# THEOREMS ON PROBABILITY – I - Quantitative Techniques for management

Theorem :P( Φ) = 0, where f is a null set.

Proof: For a sample space S of an experiment, we can write SU Φ = S .

Taking probability of both sides, we have

P(S U Φ) = P(s).

Since S and f are mutually exclusive, using axiom III, we can write

P(S) + P(f ) =P(S). Hence,P( Φ) = 0 .

Theorem :

P(Ā) 1 P(A), where Ā is compliment of A.

Proof: Let A be any event in the sample space S. We can write

A U Ā = S or P(A U Ā) = P(S)

Since A and Ā are mutually exclusive, we can write P(A) + P(Ā) = P(S) =1. Hence, P(Ā) =1- P(A) .

Theorem : For any two events A and B in a sample space S

P(Ā ∩ B) = P(B) - P(A ∩ B)

Proof: From the Venn diagram, we can write B = (Ā ∩ B)U(A ∩ B) or Since (A ∩ B) and (A ∩ B) are mutually exclusive,

we have

P(B) = P(Ā ∩ B) + P(A ∩ B)
or P(Ā ∩ B) = P(B) - P( A∩ B) .

Similarly, it can be shown that

P(Ā ∩ B) = P(A) - P(A ∩ B)

P(A U B) = P(A) + P(B) - P(A ∩ B)

Proof: From the Venn diagram, given above, we can write

A U B = A U(Ā ∩ B) or P(AU B) = P(A U(Ā ∩ B))

Since A and (Ā ∩ B) are mutually exclusive, we can write

P(A U B) = P( A) + P(Ā ∩ B)

Substituting the value of P(Ā ∩ B) from theorem 3, we get

P(A U B) = P(A) + P(B) - P(A ∩ B)

Remarks :

1. If A and B are mutually exclusive, i.e., A ∩ B = Φ , then according to theorem , we have P(A ∩ B) = 0 . The addition rule, in this case, becomes P(AU B) = P(A) + P(B) ,which is in conformity with axiom III.
2. The event AUB denotes the occurrence of either A or B or both. Alternatively, it implies the occurrence of at least one of the two events.
3. The event A∩B is a compound event that denotes the simultaneous occurrence of the two events.
4. Alternatively, the event AUB is also denoted b A + B and the event A∩B by AB.

Corollaries:

1. From the Venn diagram, we can write P(A U B) = 1- P(Ā∩ B) , where P(Ā∩ B) is the probability that none of the events A and B occur simultaneously.
2. P(exactly one of A and B occursg = P[(A ∩ B)u(Ā ∩ B)]
= P(A ∩B) + P(Ā ∩ B) Since (A∩ B)U(Ā ∩ B) = Φ
= P(A) - P(A∩ B) + P(B) - P(A∩ B) (using theorem 3) = P(AU B) - P(A∩ B) (using theorem 4)
3. The addition theorem can be generalised for more than two events. If A, B and C are three events of a sample space S, then the probability of occurrence of at least one of them is given by
= P(AUBUC)=P[AU(BUC)]= P(A)+p(BUC)-P[A ∩ (BUC)]
= P(A)+P(BUC)-P[(A ∩ B)U (A ∩ C)]

Applying theorem 4 on the second and third term, we get

=P(A)+P(B)+P(C)-P(A ∩ B)-P(A ∩ C)-P(B ∩ C)+P(A ∩ B ∩ C)

Alternatively, the probability of occurrence of atleast one of the three events can also be written as

P(AU BUC) = 1- P(Ā ∩ BC)

If A, B and C are mutually exclusive, then equation (1) can be written as

P(AUBUC) = P(A)+P(B)+P(C)

If A1, A2, ...... An are n events of a sample space S, the respective equations (1),(2) and (3) can be modified as

P(A1 U A2....U An)= Σ P(Ai)-ΣΣP(Ai∩ Aj)+ΣΣΣ P(Ai∩ Aj∩ Ak)+(-1)nP(A1∩ A2∩ ...∩ An)(i ≠j≠ k,) P(A1U A2....U An)=1- P(A1A2∩....∩An)
P(A1 U A2....U An)=Σi=1n P(Ai)

(if the events are mutually exclusive)

