Learning Outcomes:
At the end of the lesson, students should be
able to:
(a) understand independent events
(b) find the probability of independent
events.
Let A is the event of the ‘ result of flipping a
fair coin’ and B is the event of the
‘ number obtained when a pair of dice is
rolled’.
These two events are unrelated.
Obtaining a head on the coin will not
influence the outcome of the dice.
Such events are said to be independent.
From the probability rule for conditional
events,
P( A  B)
P( A | B) 
P( B)
Then, we have
P( A  B)  P( B)  P( A | B)
Definition
If A and B are independent events, it means
that the outcome of one event does not affect
the outcome of the other, then
P ( A | B )  P ( A)
and P ( B | A)  P ( B )
Thus,
P( A  B)  P( A | B)  P( B)
 P ( A)  P ( B )
P( A  B )  P( A)  P( B )
if A and B are two
independent events
Remark
If A and B are independent events, then
(A and B’), (A’ and B), and (A’ and B’) are
independent events too i.e.
P ( A ' B ')  P ( A ')  P ( B ')
Example 1:
Suppose two events A and B are independent.
Given P(A) = 0.4 and P(B) = 0.25.
Calculate:
a)P( A  B)
b) P ( A  B )
A and B are
independent
Solution:
a ) P ( A  B )  P ( A)  P ( B )
 (0.4)(0.25)
 0.1
A and B are
independent
b ) P ( A  B )  P ( A)  P ( B )  P ( A  B )
 P ( A)  P ( B )  P ( A)  P ( B )
 0.4  0.25  0.1
 0.55
Example 2
A, B and C are three events such that A and
B are independent whereas A and C are
mutually exclusive .
Given P(A) = 0.4 , P(B) = 0.2 , P(C) = 0.3 and
P(B ∩ C ) = 0.1 . Find
(a) P(A U B)
(b) P( C | B )
(c) P( C | A’)
A and B are
independent
Solution:
a) P(A U B) = P ( A)  P ( B )  P ( A  B )
 P ( A)  P ( B )  P ( A)  P ( B )
 0.4  0.2  (0.4)(0.2)
 0.52
b) P( C | B )  P (C  B )
P(B)
0.1

 0.5
0.2
c) P( C | A’)  P (C  A ')
P ( A ')
P (C )

1  P ( A)
A
B
C
0.3

1  0.4
 0.5
C  A'  C
Example 3
A mathematics puzzle is given to three
students Amin, Ali and Abu. From the past
experience, known that the probabilities
Amin, Ali and Abu will get the correct
solutions are 0.65, 0.6 and 0.55 respectively.
If three of them attempt to solve the puzzle
without consulting each other, find the
probability that:
a) the puzzle will be solved correctly by all of
them.
b) only one of them will get the correct solution.
Solution:
Let A= the event that Amin answers correctly
B = the event that Ali answers correctly
C = the event that Abu answers correctly
P(A) = 0.65, P(B) = 0.60 and P(C)=0.55
a)The event that the puzzle will be solved correctly
by all of them is the event
A B C
P ( A  B  C )  P ( A)  P ( B )  P (C )
 (0.65)(0.60)(0.55)
 0.21
b) The events that only one of them will get the
correct solution will occur if one of the events
( A  B ' C ') , ( A ' B  C ') or P ( A ' B ' C )
occurs and all these events are mutually exclusive.
Thus,
P ( A  B ' C ')  P ( A ' B  C ')  P ( A ' B ' C )
 P  A  P  B '   P  C '   P  A '   P  B   P  C '  
P  A '  P  B '  P  C 
 (0.65)(0.4)(0.45)  (0.35)(0.6)(0.45) 
(0.35)(0.40)(0.55)
 0.29
Example 4
There are 60 students in a certain college, 27 of
them are taking Mathematics, 20 are taking
Biology and 22 are taking neither Mathematics nor
Biology.
a)Find the probability that a randomly selected
student takes
(i) both Mathematics and Biology.
(ii) Mathematics only.
b) A student is selected at random. Determine
whether the event ‘ taking Mathematics’ is
statistically independent of the event ‘ taking
Biology’ .
Solution:
M
27-x
x
20-x B
22
Let B= the event of taking Biology
M= the event of taking Mathematics
x = the number of students taking both
Biology and Mathematics
Total number of students ,n(S) =60
a)
27 - x + x + 20 - x + 22 = 60
x=9
(i)
(ii)
P( B  M ) 
9
60
3

20
P ( M  B ')  18
60
3

10
b)
P( M  B) 
3
20
27
9
20 1
P( M ) 

, P ( B) 

60 20
60 3
P( M ) 
9  1 

P( B)     
 20   3 
3

20
P( M  B)  P( M )  P( B)
Hence, the two events are independent.
Example 5
The probability that Roy is late for
college on any day is 0.15 and is
independent of whether he was late on the
previous day. Find the probability that he
a) is late on Monday and Tuesday
b) arrives on time on one of these days
Solution:
0.15
0.85
0.15
LATE
0.85
LATE
ON TIME
0.15
LATE
0.85
ON TIME
ON TIME
Monday
Tuesday
a) P( late on Monday and Tuesday )
= (0.15)(0.15)
=
0.0225
b) P( arrives on time on one of these days )
= (0.15) (0.85) + (0.15)(0.85)
= 0.1275 + 0.1275
= 0.2550
EXERCISE
1. D and E are two independent events such that
P(D) = 0.3 and P(E) = 0.4 . Find
a)
P(DE)
Answer : 0.12
b)
P(DE)
Answer : 0.58
2. Three events, A , B and C are such that A and B
are mutually exclusive and
P(A)=0.3, P(C) = 0.4 , P(AB) = 0.5 and
P(BC) = 0.54.
Calculate P(B) and P(BC) Answer : 0.2 , 0.06
Determine whether or not B and C are
independent events.
Answer : B and C are independent events
CONCLUSION
The Independent Probability
P( A | B )  P( A)
P( B | A)  P( B )
P( A  B )  P( A)  P( B )