible outcomes can be shown in a sample space diagram.

Introduction to probability

6 7 8 9 10 11 12

5 6 7 8 9 10 11

4 5 6 7 8 9 10

3 4 5 6 7 8 9

2 3 4 5 6 7 8

1 2 3 4 5 6 7

1 2 3 4 5 6First throw

Seco

nd th

row

P(total = 6) = 5

36

There are 36 equally likely outcomes.

Example: A dice is thrown twice and the scores obtained are added together. Find the probability that the total score is 6.

5 of the outcomes result in a total of 6.

This notation means “probability that the total = 6”.

Venn diagramsVenn diagrams can be used to represent probabilities.

The outcomes that satisfy event A can be represented by a circle.

A

The outcomes that satisfy event B can be represented by another circle.

B

The circles can be overlapped to represent outcomes that satisfy both events.

In Venn diagrams representing mutually exclusive events, the

circles do not overlap.

Two events A and B are called mutually exclusive if they cannot occur at the same time.

A BIf A and B are mutually exclusive, then:

Addition properties

This symbol means ‘union’ or ‘OR’

However the events “the card is a club” and “the card is a queen” are not mutually exclusive.

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.

P( ) = P( ) + P( )A B A B

This addition rule for finding P(A B) is not true when A and B are not mutually exclusive.

P(A B) = P(A) + P(B) – P(A B) P(A B) = P(A) + P(B) – P(A B)

Addition properties

The more general rule for finding P(A B) is:

This symbol means ‘intersect’ or ‘AND’

Venn diagrams can help you to visualize probability calculations.

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.

Addition properties

1P( ) =

4C

4 1P( ) = =

52 13Q

1P( ) =

52C Q

Card is a club = event CCard is a queen = event Q

1 1 1 4P( ) = + – =

4 13 52 13C QSo,

This area represents the 12 clubs that are not queens.

This represents the queen of clubs.

This represents the other 3 queens.

Example 2: If P(A ′ B ) = 0.1, P(′ A) = 0.45 and P(B) = 0.75, find P(A B).

We can deduce that:

P(A B) = 1 – 0.1 = 0.9

A B

0.1

Using the formula, P(A B) = P(A) + P(B) – P(A B), we get:

0.9 = 0.45 + 0.75 – P(A B)

0.9 = 1.2 – P(A B)

Addition properties

P(A ′ B ) is the unshaded area in the Venn Diagram.′

So, P(A B) = 0.3

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 ).′

a) 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)

Addition properties

C D

x

So, P(C) = and P(D) =

Question (continued):

b) P(C D ) corresponds to the unshaded area in this ′Venn diagram.

We see that:

…as P(C) = 3P(C D)

Addition properties

C D

Do exercise 4A questions 1,2 and 3.Do worksheet – this uses the notation we’ve just seen!

= P(C ) – P(C D)

= 3x – 0.15

= 0.45 – 0.15

= 0.3