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Unit 8 Events and Probability
Unit 8 Events and
Probability
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ITD1111 Discrete Mathematics & Statistics
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Unit 8 Events and Probability
Events and Probability
‪ What is the probability that a person will
win a lottery where 6 numbers are chosen
from the first 46 positive integers?
‪ The theory of probability arises in the
study of gambling games.
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Unit 8 Events and Probability
8.1 Finite Probability
‪ An experiment is a procedure that yields
one set of possible outcomes.
‪ The sample space is the set of possible
outcomes.
‪ An event is a subset of the sample space.
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Unit 8 Events and Probability
‪ Each possible outcome is called a sample
point and the set of all possible outcomes
is the possibility space S.
‪ If the possibility space is finite, then the
number of sample points in S is denoted
by n(S).
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Unit 8 Events and Probability
‪ Also n(E) denote the number of sample
points in an event E.
‪ Clearly n(E)  n(S).
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Unit 8 Events and Probability
‪ Consider an example for one throw of a
fair die the possibility space
‪
S ={1, 2, 3, 4, 5, 6} and n(S) =6.
‪ Let E1 be the event that the number is even,
then
‪
E1 ={2, 4, 6} and n(E1) =3.
‪ Let E2 be the event that the number is
greater than 2, then
‪
E2 ={3, 4, 5, 6} and n(E2) =4.
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Unit 8 Events and Probability
Laplace’s definition of
probability:
‪ If the possibility space S consists of a
finite number of equally likely outcomes,
‪ then the probability of an event E, written
P(E) is defined as
n E 
PE  
n S 
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Unit 8 Events and Probability
‪ Refer to the previous example,
nE1  3 1
PE1  
 
n S  6 2
‪ and
n  E2  4 2
P  E2  
 
n S  6 3
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Unit 8 Events and Probability
8.2 Certain Events and
Impossible Events
‪ Suppose there are
‪
n sample points in the possibility space
‪ and
‪
r sample points in an event E,
‪ so that
‪
n(S) = n and n(E) = r.
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Unit 8 Events and Probability
‪ The probability of the event E occurs is
n E  r
PE  

n S  n
‪ Since
0  r  n,
‪ then
r
0
 1,
n
‪ hence
0  P(E)  1.
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Unit 8 Events and Probability
‪ That is the probability of an event E is
between 0 and 1 inclusive.
‪ If P(E) = 0 then the event cannot happen,
i.e. impossible event.
‪ If P(E) =1 then the event is certain to
happen, i.e. certain event.
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Unit 8 Events and Probability
‪ For example,
‪ if a coin with both heads is tossed,
‪ the following probabilities will be
obtained.
‪
P(a tail is obtained) = 0
‪
P(a head is obtained) = 1
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Unit 8 Events and Probability
8.3 Complementary Events
‪ Let E be an event in a sample space S.
‪ The probability of the event ,E
‪ the complementary event E, is given by
PE   1  PE 
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Unit 8 Events and Probability
Example 8.3-1
‪ A sequence of 8 bits is randomly generated.
‪ What is the probability that at least one of
these bits is 1?
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Unit 8 Events and Probability
Solution 8.3-1
‪ The possibility space S is the set of all bit
strings of length 8,
‪ so that n( S ) = 28.
‪ Let E be the event that at least one of the 8
bits is 1,
‪ then E is the event that all the bits are 0s
‪ and n( E ) = 1.
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Unit 8 Events and Probability
It follows that
nE 
1
1
255
PE   1  PE   1 
 1 8  1

n S 
2
256 256
‪ Hence, the probability that the bit string will
contain at least one 1 bit is
255
256
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Unit 8 Events and Probability
8.4 The Probability of
Union
‪ Let E1 and E2 be the events in the sample
space S
‪ such that P(E1)  0 and P(E2)  0.
‪ Then
P( E1 or E2 )
=P( E1 ) + P( E2 ) - P( E1 and E2 ).
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Unit 8 Events and Probability
‪ Note‪that‪‘E1 or E2’‪means‪
‪
E1 occurs,
‪
or E2 occurs,
‪
or both E1 and E2 occur.
In set notation
‪ P( E1 E2 ) =P( E1 ) + P( E2 ) - P( E1E2 ).
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Unit 8 Events and Probability
Example 8.4-1
‪ In a class of 40 students,
‪ 6 out of 15 boys and
‪ 13 out of 25 girls wear glasses.
‪ What is the probability that a student
chosen at random from the class is
‪ a boy or
‪ someone who wears glasses?
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Unit 8 Events and Probability
Solution 8.4-1
‪ Let B be the event that the student chosen
is a boy and
‪ let W be the event that the student chosen
wears glasses.
‪ Since
P( B  W ) = P( B ) + P(W) - P( B  W )
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Unit 8 Events and Probability
15 19 6


