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Mathematics
Session
Probability -2
Session Objectives
 Addition Theorem on Probability
 Conditional Probability
 Multiplication Theorem on Probability
 Independent Events
 Class Exercise
Addition Theorem For 2 Events
If A and B be are events in a sample space S. Then the probability
of occurrence of at least one of the events A and B is given by
P  A  B =P  A  +P B -P  A  B
If A and B are two mutually exclusive events i.e. A  B = 
Then, P  A  B =P  A  +P B
Addition Theorem For 3 Events
If A, B and C are three events in the sample space S. Then,
P  A  B  C  =P  A  +P B  +P C  -P  A  B  -P B  C 
-P  C  A  +P  A  B  C 
If A, B and C are three mutually exclusive events, then
P  A  B  C =P  A  +P B +P C
Addition Theorem For n Events
If A1, A2, A3,... An are mutually exclusive events, then
P  A1  A2  A3 ...  An  =P  A1  +P  A2  +...+P  An 
Event A or not A
The events ‘A’ and ‘not A’ are mutually exclusive events as each
outcome of the experiment is either favourable to ‘A’ or ‘not A’.
Therefore, the event ‘A or not A’ is a certain (or sure) event whose
probability is 1.
P  A or not A  =P  A  A' =1
 P  A  +P  A' =1
 P  A  =1-P  A' or P  A' =1-P  A 
Example-1
Given two mutually exclusive events A and B such P  A  = 1 and
2
P B  =
1
, find P(A or B).
3
Solution:
A and B are two mutually exclusive events.
P  A  B =P  A  +P B
=
1 1 5
+ =
2 3 6
Addition theorem
Example-2
An integer is chosen at random from the first 200 positive integers.
Find the probability that the integer is divisible by 6 or 8.
Solution: Let S be sample space. Then,
S = {1, 2, 3, …200}, n(S) = 200
Let A : event that the number is divisible by 6.
A = {6, 12, 18 ... 198},
n(A) = 33
Let B : event that number is divisible by 8.
B = {8, 16, 24 ... 200},
n(B) = 25
Solution Cont.
(A  B) : event that the number is divisible by 6 and 8.
A  B = {24, 48, ... 192}, n(A  B) = 8
A  B : event that the number is divisible either by 6 or 8.
P  A  B =P  A  +P B -P  A  B
=
33
25
8
+
200 200 200
=
50
1
=
200 4
Example-3
Find the probability of drawing a black king and probability of drawing
a black card or a king from a well-shuffled pack of 52 cards.
Solution: Let A and B be two events such that
A : drawing a black card,
B : drawing a king
Total number of cards n(S) = 52
Number of black cards n(A) = 26
Number of kings n(B) = 4
Solution (Cont.)
Number of black kings n A  B =2
P A =
P B =
n A 
n  S
nB
n  S
=
=
26 1
=
52 2
4
1
=
52 13
P  drawing a black king =P  A  B  =
n A  B
n  S
=
2
1
=
52 26
Solution (Cont.)
and P(drawing a black card or a king)
P  A  B =P  A  +P B -P  A  B
1 1 1
7
= +
=
2 13 26 13
Example-4
Find the probability of getting an even number on the first die or a
total of 8 in a single through of two dice.
(CBSE 2004)
Solution: Let S be sample, then n(S) = 36
Let A and B be two events such that
A = getting an even number on first die, B = getting a total of 8
 A  B = getting an even number on first die and a total of 8
Solution Cont.
A ={(2, 1), (2, 2), …, (2, 6), (4, 1), (4, 2), …, (4, 6),
(6, 1), (6, 2), …, (6, 6)},
B = {(2, 6), (6, 2), (3, 5), (5, 3), (4, 4)}
and A  B = 2, 6  ,  6, 2  ,  4, 4 
P  A  =
18
5
3
, P B  =
and P  A  B  =
36
36
36
Solution (Cont.)
Required probability =P  A  B =P  A  +P B -P  A  B

18 5
3
+

36 36 36

20 5

36 9
Example-5
Two dice are tossed together. Find the probability that the sum of
the numbers obtained on the two dice is neither a multiple of 3 nor
a multiple of 4.
(CBSE 1996)
Solution: Let S be sample, then n(S) = 36
Let A and B be two events such that
A = the sum of the numbers obtained on the two dice is a multiple of 3.,
B = the sum of the numbers obtained on the two dice is a multiple of 4
Solution (Cont.)
Then, A = {(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2),
(4, 5), (5, 1), (5, 4), (6, 3), (6, 6)},
B = {(1, 3), (2, 2), (2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2), (6, 6)}
and A  B =  6, 6 
n A  =12, nB =9 and n A  B =1
P A =
12
9
1
, P B  =
and P  A  B  =
36
36
36
Solution (Cont.)
P  sum is multiple of 3 or 4 =
P  A  B  =P  A  +P B  -P  A  B 
12
9
1
20
=
+
=
36 36 36 36
P sum is neither multiple of 3 nor 4
=1-
20 16 4


36 36 9
Conditional Probability
In a random experiment, if A and B are two events, then the
probability of occurrence of event A when event B has already
occurred and P B  0, is called the conditional probability and it
is denoted by P  A/B .
Cont.
P  A/B  =
Number of outcomes favourable to A which are also favourable to B
Number of outcomes favourable to B
 P  A/B  =
P  A  B
P B
Similarly, P B/A  =
, P B   0
P  A  B
P A
, P  A  0
Multiplication Theorem on Probability
If A and B are two events associated with a random experiment, then
P  A  B =P  A/B×P B, if P B  0
or P  A  B =P  A/B ×P  A  , if P  A   0
Example-6
If A and B are two events such that P  A  = 0.5, P B  = 0.6
and P  A  B  = 0.8. Find P  A / B  and P B / A .


