Introduction to Probability

“What are the chances of
that happening?”
Theory of Probability

Weather forecast predicts 80% chance of rain.

You know based on experience that going
slightly over speed increases your chance of
passing through more green lights.

You buy a ticket for the Euro Millions because
you figure someone has to win.

A restaurant manager thinks about the
probability of how many customers will come in
in order to prepare accordingly.

17th Century Gambling
Chevalier de Mere gambled frequently to
increase his wealth.

He bet on a roll of a die that at least one six
would appear in four rolls.

Tired of this approach he decided to change his
bet to make it more challenging.

He bet that he would get a total of 12, a double
6, on 24 rolls of two dice.

17th Century Gambling
Chevalier de Mere gambled frequently to
increase his wealth.

He bet on a roll of a die that at least one six
would appear in four rolls.

Tired of this approach he decided to change his
bet to make it more challenging.

He bet that he would get a total of 12, a double
6, on 24 rolls of two dice.

THE OLD METHOD WAS MOST PROFITABLE!

Correspondence leads to
Theory
He asked his friend Blaise Pascal why this was
the case.

Pascal worked it out and found that the
probablity of winning was only 49.1% with the
new method compared to 51.8% using the old
approach!

Pascal wrote a letter to Pierre De Fermat, who
wrote back.

They exchanged their mathematical principles
and problems and are credited with the founding
of probability theory.

Probability

is really about dealing with the unknown in a
systematic way, by scoping out the most likely
scenarios, or having a backup plan in case those
most likely scenarios don’t happen.

Life is a sequence of unpredctable events, but
probability can be used to help predict the
likelihood of certain events occuring.

Careers using the
“Chance Theory”
Actuarial Science Industrial Statistics

Atmospheric Science Medicine

Bioinformatics Meteorology/Atmospheric Science

Biostatistics Nurses/Doctors

Ecological/Environmental Statistics Pharmaceutical Research

Educational Testing and Measurement Public Health

environmental Health Sciences Public Policy

Epidemiology Risk Analysis

Government Risk Management and Insurance

Financial Engineering/Financial Mathematics/ Social Statistics
Mathematical Finance/Quantitative Finance

First you gotta
learn the rules
and terms!

Problem:
A spinner has four
equal sectors colored
yellow, blue, green
and red.

What are the chances
on landing on blue
after spinning the
spinner?

What are the chances
of landing on red?

Terms
Defintion Example
An EXPERIMENT is a situation
involving chance or probability
What color would we land on?
that leads to results called
outcomes.
A TRIAL is the act of doing an
Spinning the spinner.
experiment in P!
The set or list of all possible
Possible outcomes are Green,
outcomes in a trial is called the
Blue, Red and Yellow.
SAMPLE SPACE.
An OUTCOME is one of the
Green.
possible results of a trial.

An EVENT is the occurence of One event in this experiment is
one or more specific outcomes landing on blue.

Getting the rules!
The P! of an outcome is the percentage of times
the outcome is expected to happen.

Every P! is a number (a percentage) between 0%
and 100%. [Note that statisicians often express
percentages as proportions-no. 0 and 1]

If an outcome has P! of 0% it can NEVER
happen, no matter what. If an outcome has a P!
of 100% it will ALWAYS happen. Most P! are
neither 0/100% but fall somewhere in between.

The sum of all the Probabilities of all possible
outcomes is 1 (or 100%).

PROBABILITY SCALE
IN WORDS
RANGE IN NUMBER RANGE IN PERCENTAGE
0-1 0% - 100%

Want to have a guess?
0.5
0 1

A B C D E
5 EVENTS (A,B,C,D AND E ) ARE SHOWN ON A P! SCALE.
COPY AND COMPLETE THE FOLLOWING TABLE:
Probability Event
Fifty-fifty C
Certain
Very unlikely
Impossible
Very likely

Back to the Spinner
AFTER SPINNING THE SPINNER, WHAT IS THE PROBABILITY OF LANDING ON
EACH COLOR?

Back to the Spinner
EACH COLOR?

RED?

Back to the Spinner
EACH COLOR?

RED?

BLUE?

Back to the Spinner
EACH COLOR?

RED? GREEN?

BLUE?

Back to the Spinner
EACH COLOR?

RED? GREEN?

BLUE? ORANGE?

Back to the Spinner
EACH COLOR?

RED? GREEN?

BLUE? 1/4!!! ORANGE?