4. The probability of occurrence of at least two of the three events can be written as
P[(A ∩ B) U(B ∩ C)U (A ∩ C)]=P(A ∩ B)+ p(B ∩ C) +P(A ∩ C)- 3P(A ∩B∩C)+P(A ∩B∩C) = P[(A ∩ B) U(B ∩ C)U (A ∩ C)]=P(A ∩ B)+ p(B ∩ C) +P(A ∩ C)- 2P(A ∩B∩C)
5. The probability of occurrence of exactly two of the three events can be written as
P[(A ∩ B ∩ C) U (A ∩ B∩ C)U(A∩ B ∩ C)]= P[( A ∩ B)U (B ∩ C) U(A ∩ C)]- P(A∩B∩C)
6. The probability of occurrence of exactly one of the three events can be written as
P (A ∩ BC)U(A∩ B ∩C)U(AB∩C) = P(at least one of the three events occur)- P(at least two of the three events occur). = P(A) + P(B) + P(C) - 2P(A ∩B) - 3P(B ∩C) - 2P(A ∩C) + 3P(A∩B∩C).

Example : In a group of 1,000 persons, there are 650 who can speak Hindi, 400 can speak English and 150 can speak both Hindi and English. If a person is selected at random, what is the probability that he speaks (i) Hindi only, (ii) English only, (iii) only one of the two languages, (iv) at least one of the two languages?

Solution: Let A denote the event that a person selected at random speaks Hindi and B denotes the event that he speaks English.

Thus, we have n(A) = 650, n(B) = 400, n(A∩B) = 150 and n(S) = 1000, where n(A), n(B), etc. denote the number of persons belonging to the respective event.

(i) The probability that a person selected at random speaks Hindi only, is given by

P(A ∩B)= n(A)/n(S)-n(A ∩ B)/n(S)= 650/1000-150/1000= 1/2

(ii) The probability that a person selected at random speaks English only, is given BY

P(Ā ∩ B)= n(B)/n(S)-n(A ∩ B)/n(S)= 400/1000-150/1000=1/4

(iii)The probability that a person selected at random speaks only one of the languages, is given by

P[(A ∩ B))U(Ā ∩ B)]= P(A)+P(B)- 2P(A ∩ B) = n(A)+n(B)-2 n (A ∩ B)/n(S)= 650+400-300/1000=3/4

(iV)The probability that a person selected at random speaks at least one of the languages, is given by

P(A U B)= 650+400-150/1000= 9/10

Alternative Method: The above probabilities can easily be computed by the following nine-square table : From the above table, we can write

(i) p(A ∩ B) = 500/1000=1/2
(ii) p(A ∩ B) = 250/1000=1/4
(iii) P[(A ∩ B)U(Ā ∩ B)] = 500+200/1000= 3/4
(iv) P(AU B) = 150+500+250/1000= 9/10

This can, alternatively, be written as

P(AU B) = 1 - P(Ā ∩ B)= 9/10.

Example : What is the probability of drawing a black card or a king from a wellshuffled pack of playing cards?

Solution: There are 52 cards in a pack,

∴ n(S) = 52.

Let A be the event that the drawn card is black and B be the event that it is a king.

We have to find P(AU B) .

Since there are 26 black cards, 4 kings and two black kings in a pack, we have

n(A) = 26, n(B) = 4 and n(AUB) = 2 Thus, P(AU B) = 26+4-2/52=7/13

Example : A pair of unbiased dice is thrown. Find the probability that (i) the sum of spots is either 5 or 10, (ii) either there is a doublet or a sum less than 6.

Solution:

Since the first die can be thrown in 6 ways and the second also in 6 ways, therefore, both can be thrown in 36 ways (fundamental principle of counting). Since boththe dice are given to be unbiased, 36 elementary outcomes are equally likely.

(i) Let A be the event that the sum of spots is 5 and B be the event that their sum is 10. Thus, we can write

A = {(1, 4), (2, 3), (3, 2), (4, 1)} and B = {(4, 6), (5, 5), (6, 4)}

We note that

(A∩B) = Φ ,

i.e. A and B are mutually exclusive.
∴ By addition theorem, we have

P(AU B) = P(A) + P(B) = 4/36+3/36 = 7/36

(ii) Let C be the event that there is a doublet and D be the event that the sum is less than 6. Thus, we can write

C = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} and
D = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

Further, (C ∩D) = {(1, 1), (2, 2)}

By addition theorem, we have P(CU D) = 6/36+10/36 -2/36 = 7/18

Alternative Methods:
(i) It is given that n(A) = 4, n(B) = 3 and n(S) = 36. Also n(A∩B) = 0 . Thus, the corresponding nine-square table can be written as follows : From the above table, we have P(A∪B) = 1 -
29
36
=
7
36

(ii) Here n(C) = 6, n(D) = 10, n(C∩D) = 2 and n(S) = 36. Thus, we have Thus, P(C∪D) = 1- P(CD = 1 -
22
36
=
7
18

Example: Two unbiased coins are tossed. Let A1 be the event that the first coin shows a tail and A2 be the event that the second coin shows a head. Are A1 and A2 mutually exclusive? Obtain P(A1∩A2) and P(A1∪A2). Further, let A1 be the event that both coins show heads and A2 be the event that both show tails. Are A1 and A2 mutually exclusive? Find P(A1∩A2) and P(A1∪A2). .

