2. SPECIAL ADDITION LAW OF PROBABILITY
If A and B are two mutually exclusive events then the probability that anyone of them occur
is given by sum of their respective probabilities. That is, P (A B) = P(A) + P(B).
S
A B
m2
m1
n
Let there are n equally likely, mutually exclusive and
collectively exhaustive outcomes in a sample space S.
Out of them, m1 outcomes are favorable to an event A
and m2 outcomes are favorable to an event B.
Therefore, P(A) =
m1
n
and P(B) =
m2
n
Since A and B are mutually exclusive (disjoint sets), therefore, all the outcomes in A and B
are distinct. Then obviously, the number of outcomes contained in A B is m1 + m2. Hence
P(A B) =
m1 + m2
n
=
m1
n
+
m2
n
= P(A) + P(B)
Corollary: If A1, A2,……..,An are n mutually exclusive events, then
P(A1 A2 …… An) = P(A1) + P(A2) + ……….+P(An).
Amjad Ali
2
3. PROBLEM 31
Two dice are rolled. Find the probability that sum of numbers is either 5 or 11.
The sample space is
S ={(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
n(S) = 36
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
A: Sum is numbers is 5
A = {(1,4), (2,3), (3,2), (4,1)} n(A) = 4
P(A) =
n(A)
n(S)
=
𝟒
𝟑𝟔
B: Sum of numbers is 11
B = {(5,6), (6,5)} n(B) = 2
P(B) =
n(B)
n(S)
=
𝟐
𝟑𝟔
A and B are mutually exclusive
P(A B) = P(A) + P(B)
= 4/36 + 2/36 = 6/36
Amjad Ali
3
4. PROBLEM 32
Three horses A, B and C are in a race. A is twice as likely to win as B and B is twice as likely
to win as C, then what are their respective chances of winning
A: Horse A wins the race
B: Horse B wins the race
C: Horse C wins the race
Let C’s chance of winning the race are k, then
P(C) = k
P(B) = 2 k
P(A) = 4 k
Since A, B and C are mutually exclusive and collectively exhaustive events; therefore
P (A) + P (B) + P (C) = 1
k + 2k + 4k = 1
7k = 1
k = 1/ 7
Hence P(C) = 1/ 7
P(B) = 2/ 7
P(A) = 4/7
4
5. 5
PROBLEM 33
Explain what is wrong with each of the following statements
(i) An investment counselor claims the probability that a stock price will go up is 0.60,
remains unchanged is 0.38, or go down is 0.25
(ii) If two coins are tossed, there are three possible outcomes: 2 heads, one head and one
tail, and two tails. Hence probability of each of these outcomes is 1/3
(iii) The probabilities that a certain truck driver would have no, one and two or more
accidents during the year are 0.90, 0.02 and 0.09.
(iv) P(A) =
𝟐
𝟑
. P(B) =
𝟏
𝟒
. P(C) =
𝟏
𝟔
for the probabilities of three mutually exclusive events A, B
and C.
(i) A: Stock price will go up P(A) = 0.60
B: Stock price will remain unchanged P(B) = 0.38
C: Stock price will go down P(C) = 0.25
Since A, B and C are mutually exclusive and collectively exhaustive events; therefore
sum of their probabilities should be equal to 1.
P(A) + P(B) + P(C) = 0.60 + 0.38 + 0.25 = 1.23 which is greater than 1
(ii) When two coin are tossed, S = {HH, HT, TH, TT}
A: getting two heads
A = {HH}
P(A) = 1/ 4
6. 6
B: getting one head and one tail
B = {HT, TH} P(B) = 2/ 4
C: getting two tails
C = {TT} P(C) = 1/ 4
(iii) A: there is no accident P(A) = 0.90
B: there is one accident P(B) = 0.02
C: There are two or more accidents P(C) = 0.09
Since A, B and C are mutually exclusive and collectively exhaustive events; therefore
sum of their probabilities should be equal to 1.
P(A) + P(B) + P(C) = 0.90 + 0.02 + 0.09 = 1.01 which is greater than 1
(iv) P(A) =
𝟐
𝟑
. P(B) =
𝟏
𝟒
. P(C) =
𝟏
𝟔
Here P(A) =
𝟏
𝟔
Also P(A) =
𝟐
𝟑
. P(B) or
𝟐
𝟑
. P(B) =
𝟏
𝟔
or P(B) =
𝟏
𝟔
.
𝟑
𝟐
=
𝟏
𝟒
Also P(A) =
𝟏
𝟒
. P(C) or
𝟏
𝟒
. P(C) =
𝟏
𝟔
or P(C) =
𝟏
𝟔
.
𝟒
𝟏
=
𝟐
𝟑
If A, B and C are mutually exclusive and collectively exhaustive events; then sum of their
probabilities should be equal to 1.