In order to measure the P!
of an event
mathematicians have
developed a method to do
this!

Probability of an Event
THE P! OF AN EVEN A OCCURING

P(A) = THE NUMBER OF WAYS EVENT A CAN OCCUR

THE TOTAL NUMBER OF POSSIBLE OUTSOMES

Enter P(not A)
Complement of an Event

Probability of an Event
Not Happening

If A is any event, then ‘not A’ is the event that A
does not happen.

Clearly A and ‘not A’ cannot occur at the same
time.

Either A or ‘not A’ must occur.

Thus we have the following relationship between
the probabilities of A and ‘not A’:

P(A) + P(NOT A) = 1

OR

P(NOT A) = 1 - P(A)

Let’s understand!

A spinner has 4 equal sectors colored yellow, blue, green and red. What
is the probability of landing on a sector that is not red after spinning this
spinner?

Sample Space: {yellow, blue, green, red}

Probability:
The probability of each outcome in this experiment is one fourth. The
probability of landing on a sector that is not red is the same as the
probability of landing on all the other colors except red.

P(not red) = 1/4 + 1/4 + 1/4 = 3/4

Using the rule!

P(not red) = 1 - P(red)

P(not red) = 1 - 1/4 = 3/4

You try please!

A single card is chosen at random from a
standard deck of 52 playing cards. What is the
probability of choosing a card that is not a club?

You try please!


P(not club) = 1 = P(club)

You try please!



1 - 13/52

You try please!



1 - 13/52

= 39/52 or 3/4

Notes:

P(not A) can be also written as P(A’) or P(A).

It is important NOT to count an outcome twice in
an event when calculating probabilities.

In Q’s on probability, objects that are identical
are treated as different objects.

The phrase ‘drawn at random’ means each
object is equally likely to be picked.

‘Unbiased’ means ‘fair’.

‘Biased’ means ‘unfair’ in some way.

Conditional P!
With this you are normally given some prior
knowledge or some extra condition about the
outcome.

This usually reduces the size of the sample
space.

EG. In a class - 21boys&15 girls. 3boys&5girls
wear glasses.

Conditional P!
outcome.

space.

wear glasses.
A PUPIL PICKED AT RANDOM FROM THE CLASS IS WEARING GLASSES.
WHAT IS THE P! THAT IT IS A BOY?

Conditional P!
outcome.

space.

wear glasses.
WE ARE CERTAIN THAT THE PUPIL PICKED WEARS CLASSES. THERE ARE 8 PUPILS
THAT WEAR GLASSES AND 3 OF THOSE ARE BOYS

Conditional P!
outcome.

space.

wear glasses.
P(WHEN A PUPIL WHO WEARS GLASSES IS PICKED, THE PUPIL IS A BOY)

Conditional P!
outcome.

space.

wear glasses.
P(WHEN A PUPIL WHO WEARS GLASSES IS PICKED, THE PUPIL IS A BOY)

= 3/8

Combining two events

There are many situations where we have to
consider two outcomes. In these situations, all
the possible outcomes, the sample space can
be represented on a diagram - often called a
two-way table!

Example
Two fair six-sided dices, one red and one blue,
are thrown. What is the P! of getting two equal
scores or of the scores adding to 10?

Solution:

Example

Solution:
6 X X
5 X
4 X X
3 X
2 X
1 X
1 2 3 4 5 6

Example

Solution:
36 POSSIBLE OUTCOMES
6 X X
5 X
4 X X
3 X
2 X
1 X
1 2 3 4 5 6

Example

Solution:
6 X X P(TWO EQUAL SCORES OR A TOTAL OF 10)
5 X = 8/36 = 2/9
4 X X
3 X
2 X
1 X
1 2 3 4 5 6

Example

Solution:
6 X X P(TWO EQUAL SCORES OR A TOTAL OF 10)
5 X = 8/36 = 2/9
4 X X
3 X
2 X NOTE: (5,5) IS NOT COUNTED TWICE!
1 X
1 2 3 4 5 6

Relative Frequency
Experimental Probability

Some P! cannot be
calculated by just looking
at the situation!

For example you cannot work out the P! of
winning a football match by assuming that win,
draw or loose are equally likely!

But we can look at previous results in similar
matches and use these results to estimate the P!
of winning!

Example 1
The Blues and Naoimh Martin Gaelic Teams are
playing a match tonight and you want to know
what is the P! that the Blues will win?