Solution: The sample space of the experiment is S = {(H, H), (H, T), (T, H), (T, T)}
(i) A1 = {(T, H), (T, T)} and A2 = {(H, H), (T, H)}

Also A1∩A2 = {(T, H)}, Since A1∩ A2≠ Ø , A1 and A2 are not mutually exclusive. Further, the coins are given to be unbiased, therefore, all the elementary events are equally likely.

P(A1)=
1
2
, P(A2)=
1
2
, P(A1∩A2)=
1
4
P(A1∪A2)=
3
4

(ii) When both the coins show heads; A1 = {(H, H)},When both the coins show tails; A2 = {(T, T)}
Here A1∩ A2= Ø ,.....A1 and A2 are mutually exclusive. Thus, P (A1∪ A2)=1/2

Alternatively, the problem can also be attempted by making the following nine square tables for the two cases : Theorem : Multiplication or Compound Probability Theorem: A compound event is the result of the simultaneous occurrence of two or more events. For convenience, we assume that there are two events, however, the results can be easily generalised. The probability of the compound event would depend upon whether the events are independent or not. Thus, we shall discuss two theorems; (a) Conditional Probability Theorem, and (b) Multiplicative Theorem for Independent Events.

(a) Conditional Probability Theorem: For any two events A and B in a sample space S, the probability of their simultaneous occurrence, is given by

P(A∩B)=P(A)P(B/A)=P(B)P(A/B)

or equivalently =P(B)P(A/B) Here, P(B/A) is the conditional probability of B given that A has already occurred. Similar interpretation can be given to the term P(A/B).

Proof: Let all the outcomes of the random experiment be equally likely. Therefore,

P(A∩B)=
n(A∩B)
n(S)
=
no. of elements in (A∩B)
no. of elements in sample space

For the event B/A, the sample space is the set of elements in A and out of these the number of cases favourable to B is given by n(A∩B)

P(B/A) =
n(A∩B)
n(A)

If we multiply the numerator and denominator of the above expression by n(S), we get

P(B/A) =
n(A∩B)
n(A)
x
n(S)
n(S)
=
P(A∩B)
P(A)

or P(A∩B) = P(A).P(B/A).

The other result can also be shown in a similar way.

Note: To avoid mathematical complications, we have assumed that the elementary events are equally likely. However, the above results will hold true even for the cases where the elementary events are not equally likely.

(b) Multiplicative Theorem for Independent Events: If A and B are independent, the probability of their simultaneous occurrence is given by P(A∩B)= P(A).P(B).

Proof: We can write A = (A∩B)∪(A∩B).
Since(A∩B)and(A∩B) are mutually exclusive, we have
P(A) = P(A ∩ B) + P(A ∩ B)
= P(B).P(A/B) + P(B)).P(A/B))
If A and B are independent, then proportion of A's in B is equal to proportion of A's in B’s, i.e., P(A/B)P(A/B).

Thus, the above equation can be written as

n(B) =
600x30
100
+
400x5
100
= 200

Substituting this value in the formula of conditional probability theorem, we get P(A∩B) = P(A).P(B).

Corollaries:

1. If A and B are mutually exclusive and P(A).P(B) > 0, then they cannot be independent since P(A ∩B) = 0 .
2. If A and B are independent and P(A).P(B) > 0, then they cannot be mutually exclusive since P(A∩B) > 0
1. Generalisation of Multiplicative Theorem : If A, B and C are three events, then
P(A∩B∩C)=P(A).P(B/A).P[(C/A∩B)]
Similarly, for n events A1, A2, ...... A.n, we can write
P(A1∩A2∩...............∩An)=P(A1)P(A2/A1).P(A3/A1∩A2)...........P(An/A1∩A2∩.......∩An-1)
Further, if A1, A2, ...... An are independent, we have
P(A1∩A2∩.......∩An)=P(A1).P(A2).........P(An).
2. If A and B are independent, then A and B , A and B, A and B are also independent.
we can write P(A∩B) = P(A)-P(A∩B)
P(A)-P(A).P(B)=P(A)[1-P(B)]=P(A).P(B)
which shows that A and B are independent.
3. The probability of occurrence of at least one of the events A1, A2, ...... An is given by P(A1∪A2∪...... ∪An)=1-P(Ā1∩Ā2∩.......∩Ān)
If A1, A2, ...... An are independent then their compliments will also be independent, therefore, the above result can be modified as
P(A1∪A2∪...... ∪An)=1-P(Ā1).P(Ā2)........P(Ān)