Solution: Given that P(A) = 0.5, P(B) = 0.6 and P A  B =0.8
By addition theorem
P  A  B =P  A  +P B -P  A  B
 0.8=0.5+0.6-P  A  B
 P  A  B =1.1- 0.8=0.3
Solution Cont.
By multiplication theorem
P  A/B  =
P  A  B
P B
and P B/A  =
=
0.3 1
=
0.6 2
P  A  B
P A
=
0.3 3
=
0.5 5
Example –7
A die is rolled twice and the sum of the numbers appearing on them
is observed to be 7. What is the conditional probability that the number
2 has appeared at least once.
(CBSE 1990, 2003)
Solution: Let A and B be two events such that
A = getting number 2 at least once
B = getting 7 as the sum of the numbers on two dice
A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),
(1, 2), (3, 2), (4, 2), (5, 2), (6, 2)}
Solution (Cont.)
B = {(2, 5), (5, 2), (6, 1), (1, 6), (3, 4), (4, 3)}
11
6
2
P A =
, P B  =
and P  A  B  =
36
36
36
Required probability =P  A / B  =
2
1
36
=
=
6
3
36
P  A  B
P B
Independent Events
Two events A and B are said to be independent if occurrence of one
does not affect the occurrence of other; then
P  A/B =P  A  and P B/A  =P B
P  A  B =P  A ×P B, where A and B are independent events.
Example –8
The probability of student A passing an examination is 2
9
and the probability of student B passing is
5
. Find the probability of
9
(i) only A passing the examination
(ii)
only one of them passing the examination
Solution: (i) Let A : event of A passing the examination =
P(A') = 1 – P(A) =1-
2 7
=
9 9
2
9
Solution Cont.
Let B : event of B passing the examination 
P(B’) = 1 – P(B) =1-
5
9
5 4
=
9 9
i P only A passing the examination =P  A  B' =P  A   P B'
[As A and B' are independent events]
2 4
8
= × =
9 9 81
Solution Cont.
(ii)
P(only one passing the examination)
= P   A  B'   A'  B  
= P   A  B'   A'  B  
=P  A ×P B' +P  A'×P B
[As A and B' are independent events and A’ and B are also
independent events]
=
8 7 5
8 35 43
+ × =
+
=
81 9 9 81 81 81
Example –9
Three bags contain 7 white 7 red, 5 white 8 red, and 8 white 6 red
balls respectively. One ball is drawn at random from each bag. Find
the probability that the three balls drawn are of the same colour.
Solution:
W1 : White ball from bag I
W2 : White ball from bag II
W3 : White ball from bag III
R1 : Red ball from bag I
R2 : Red ball from bag II
R3 : Red ball from bag III
Solution (Cont.)
P(Three balls are of same colour)
= P   W1  W2  W3   R1  R2  R3  
=P   W1  W2  W3   R1  R2  R3  
=P W1  ×P W2 ×P W3  +P R1 ×P R2 ×P R3 
[As W1, W2 and W3 are independent events and R1, R2 and R3 are also
independent events]
5
8   7
8
6  22
 7
=  × × + × ×  =
 14 13 14   14 13 14  91
Example –10
A speaks truth in 60% of the cases and B in 90% of the cases. In what
percentage of the cases they are likely to contradict each other in
stating the same fact?
Solution:
Let A : event of A speaking the truth.
B : event of B speaking the truth.
P A =
P B  =
60
40
, P  A' =
100
100
90
10
, P B' =
100
100
(CBSE 2003)
Solution (Cont.)
P(A and B contradicting each other)


=P A  B'   A'  B



=P A  B' +P  A'  B 
=P  A  ×P B' ++P  A'×P B
[As A and B' are independent of each other and A' and B are
also independent of each other.
10   40
90  21
 60
=
×
+
×
=



 100 100   100 100  50
Example –11
The probabilities of A, B and C solving a problem are
1 1
1
, and
2 3
4
respectively. If the problem is attempted by all simultaneously,
find the probability of exactly one of them solving it.
Solution: P  A  =
P B  =
P C  =
1
1 1
 P  A' =1- =
2
2 2
1
1 2
 P B' =1- =
3
3 3
1
1 3
 P C' =1- =
4
4 4
(CBSE 2004)
Solution (Cont.)
Required probability
= P  A  B'  C'   A'  B  C'   A'  B'  C  
=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
[As A, B’ and C’ are independent events; A’, B and C’ are independent
events; A’, B’ and C are also independent events]
1 2 3 1 1 3 1 2 1
= × × + × × + × ×
2 3 4 2 3 4 2 3 4
=
6+3+2 11
=
24
24
Thank you