They have played each other 50 times before.
The Blues won 35 of those games and there was
also 5 draws!

So we can say so far the Blues have won 35/50
games or 7/10!

The fraction isn’t the P! of the Blues winning but
an estimate!

We say that the relative frequency og the Blues
winning is 7/10.

Example 2
Matthew decides to see what the P! is that
buttered toast lands buttered side down when
dropped.

He drops 50 pieces of buttered toast.

30 pieces land buttered side down.

His relative frequency is 30/50=3/5.

Therefore he would estimate that the P! of
landing buttered side down is 3/5.

Definition:

Relative Frequency is a good estimate of how
likely an event is to occur, provided that the
number of trials is sufficiently large.

Experiment - Formula
The relative frequency of an event in an
experiment is given by:


P(E) = RELATIVE FREQUENCY OF AN EVENT=


NO. OF SUCCESSFUL TRIALS


NO. OF SUCCESSFUL TRIALS
NO. OF TRIALS

How many times you expect
a particular outcome to
happen in an experiment.
The expected number of outcomes is calculated
as follows:

EXPECTED NO. OF OUTCOMES = (RELATIVE FREQUENCY) X (NO. OF TRIALS)

OR

EXPECTED NO. OF OUTCOMES = P(EVENT) X (NO. OF TRIALS)

Example
Sarah throws her fair six sided die a total of
1,200 times. Find the expected number of times
the number 3 would appear.

Solution:
IF THE DIE IS FAIR, THEN THE P! OF A SCORE OF 3 WOULD BE 1/6!

THUS THE EXPECTED NO OF 3’S
= P(EVENT) X (NO. OF TRIALS)
= 1/6 X 1200
=200.

Example 2
A spinner numbered 1-5 is biased. The P! that
the spinner will land on each of the numbers 1 to
5 is given in the P! distribution table below.

Number 1 2 3 4 5

Probability 0.25 0.2 0.25 0.15 B

(I) Write down the value of B.

(ii) If the spinner is spun 200 times, how many
fives would expect?

Solution:

(i) Since one of the no. 1-5 must appear, the sum
of all the P! must add to 1!
Therefore 0.25 + 0.2 + 0.25 + 0.15 + B = 1

0.85 + B =1 thus B = 0.15

Expected no. of 5’s.
= P(5) X (no. of trials)

= 0.15 X 200 = 30

Combined Events
If A and B are two different events of the same
experiment, then the P! that the two events, A or
B, can happen is given by:
P(A OR B) = P(A) + P(B) - P(A AND B)

REMOVES DOUBLE COUNTING

It is often called the ‘or’ rule!

It is important to remember that P(A or B) means
A occurs, or B occurs, or both occur. By
subtracting p(A and B), the possibility of double
counting is removed.

Mutually Exclusive
Events
Events A and B are said to be mutually exclusive
events if they cannot occur at the same time.

Consider the following event of drawing a single
card from a deck of 52 cards. Let A be the even
a king is drawn and B the event a Queen is
drawn. The single card drawn cannot be a King
and a Queen. The events A and B are said to be
mutually exclusive events.

If A and B are mutually exclusive events, then
P(A ∩ B) = 0. There is no overlap of A and B.

For mutually exclusive events,
P(A ∪ B) = P(A) + P(B).

Example 1
A and B are two events such that P(A ∪ B) = 9/10,
P(A) = 7/10 and P(A ∩ B) = 3/20

Find: (i) P(B) (ii) P(B’) (iii) P[(A ∪ B)’]

Solution: U
(I) P(A ∪ B) = P(A) + P(B) - P(A ∩ B) A B
9/10 = 7/10 + P(B) - 3/20
P(B) = 9/10 - 7/10 + 3/20 11/20 3/20 2/5

= 7/20
1/10
(II) P(B’) = 1 - P(B)
= 1 - 7/20
= 13/20 (7/10 - 3/20 = 11/20) (7/20 - 3/20 = 2/5)

(III) P[(A ∪ B)’] = 1 - P(A ∪ B)
= 1 - 9/10
= 1/10

Example 2

An unbiased 20-sided die, numbered 1 to 20, is
thrown.

(i) What is the P! of obtaining a no. divisible by 4
or 5?

(ii) Are these events mutually exclusive?