Pair-wise and Mutual Independence

Three events A, B and C are said to be mutually independent if the following conditions are simultaneously satisfied :

P(A∩B) = P(A).P(B), P(B∩C) = P(B).P(C), P(A∩C) = P(A).P(C) and P(A∩B∩C) = P(A).P(B).P(C)

If the last condition is not satisfied, the events are said to be pair-wise independent. From the above we note that mutually independent events will always be pair-wise independent but not vice-versa.

Example: Among 1,000 applicants for admission to M.A. economics course in a University, 600 were economics graduates and 400 were non-economics graduates; 30% of economics graduate applicants and 5% of non-economics graduate applicants obtained admission. If an applicant selected at random is found to have been given admission, what is the probability that he/she is an economics graduate?

Solution: Let A be the event that the applicant selected at random is an economics graduate and B be the event that he/she is given admission We are given
n(S) = 1000, n(A) = 600, n(Ā) = 400

Also, n(B)=
600x30
100
+
400x5
100
=200
And, n(A∩B)=
600x30
100
=180
Thus, the required probability is given by P(A/B)=
n(A∩B)
n(B)
=
180
200
=
9
10

Alternative Method: Writing the given information in a nine-square table, we have From the above table we can write P(A/B) =
180
200
=
9
10

Example : A bag contains 2 black and 3 white balls. Two balls are drawn at random one after the other without replacement. Obtain the probability that (a) Second ball is black given that the first is white, (b) First ball is white given that the second is black.
Solution: First ball can be drawn in any one of the 5 ways and then a second ball can be drawn in any one of the 4 ways. Therefore, two balls can be drawn in 5 x 4 = 20 ways. Thus, n(S) = 20.

(a) Let A1 be the event that first ball is white and A2 be the event that second is black. We want to find P (A2/A1).

First white ball can be drawn in any of the 3 ways and then a second ball can be drawn in any of the 4 ways, ∴ n(A1) = 3 x 4 = 12.

Further, first white ball can be drawn in any of the 3 ways and then a black ball can be drawn in any of the 2 ways, ∴ n (A1∩A2) = 3 x 2 = 6

Thus ,P(A2/A1) =
n (A1∩A2)
n(A1)
=
6
12
=
1
2

(b) Here we have to find P (A1/A2). The second black ball can be drawn in the following two mutually exclusive ways:

1. First ball is white and second is black or
2. both the balls are black.

Thus, n(A2) = 3 x 2 + 2 x 1 = 8,

∴ P(A1/A2) =
n (A1∩A2)
n(A1)
=
6
8
=
3
4

Alternative Method: The given problem can be summarised into the following ninesquare table: The required probabilities can be directly written from the above table

Example : An urn contains 3 red and 2 white balls. 2 balls are drawn at random. Find the probability that either both of them are red or both are white.
Solution:Let A be the event that both the balls are red and B be the event that both the balls are white. Thus, we can write

n(S) = 5C2 =10,n(A) = 3C2=3, n(B) = 2C2, also n(A∩B)=0
The required probability is P(A∪B)=
n(A)+n(B)
n(S)
=
3+1
10
=
2
5

Example : A bag contains 10 red and 8 black balls. Two balls are drawn at random. Find the probability that (a) both of them are red, (b) one is red and the other is black.
Solution: Let A be the event that both the balls are red and B be the event that one is red and the other is black.

Two balls can be drawn from 18 balls in 18C2 equally likely ways.

n(S)=18C2
=
18!
2!16!
=153

(a) Two red balls can be drawn from 10 red balls in 10C2 ways.

n(S)=10C2
=
10!
2!8!
=45
Thus, P(A) =
n(A)
n(s)
=
45
153
=
5
17

(b) One red ball can be drawn in 10C1ways and one black ball can be drawn in 8C1ways

n(B) =10C1X 8C1=10 X 8 =80
Thus,P(B)=
80
153

Example: Five cards are drawn in succession and without replacement from an ordinary deck of 52 well shuffled cards :

1. What is the probability that there will be no ace among the five cards?
2. What is the probability that first three cards are aces and the last two cards are kings?
3. What is the probability that only first three cards are aces?
4. What is the probability that an ace will appear only on the fifth draw?