P(A) + P(B) + P(C) =
𝟏
𝟔
+
𝟏
𝟒
+
𝟐
𝟑
=
𝟏𝟑
𝟏𝟐
which is greater than 1
7. GENERAL ADDITION LAW OF PROBABILITY
The If A and B are any two events defined in a sample space S then the probability that at
least one of them occur is given by sum of their respective probabilities minus probability of
their joint occurrence. That is, P (A B ) = P(A) + P(B) – P(A B)
S
A B
A B
B ഥ
𝐀
A ഥ
𝐁
From the Venn diagram, the event A B may be written
as the union of two events A and B ഥ
𝐀 which are
mutually exclusive.
A B = A (B ഥ
𝐀)
P(A B) = P[A (B ഥ
𝐀)]
P(A B) = P(A) + P(B ഥ
𝐀)………(i)
We have,
B = (A B) (B ഥ
𝐀)
P(B) = P[(A B) (B ഥ
𝐀)]
Events (A B) and (B ഥ
𝐀) are mutually exclusive, therefore,
P(B) = P(A B) + P(B ഥ
𝐀)
P(B ഥ
𝐀) = P(B) – P(A B). Put in equation (i), we have
P(A B) = P(A) + P(B) – P(A B)
This is called general law of addition of probability
A Special Case: When two events A and B are mutually exclusive, then P(AB) = 0,
hence, P(A B) = P(A) + P(B). That is, probability that anyone of them occur is given by
sum of their respective probabilities.
Amjad Ali
7
8. PROBLEM 34
An integer is chosen from first 50 positive integers. What is the probability that it is divisible
by 6 or by 8?
Sample Space is
S = {1, 2, 3, ……, 50} ; n(s) = 50
A: Selected number is divisible by 6
A = {6, 12, 18, 24, 30, 36, 42, 48} n(A)= 8
P(A) =
n(A)
n(S)
=
𝟖
𝟓𝟎
B: Selected number is divisible by 8
B = {8, 16, 24, 32, 40, 48} n(B)= 6
P(B) =
n(B)
n(S)
=
𝟔
𝟓𝟎
A B: Selected number is divisible by both 6 and 8
A B = {24, 48} n(A B)= 2
P A ∩ B =
𝐧(A ∩ B)
n(S)
=
𝟐
𝟓𝟎
Since A and B are not mutually exclusive
P(A B) = P(A) + P(B) – P(A B)
=
𝟖
𝟓𝟎
+
𝟔
𝟓𝟎
−
𝟐
𝟓𝟎
=
𝟏𝟐
𝟓𝟎
Amjad Ali
8
9. PROBLEM 35
Two dice are rolled. If A and B, respectively, the events that the sum of points is 8 and both
dice should give odd numbers, then find P(A B)
The sample space is
S ={(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} n(S) = 36
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
A: Sum of numbers is 8
A = {(2,6),(3,5),(4,4),(5,3),(6,2)};
n(A) = 5 P(A) = 5/36
B: Sum numbers is odd
B = {(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(3,6),
(4,1),(4,3),(4,5),(5,2),(5,4),(5,6),(6,1),(6,3),(6,5)};
n(B) = 9 P(B) = 9/ 36
A B = {(3,5), (5,3)};
n(A B) = 2 P(A B) = 2/ 36
Since A and B are not mutually exclusive
P(AB) = P(A) + P(B) –P(AB) = 5/36 + 9/36 – 2/36 = 12/36 Amjad Ali
9
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
10. PROBLEM 36
Two good dice are rolled simultaneously. Let A denote the event that the sum shown is 8 and
B the event that the two dice show the same number. Find P(A), P(B), P(A B) and P(A B)
The sample space is
S ={(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} n(S) = 36
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
A: Sum of number is 8
A = {(2,6), (3,5), (4,4), (5,3), (6,2)}; n(A) = 5
P(A) = 5/ 36
B: Both dice show the same number
A = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}; n(B) = 6
P(B) = 6/ 36
A B = {(4, 4)}; n(A B) = 1
P(A B) = 1/ 36
Since A and B are not mutually exclusive
P(A B) = P(A) + P(B) – P(A B)
= 5/36 + 6/ 36 – 1/ 36 = 10/ 36 Amjad Ali
10
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
11. PROBLEM 37
A is card drawn from an ordinary deck. If A is the event of getting a red card and B is the event
that the card is greater than 2 but less than 9. Find P(A), P(B), P(A and B), and P(A or B).
S ={ HA H2 H3 H4 H5 H6 H7 H8 H9 H10 HJ HQ HK
DA D2 D3 D4 D5 D6 D7 D8 D9 D10 DJ DQ DK
SA S2 S3 S4 S5 S6 S7 S8 S9 S10 SJ SQ SK
CA C2 C3 C4 C5 C6 C7 C8 C9 C10 CJ CQ CK}
A: Selecting a red card
A = { HA, H2,H3,H4,H5,H6,H7,H8, H9,H10,HJ,HQ,HK
DA, D2, D3,D4,D5,D6,D7,D8,D9,D10, DJ,DQ,DK }; n(A) = 26
P(A) = 26/ 52
B: Selecting a card numbered greater than 2 but less than 9
B = { H3,H4,H5,H6,H7,H8, D3,D4,D5,D6,D7, D8,
S3, S4, S5, S6, S7, S8, C3,C4, C5,C6, C7, C8}; n(B) = 24
P(B) = 24/ 52
A B = {H3,H4,H5,H6,H7,H8,D3,D4,D5,D6,D7, D8}; n(A B)= 12
P (A B) = 12/ 52
Since A and B are not mutually exclusive
P(A B) = P(A) + P(B) – P(A B)
= 26/ 52 + 24/ 52 – 12/ 52 = 28/ 52 = 7/13 Amjad Ali
11
12. PROBLEM 38
A class contains 10 men and 20 women, of which half of the men and half of the women
have brown eyes. Find the probability that a person chosen at random is a man or has
brown eyes.
