© Boardworks Ltd 20051 of 32 © Boardworks Ltd 2005 1 of 32 AS-Level Maths: Statistics 1 for...

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AS-Level Maths: Statistics 1for Edexcel

S1.3 Probability

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Probability

Probability basics and notation

Estimating probability

Addition properties

Independent events

Conditional probability


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?

Are interest rates likely to go up?

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

Sec

on

d t

hro

w

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”.


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Addition properties

Independent events




Some probabilities are less simple. It is not always possible to calculate how likely each outcome is.


number of times event occursnumber of times experiment is repeated

You can increase the accuracy of the relative frequency as an estimate of probability, by increasing

the number of times you repeat the experiment.

This is referred to as the relative frequency.

However, the probability of an event happening can be estimated experimentally, by repeating an experiment over and over again. The probability is estimated using:


Venn diagrams

Venn 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.


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Addition properties

Independent events


Addition properties


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) = xP(C D) = P(C) + P(D) – P(C D)

So, 0.52 = 3x + 1.5x – x

Therefore x = P(C D) = 0.52 ÷ 3.5 = 0.149 (3 s.f.)

Addition properties

C D

x


Question (continued):

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

We see that:

P(C D′) = P(C′ D′) + P(C)

= (1 – 0.52) + 3 × 0.149

= 0.48 + 0.447

= 0.927 (3 s.f.)

…as P(C) = 3P(C D)

Addition properties

C D


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Addition properties

Independent events


Independent events


Two events are said to be independent if the occurrence of one has no effect on the probability of the second occurring.

If A and B are independent, then:

Independent events

P(A B) = P(A) × P(B)P(A B) = P(A) × P(B)

For example, if a coin and a die are both thrown, then the events “the coin shows a head” and “the die shows

an odd number” are independent events.


Example: A and B are independent events. P(A) = 0.7 and P(B) = 0.4.

a) Find P(A B).

b) Find P(A′ B).

A B

Independent events

a) As A and B are independent,

P(A B) = P(A) × P(B)

= 0.7 × 0.4 = 0.28

b) P(A′ B) is the shaded region in the Venn diagram.

So, P(A′ B) = P(B) – P(A B)

= 0.4 – 0.28 = 0.12


Tree diagrams are sometimes a useful way of finding probabilities that involve a succession of events.

Independent events

Example: A bag contains 6 green counters and 4 blue counters. A counter is chosen at random from the bag and then replaced. This is repeated two more times.Find the probability that the 3 counters chosen are:

a) all greenb) not all the same colour.


G

GG

G

GG

G

B

B

B

B

B

B

B

0.6

0.4

0.4

0.6

0.6

0.4

0.6

0.6

0.6

0.6

0.4

0.4

0.4

0.4

a) P(GGG) = 0.6 × 0.6 × 0.6

Independent events

= 0.216


b) To find the probability that there will be at least one of each colour, we can find 1 – P(GGG) – P(BBB)P(BBB) = 0.4 × 0.4 × 0.4 = 0.064.

Independent events

G

GG

G

GG

G

B

B

B

B

B

B

B

0.6

0.4

0.4

0.6

0.6

0.4

0.6

0.6

0.6

0.6

0.4

0.4

0.4

0.4

Therefore, the answer is 1 – 0.216 – 0.064 = 0.72


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Addition properties

Independent events




However, the probability of event B happening might depend on whether A has happened or not.

For example, if blue and green counters are pulled from a bag twice and not replaced, then the probability of pulling out a green counter on the second try will depend on what colour was pulled out on the first try.


The probability that event A will happen, given that event B has happened, is written

P(A | B)P(A | B)

This is a conditional probability.


D

M

D

D

M

M

Example: A bag contains 8 dark chocolates and 4 milk chocolates. One chocolate is taken out and eaten. A second chocolate is then taken. Find the probability that:

a) two milk chocolates are taken.

b) the two chocolates are of different types.

2

3


1

3

7

11

4

11

8

11

3

11

a) P(M M) = 1 3 1

3 11 11

b) P(D M) = 2 4 8

3 11 33

P(M D) = 1 8 8

3 11 33

8 8

33 33

16

33


Examination-style question: A man has 2 shirts (one white and one blue) and 2 ties (red and silver).

If he wears the white shirt, he chooses the red tie with probability 0.4.

If he wears the blue shirt, he chooses the red tie with probability 0.75.

The probability that he wears the white shirt is 0.7.


a) Find the probability that he wears the red tie.

b) Given that he is wearing a red tie, find the probability that he picked the blue shirt.


W

B

R

R

S

S

a) P(red tie) = P(W R) + P(B R)

= (0.7 × 0.4) + (0.3 × 0.75)

= 0.28 + 0.225 = 0.505

Shirt Tie

0.7

0.3

0.4

0.6

0.75

0.25



So,

b) Recall the formula for conditional probability:


P( ) 0.225P( | )

P( ) 0.505

B RB R

R

P( )P( | )

P( )

A BB A

A

= 0.446 (3 s.f.)


Example: A disease affects 1 in 500 people.

A diagnostic test for the disease records a positive result 99% of the time when the disease is present (this is called the sensitivity of the test).

The test records a negative result 95% of the time when the disease in not present.

The test results are always either positive or negative.

Find the probability that a person has the disease, given that the test result is positive.

The example below demonstrates an important application of probability in the field of medicine.



D

D′

+ve

+ve

–ve

–ve

Disease Test

0.002

0.998

0.99

0.01

0.05

0.95

P(D +ve) = 0.002 × 0.99 = 0.00198

P(D′ +ve) = 0.998 × 0.05 = 0.0499


P( +ve) 0.00198P( |+ve) (3 s.f.)

P(+ve) 0.0.0382

05188

DD

Therefore, P(+ve) = 0.05188

So,

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