Event A : The score on the second die > the score on the 1st die.
i.e. A = { (1, 2), (1, 3), (1, 4), (1, 5), (1, 6) (2, 3), (2, 4), (2,
5), (2, 6) (3, 4), (3, 5), (3, 6) (4, 5), (4, 6) (5, 6)}
\
n (A) = 15
Therefore, p (A) =
Example A coin is tossed three times.
Find the probability of getting at least one head.
Solution : Experiment : A coin is tossed three
times.
Sample space : S = {(H H H), (H H T), (HTH), (HTT), (THT), (TTH), (THH), (TTT) }
Þ
n (S) = 8
Event A : getting at least one head
so that A : getting no head at all
= { (TTT) Þ n ()
= 1
P ()
=
Therefore, P (A) = 1  P ( A ) =
Example A ball is drawn at random from
a box containing 6 red balls, 4 white balls and 5 blue balls. Determine
the probability that the ball drawn is (i) red (ii) white (iii)
blue (iv) not red (v) red or white.
Solution : Let R, W and B denote the
events of drawing a red ball, a white ball and a blue ball respectively.
(i)
[next page]

Index
7. 1 Introduction
7.2 Trial
7.3 Sample Space
7. 4 Definition of Probability
7. 5 The Laws of Probability
7. 6 Conditional Probability
7. 7 Theoretical Distribution
7. 8 Binomial Distribution
7. 9 Normal Distribution
Chapter 8