Men Women Total
Brown
Eyes
5 10 15
Not Brown
Eyes
5 10 15
Total 10 20 30
A: Person is a man n(A) = 10
P(A) = 10/30
B: Person has brown eyes n(B) = 15
P(B) = 15/30
A B: Person is a man with brown eyes
n(A B) = 5
P (A B) = 5/30
P(A B) = P(A) + P(B) – P(A B)
= 10/30 + 15/30 – 5/30 = 2/3
A and B are not mutually
exclusive
Amjad Ali
12
13. PROBLEM 39
A drawer contains 50 bolts and 150 nuts. Half of the bolts and half of the nuts are rusted. If
one item is selected at random, what is the probability that it is rusted or is a bolt?
Bolts Nuts Total
Rusted 25 75 100
Not Rusted 25 75 100
Total 50 150 200
A: Item is rusted n(A) = 100
P(A) = 100/200
B: Item is bolt n(B) = 50
P(B) = 50/200
A B: Item is a rusted bolt n(A B) = 5
P(A B) = 25/200
P(A B) = P(A) + P(B) – P(A B)
= 100/200 + 50/200 – 25/200
= 125/200
A and B are not mutually
exclusive
Amjad Ali
13
14. PROBLEM 40
If A and B are any two events defined in a sample space S, show that
P[(A ഥ
𝐁) (B ഥ
𝐀] = P(A) + P(B) – 2 P (A B)
S
A B
A B
B ഥ
𝐀
A ഥ
𝐁
We have
A = (A ഥ
𝐁) (A B)
P(A) = P[(A ഥ
𝐁) (A B)]
Events (A ഥ
𝐁) and (A B) are mutually exclusive,
therefore,
P(A) = P(A ഥ
𝐁) + P(A B)
P(A ഥ
𝐁) = P(A) – P(A B)
We have
B = (A B) (B ഥ
𝐀)
B = (A B) (ഥ
𝐀 B)
P(B) = P[(A B) (ഥ
𝐀 B)]
Events (A B) and (ഥ
𝐀 B) are mutually exclusive, therefore,
P(B) = P(A B) + P(ഥ
𝐀 B)
P(ഥ
𝐀 B) = P(ഥ
𝐀) – P(A B)
Now the events A ഥ
𝐁 and ഥ
𝐀 B are mutually exclusive,
P[(A ഥ
𝐁) (ഥ
𝐀 B)] = P[(A ഥ
𝐁) + P ( ഥ
𝐀 B)]
= P(A) P(A B) + P(B) P(A B)
= P(A) + P(B) 2 P(A B) Amjad Ali
14
15. PROBLEM 41
Of the households in Lahore 45% have a freezer and 65% have a color T.V set. Given that
30% of the households have both a freezer and color T.V set, calculate the probability that
household has either a freezer or a color T.V. set but not both.
A: Household has a freezer
P(A) = 0.45
B: Household has a color T.V. set
P(B) = 0.65
A B: Household has a both freezer and color T.V.
P(A B) = 0.30
P[(A ഥ
𝐁) (ഥ
𝐀 B)] = P(A) + P(B) 2 P(A B)
= 0.45 + 0.65 2(0.30) = 0.50
Amjad Ali
15
16. PROBLEM 42
For three events A, B, and C, which are not mutually exclusive events, the probability that at
least one of them occur is given by
P(A B C) = P(A) + P(B) + P(C) P(A B) P(B C) P(A C) + P(A B C)
S
A B
C
Let A B C = A (B C)
= A K where K = B C
Then P(A B C) = P(A K)
= P(A) + P(K) P(A K) -------- (i)
Now P(K) = P(B C)
= P(B) + P(C) P(B C)…………….(ii)
and P(A K) = P[A (B C)]
= P [(A B) P(A C)]
= P(A B) + P(A C) – P(A B C)---- (iii)
Substituting the results from (ii) and (i) in (i), we have
P(A B C) = P(A) + P(B) + P(C) – P(B C) – P(A B) – P (A C) + P (A B C)
P(A B C) = P(A) + P(B) + P(C) – P (A B) – P (B C) – P(A C) + P (A B C)
Corollary: If A, B and C are mutually exclusive events, then
P(A B) = P (B C) = P(A C) = P(A B C) = 0,
therefore, P(A B C) = P(A) + P(B) + P(C)
called addition of probability for three mutually exclusive event A, B, and C.
Amjad Ali
16