Solution:
There are 20 possible outcomes:

(i) No ÷ by 4 are 4, 8, 12, 16 or 20 ∴ P(no÷4) =
5/20 No ÷ by 5 are 5, 10, 15 or 20 ∴ P(no÷5) =
4/20 No ÷ by 4 and 5 is 20 ∴ P(no divisible by 4 and
5) = 1/20

P(no÷4/5) = P(no÷4) + P(no÷5) - P(no÷4 and 5)
= 5/20 + 4/20 - 1/20
= 8/20 = 4/5
THE NO 20 IS COMMON TO BOTH EVENTS, AND IF THE PROBABILITIES WERE SIMPLY ADDED, THEN
THE NO 20 WOULD HAVE BEEN COUNTED TWICE.

∕
(II) P(NUMBER DIVISIBLE BY 4 AND 5) = 0
∴ THE EVENTS ARE NOT MUTUALLY EXCLUSIVE!

Example 3

A bag contains five red, three blue and yellow
discs. The red discs are numbered 1, 2, 3, 4 and
5; the blue are numbered 6, 7, and 8; and the
yellow are 9 and 10. A single disc is drawn at
random from the bag. What is the P! that the
disc is blue and even? Are these mutually
exclusive?

Solution:
1 2 3 4 5 8

Solution:
1 2 3 4 5 6 8

Solution:
1 2 3 4 5 6 7 8

Solution:
1 2 3 4 5 6 7 8 9

Solution:
1 2 3 4 5 6 7 8 9 10

Solution:
1 2 3 4 5 6 7 8 9 10

Let B represent that a blue disc is chosen and E
represent that a disc with an even no is chose.

P(B or E) = P(B) + P(E) - P(B and E)

Solution:
1 2 3 4 5 6 7 8 9 10


= 3/10 + 5/10 - 2/10

Solution:
1 2 3 4 5 6 7 8 9 10


= 3/10 + 5/10 - 2/10
= 6/10 = 3/5

Solution:
1 2 3 4 5 6 7 8 9 10


= 3/10 + 5/10 - 2/10
= 6/10 = 3/5

P(B AND E) = 2/10 = 0

Solution:
1 2 3 4 5 6 7 8 9 10


= 3/10 + 5/10 - 2/10
= 6/10 = 3/5

∕
P(B AND E) = 2/10 = 0

Q. 8 Pg: 73 Active Math
Given the Venn S
diagram, write down: E F
P(E) =
0.3 0.1s 0.5
P(F) =

P(E ∩ F) = 0.1

P(E ∪ F) =

Verify that
P(E ∪ F) = P(E)+P(F)-P(E ∩ F)

Given the Venn S
P(E) = 0.4
0.3 0.1s 0.5
P(F) =

P(E ∩ F) = 0.1

P(E ∪ F) =

Verify that
P(E ∪ F) = P(E)+P(F)-P(E ∩ F)

Given the Venn S
P(E) = 0.4
0.3 0.1s 0.5
P(F) = 0.6

P(E ∩ F) = 0.1

P(E ∪ F) =

Verify that
P(E ∪ F) = P(E)+P(F)-P(E ∩ F)

Given the Venn S
P(E) = 0.4
0.3 0.1s 0.5
P(F) = 0.6

P(E ∩ F) = 0.1 0.1

P(E ∪ F) =

Verify that
P(E ∪ F) = P(E)+P(F)-P(E ∩ F)

Given the Venn S
P(E) = 0.4
0.3 0.1s 0.5
P(F) = 0.6

P(E ∩ F) = 0.1 0.1

P(E ∪ F) = 0.9

Verify that
P(E ∪ F) = P(E)+P(F)-P(E ∩ F)

Given the Venn S
P(E) = 0.4
0.3 0.1s 0.5
P(F) = 0.6

P(E ∩ F) = 0.1 0.1

P(E ∪ F) = 0.9

Verify that
P(E ∪ F) = P(E)+P(F)-P(E ∩ F) 0.9 = 0.4 + 0.6 - 0.1

Q. 18 Pg: 74
The data shows the number of boys and girls
aged either 17 or 18 on a school trip, in which
only these 50 students took part.
Boys Girls
Aged 17 18 12
Aged 18 15 5
A students is chosen at random from the party. Find
the P! that the student:

(i) Is a boy:

(ii) is aged 18:

(iii) is a girl of 18 years:

(iv) is a girl aged 17 or a boy aged 18:

