Download Introduction to Probability - Maths@Deyi

General Instructions
Dear 4E/5N Students
You are expected to do the following for the Maths module:
 Read and study the slides on Probability
 Explore the websites given
 Complete the one task given on the worksheet and submit them
to your math teacher on 5 February 2008.
Happy Enjoy-Learning!
Best Regards
Your Math Teachers
Probability
Introduction
In this e-lesson, you will learn to
solve simple probability problems in
Part One

use possibility diagrams and tree
diagrams to solve probability problems
involving combined events in Part Two

Introduction
Probability Theory was first used to solve gambling
problems.
Lotteries have always been a magnet to those who
dream of instant riches. Toto, 4-D and Singapore Sweep
are some of the favourite games of chance among
Singaporeans.
However, do you know that you have a one-in-8.1 million
chance of winning the first prize for Toto, and the odds
for striking any of the
first three prizes in the Singapore Sweep
and 4-D are one in 3 million and one in
10,000 respectively? Do you know how
to calculate the odds?
Introduction
Probability Theory has since been widely used in
areas like business, finance, science and
industry, and has become a powerful branch of
mathematics.
We often make statements involving probability
or chance in our daily life. Some examples of
these statements are:
‘It will probably rain today.’
‘It is unlikely that we will win the championship.’
‘There is a high chance that you will find him
in the canteen.’
‘It is impossible to pass the test!’
Part One
Experiments
Probability is defined as the likelihood of an
occurrence of a special event.
In probability, an experiment is an operation or
a process with a result or an outcome whose
occurrence depends on chance. Some examples
of an experiment are:
Example 1 Tossing a coin
Example 2 Tossing a dice
Sample Space
An experiment can result in several possible
outcomes. The set of all possible outcomes is
called the sample space or probability space, S.
Example 1 Tossing a coin
Possible Outcomes: Head or Tail
 S = {Head, Tail}
Example 2 Tossing a dice
Possible Outcomes: 1 or 2 or 3
or 4 or 5 or 6
 S = {1, 2, 3, 4, 5, 6}
Events
An event, E is a particular result of an
experiment. Hence, E contains some or all of the
possible outcomes in S.
Example 1 Tossing a coin
Let E be the event of getting a tail.
 E = {Tail}
Example 2 Tossing a dice
Let E be the event of getting a
number less than 5 on the dice.
 E = {1, 2, 3, 4}
Simple Probability
The probability of an event E occurring is given by
n(E)
P(E) 
n(S)
where n(E) is the number of outcomes in E and
n(S) is the total number of possible outcomes in S.
Simple Probability
Example 1 Tossing a coin
S = {Head, Tail} and n(S) = 2
E = {Tail} and n(E) = 1
1
 P(E)  .
2
Example 2 Tossing a dice
S = {1, 2, 3, 4, 5, 6} and n(S) = 6
E = {1, 2, 3, 4} and n(E) = 4
4 2
 P(E)   .
6 3
Simple Probability
The probability of any event occurring lies between
0 and 1 inclusive, i.e. 0 ≤ P(E) ≤ 1.
Do you know why?
Some important notes:
 If P(E) = 0, then the event cannot possibly occur.
 If P(E) = 1, then the event will certainly occur.

Probability of an event E not occurring
= 1 – probability of an event occurring
i.e. P(E’) = 1 – P(E)
Sample Question
Question: In an experiment, a card is drawn
from a pack of 52 playing cards.
(a) What is the total number of
possible outcomes of this
experiment?
(b) What is the probability of drawing
(i) a black card,
(ii) a green card,
(iii) a red ace,
(iv) a heart,
(v) a card which is not a heart?
Solution to Sample Question
(a) Total number of possible outcomes, n(S) = 52.
n(black card) 26 1

 .
(b)(i) P(drawing a black card) =
n(S)
52 2
n(green card) 0

 0.
(ii) P(drawing a green card) =
n(S)
52
(iii) P(drawing a red ace) =
n(red ace) 2
1