‪ Then P( B  W ) 
40 40 40
28

40
7

10
‪ Therefore, the probability that a student
chosen at random from the class is a boy or
someone who wears glasses is 0.7.
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Unit 8 Events and Probability
8.5 Mutually Exclusive
Events
‪ Let E1 and E2 be the events in the sample
space S such that
‪ E1 can occur
‪ or E2 can occur
‪ but not both E1 and E2 can occur,
‪ then the two events E1 and E2 are said to
be mutually exclusive.
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Unit 8 Events and Probability
‪ In this case
n( E1  E2 ) =0 and E1  E2 = .
‪ In set notation, when E1 and E2 are
mutually exclusive events
‪ P( E1  E2 ) = P( E1 ) + P( E2 ) and
‪ P( E1  E2 ) = 0
‪ This is known as the addition law for
mutually exclusive events.
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Unit 8 Events and Probability
Example 8.5-1
‪ Suppose that there are eight runners in a
race including John, David and Albert.
The probability that
1
‪ John wins the race is
2
1
‪ David wins the race is
4
1
‪ and Albert wins the race is
8
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Unit 8 Events and Probability
‪ Assume there are no dead heats, find the
probability that
‪ (a) John or David or Albert wins,
‪ (b) neither John nor David wins.
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Unit 8 Events and Probability
Solution 8.5-1
‪ Since we assume that only one runner can
win, the events above are mutually
exclusive.
‪ Let J be the event that John wins the race,
‪ D be the event that David wins the race
and
‪ A be the event that Albert wins the race.
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Unit 8 Events and Probability
‪ (a) Probability that John or David or
Albert wins the race is
‪
P( J or D or A )
‪
= P( J ) + P( D ) + P( A )
1 1 1
  
2 4 8
7

8
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Unit 8 Events and Probability
‪ (b) Probability that neither John nor
David wins the race is
‪
1 - P( J or D )
‪
= 1 - [ P( J ) + P( D ) ]
1 1
 1   
2 4
1

4
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Unit 8 Events and Probability
8.6 Exhaustive Events
‪ Let E1 and E2 be the events in the sample
space S
‪ such that E1  E2 =S
‪ then P( E1  E2 ) =1.
‪ The events E1 and E2 are said to be
exhaustive.
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Unit 8 Events and Probability
Example 8.6-1
‪ Two fair coins are tossed.
‪ A is the event that at least one tail is
obtained.
‪ (a) Describe an event B such that
‪
A and B are exhaustive events
only.
‪ (b) Describe an event C such that
‪
A and C are both mutually
exclusive and exhaustive.
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Unit 8 Events and Probability
Solution 8.6-1
‪ (a) The possibility space
‪
‪
‪
S = {HH, HT, TH, TT}
and the event A = {HT, TH, TT}.
Let B be the event that at least
one head is obtained, then
‪
‪
‪
B = {HH, HT, TH}.
Since AB={HH, HT, TH, TT}=S,
A and B are exhaustive events.
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Unit 8 Events and Probability
‪ (b) Let C be the event that no tail is
obtained, then C = {HH}.
‪
Since A  C={HH, HT, TH, TT} =S
‪
and A  C = ,
‪
A and C are both mutually
exclusive and exhaustive.
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Unit 8 Events and Probability
Example 8.6-2
‪ In a class of 40 students all study at least
one of the subjects computer science and
discrete mathematics.
‪ 27 attend the computer science class and
32 attend the discrete mathematics class.
‪ Find the probability that a student chosen
at random studies both computer science
and discrete mathematics.
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Unit 8 Events and Probability
Solution 8.6-2
‪ Let C be the event that the student chosen
is study computer science and
‪ let M be the event that the student chosen
is study discrete mathematics.
‪ Since all students study at least one of the
subjects computer science and discrete
mathematics, C and M are exhaustive
events.
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Unit 8 Events and Probability
‪ Then
‪ n( C  M ) = n( C ) +n( M ) - n( C M )
‪
40 = 27 +32 - n( C  M )
‪ n( C  M ) = 19
‪ Therefore the probability that a student
chosen at random studies both computer
science and discrete mathematics is
19
‪
P( C  M ) =
40
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