Q. 18 Pg: 74
Boys Girls
Aged 17 18 12
Aged 18 15 5

(i) Is a boy: 33/50

(ii) is aged 18:



Q. 18 Pg: 74
Boys Girls
Aged 17 18 12
Aged 18 15 5

(i) Is a boy: 33/50

(ii) is aged 18: 20/50 = 2/5



Q. 18 Pg: 74
Boys Girls
Aged 17 18 12
Aged 18 15 5

(i) Is a boy: 33/50

(ii) is aged 18: 20/50 = 2/5

(iii) is a girl of 18 years:5/50 = 1/10


Q. 18 Pg: 74
Boys Girls
Aged 17 18 12
Aged 18 15 5

(i) Is a boy: 33/50

(ii) is aged 18: 20/50 = 2/5

(iii) is a girl of 18 years:5/50 = 1/10

(iv) is a girl aged 17 or a boy aged 18: 27/50

Conditional Probability
In a class there are 15 male students, 5 of whom
wear glasses and 10 female students, 3 of whom
wear glasses.

We will let M = {male students}, F = {female
students} and G = {students who wear glasses}.

A student is picked at random from the class.
What is the P! that the student is female, given
that the student wears glasses?

So we write this as: P(F|G) [The P! of F, given G]

=


8 students wear glasses.

=



3 of these are girls.

=




Hence P(F|G) = 3/8

=




Hence P(F|G) = 3/8

In general the P(A|B) = =




Hence P(F|G) = 3/8

# (A ∩ B) P(A ∩ B)
In general the P(A|B) = #B = P(B)

Q 4. Pg: 79

A family has three children. Complete the outcome
space:
{GGG, GGB, ...}, where GGB means the first two are
girls and the third is a boy.

Find the P! that all 3 children are girls, given that the
family has at least two girls.
{GGG, GGB, BGG, GBG, BBG, GBB, BGB, BBB}

2/8 = 0.25

Q. 9 PG: 80

E and F are two events such that P(E) = 2/5, P(F) = 1/2,
and P(E|F) = 1/9.

Find: (i) P(E ∩ F) (ii) P(F|E) (iii) P(E ∪ F)

Q. 9 PG: 80

and P(E|F) = 1/9.


P(E|F) = P(E ∩ F) P(F)

Q. 9 PG: 80

and P(E|F) = 1/9.


P(E|F) = P(E ∩ F) P(F)
1/9 = P(E ∩ F) 1/2

Q. 9 PG: 80

and P(E|F) = 1/9.


P(E|F) = P(E ∩ F) P(F)
1/9 = P(E ∩ F) 1/2
∴ P(E ∩ F) = 1/18

Q. 9 PG: 80

and P(E|F) = 1/9.


P(E|F) = P(E ∩ F) P(F) P(F|E) = P(F ∩ E) P(E)
1/9 = P(E ∩ F) 1/2
∴ P(E ∩ F) = 1/18

Q. 9 PG: 80

and P(E|F) = 1/9.


P(F
1/9 = P(E ∩ F) 1/2
P(F|E) = 1/18 2/5
∴ P(E ∩ F) = 1/18

Q. 9 PG: 80

and P(E|F) = 1/9.


P(F
1/9 = P(E ∩ F) 1/2
P(F|E) = 1/18 2/5
∴ P(E ∩ F) = 1/18 P(F|E) = 5/36

Q. 9 PG: 80

and P(E|F) = 1/9.


P(F
1/9 = P(E ∩ F) 1/2
P(F|E) = 1/18 2/5
∴ P(E ∩ F) = 1/18 P(F|E) = 5/36

P(E ∪ F) = P(E) + P(F) - P(E ∩ F)

Q. 9 PG: 80

and P(E|F) = 1/9.


P(F
1/9 = P(E ∩ F) 1/2
P(F|E) = 1/18 2/5
∴ P(E ∩ F) = 1/18 P(F|E) = 5/36

P(E ∪ F) = P(E) + P(F) - P(E ∩ F)
2/5 + 1/2 - 1/18

Q. 9 PG: 80

and P(E|F) = 1/9.


P(F
1/9 = P(E ∩ F) 1/2
P(F|E) = 1/18 2/5
∴ P(E ∩ F) = 1/18 P(F|E) = 5/36

P(E ∪ F) = P(E) + P(F) - P(E ∩ F)
2/5 + 1/2 - 1/18
38/45

Introduction to Probability

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