.
n(S)
52 26
Solution to Sample Question
n(heart) 13 1

 .
(b)(iv) P(drawing a heart) =
n(S)
52 4
(v) P(drawing a card which is not a heart)
= 1 – P(drawing a heart)
13
1 3
 1
 1  .
52
4 4
Part Two
Possibility Diagrams and Tree Diagrams
Possibility diagrams and tree diagrams
are used to list all possible outcomes
of a sample space in a systematic and
effective manner.
These diagrams are useful for finding
the probabilities of combined events.
An example of possibility diagrams
Two coins are tossed together. The possibility
diagram below shows all the possible outcomes:
S = { HH, HT,TH,TT }
2nd coin
H
Each represents an
outcome.
T
Eg. P(getting 2 heads)
= P(HH)
H
T
1st coin
=
1
.
4
An example of tree diagrams
Two coins are tossed together. The tree diagram below
shows all the possible outcomes:
1st
2nd
coin
½
½
H
coin
½
H
½
T
½
H
Outcome
Probability
HH
P(HH) = ½  ½ = ¼
HT
P(HT) = ½  ½ = ¼
TH
P(TH) = ½  ½ = ¼
T
P(TT) = ½  ½ = ¼
TT
½ T
Each outcome is obtained by tracing along a branch from
left to right. The probability of each outcome is
obtained by multiplying the probabilities along the
respective branch. The total probability of all possible
outcomes is ¼+¼+¼+¼ = 1.
Sample Question
Question: A box contains three cards numbered 1, 3, 5.
A second box contains four cards numbered
2, 3, 4, 5. A card is chosen at random from
each box.
(a) Show all the possible outcomes of the
experiment using a possibility diagram or
a tree diagram.
(b) Calculate the probability that
(i) the numbers on the cards are the same,
(ii) the numbers on the cards are odd,
(iii) the sum of the two numbers on the
cards is more than 7.
Solution to Sample Question
(a) Using a possibility diagram:
S = { (1,2), (1,3), (1,4), (1,5), (3,2),
(3,3), (3,4), (3,5), (5,2), (5,3),
(5,4), (5,5) }
5
2nd box
n(S) = 12
4
(b)(i) P(both numbers are the same)
3
= 2  1.
12
2
6
(b)(ii) P(both numbers are odd)
1
3
1st box
5
= 6  1.
12 2
(b)(iii) P(sum > 7) =
4 1
 .
12 3
Solution to Sample Question
(a) Using a tree diagram:
(b)(i) P(both numbers are the same)
1st box
2nd box
1
4
1
3
1
3
5
2
3
4
5
2
3
4
5
2
3
4
5
= P[(3,3) or (5,5)]
=  1  1    1  1    1  1  2  2  1 .

 
 

3 4 3 4 3 4
12
(b)(ii) P(both numbers are odd)
= P[(1,3) or (1,5) or (3,3) or
(3,5) or (5,3) or (5,5)]
=  1  1 6  6  1 .


3 4 
12
2
(b)(iii) P(sum > 7)
= P[(3,5) or (5,3) or (5,4) or (5,5)]
=  1  1   4  4  1 .
3 4 
12
3
6
Websites:
http://mathforum.org/dr.math/faq/faq.prob
.intro.html
http://regentsprep.org/Regents/math/matha.cfm#a6
http://www.bbc.co.uk/schools/ks3bitesize/m
aths/handling_data/index.shtml
Assignment
Task:
Complete the Multiple Choice Questions.
Do your work on foolscap paper and show
your working clearly.
NOTE: Submit your assignment to your Math Teacher
on 05 February 2008.
References:
1. Lee, P. Y., Fan, L. H., Teh, K. S. and
Looi, C. K. (2002) New Syllabus
Mathematics 4 Singapore: Shing Lee
Publishers Pte Ltd.
2. Tay, C. H. (2003) New Mathematics
Counts for Secondary 5 Normal (Academic)
Singapore: Federal Publications.
End
of
e-Lesson