Solution:

(a)

P(there is no ace) =
48X47X46X45X44
52X51X50X49X48
= 0.66

(b)

P(first three card are aces and last two are kings) =
4X3X2X4X3
52X51X50X49X48
=0.0000009

(c)

P(only first three card are aces) =
4X3X2X48X47
52X51X50X49X48
=0.00017

(d)

(an ace appears only on 5thdraw) =
48X47X46X45X44
52X51X50X49X48
=0.0059

Example : Two cards are drawn in succession from a pack of 52 well-shuffled cards. Find the probability that

1. Only first card is a king.
2. First card is jack of diamond or a king.
3. At least one card is a picture card.
4. Not more than one card is a picture card.
5. Cards are not of the same suit.
6. Second card is not a spade.
7. Second card is not a spade given that first is a spade.
8. The cards are aces or diamonds or both.

Solution:

(a)

P(only first card in king) =
4X48
52X51
=
16
221

(b)

P(first card is a jack of diamond or a king) =
5X51
52X51
=
5
52

(c)

P(at least one card is a picture card) =
(52X51)-(40X39)
52X51
=
7
17

(d)

P(not more than one card is a picture card) =
(40X39)+(40X12)+(12X40)
52X51
=
210
221

(e)

P(cards are not of the same unit) =
52X39
52X51
=
13
17

(f)

P(second card is not a spade) =
(13X39)+(39X38)
52X51
=
3
4

(g)

P(second card is not a spade given that first is spade) =
39
51
=
13
17

(h)

P(the cards are aces or diamonds or both) =
16X15
52X51
=
20
221

Example : The odds are 9 : 7 against a person A, who is now 35 years of age, living till he is 65 and 3 : 2 against a person B, now 45 years of age, living till he is 75. Find the chance that at least one of these persons will be alive 30 years hence.

Solution:

Note: If a is the number of cases favourable to an event A and a is the number of cases favourable to its compliment event (a + α = n), then odds in favour of A are a : α and odds against A are a :α.

Obviously,P(A) =
a
a + α
andP(Ā) =
α
a + α

Let A be the event that person A will be alive 30 years hence and B be the event that person B will be alive 30 years hence.

P(A) =
7
16
andP(B) =
2
5

We have to find P(A∪B) . Note that A and B are independent.

P(A ∪ B) =
(7X5)+(2X16)-14
80
=
53
80
P(A∪B)=
80-27
80
=
53
80

Example : If A and B are two events such that

P(A)=
2
3
P(Ā∩B)=
1
6
P(A∩B)=
1
3
find P(B),P(A∪B),P(A/B),P(B/A),P(Ā∪B),P(Ā∩B)and P(B)

Also examine whether the events A and B are : (a)Equally likely, (b)Exhaustive, (c)Mutually exclusive and (d)Independent

Solution: The probabilities of various events are obtained as follows :

P(A)=P(Ā∩B)+P(A∩B)=
9
18
=
1
2
P(A∪B)=
2
3
+
1
2
-
1
3
P(A⁄B)=
P(A∩B)
P(B)
=
1
3
x
2
1
=
2
3
P(B⁄A)=
P(A∩B)
P(A)
=
1
3
x
3
2
=
1
2
P(Ā∪B)=P(Ā)+P(B)-P(Ā∩B)
1
3
+
1
2
-
1
6
=
2
3
P(Ā∩B=1-p(A∪B)=1-
5
6
=
1
6
P(B=1-p(B)=1-
1
2
=
1
2
1. Since P(A) ≠P(B), A and B are not equally likely events.
2. Since P(A∪B) ≠1, A and B are not exhaustive events.
3. Since P(A∩B)≠0 ,A and B are not mutually exclusive.
4. Since P(A)P(B) = P(A∩B),A and B are independent events.

Example: Two players A and B toss an unbiased die alternatively. He who first throws a six wins the game. If A begins, what is the probability that B wins the game?
Solution: Let Ai and Bi be the respective events that A and B throw a six in Ith toss, i = 1, 2, .... . B will win the game if any one of the following mutually exclusive events occur:

A1B or A1B1A2B2 or A1B1A2B2A3B3,etc...
Thus P(B wins)=
5
6
x
1
6
+
5
6
x
5
6
x
5
6
x
1
6
+
5
6
x
5
6
x
5
6
x
1
6
x
5
6
+...........
=
5
36
[1+
52
62
+
54
64
+...........]
=
5
36
x
1
1-(5⁄6)2
=
5
11

Example: A bag contains 5 red and 3 black balls and second bag contains 4 red and 5 black balls.

1. If one ball is selected at random from each bag, what is the probability that both of them are of same colour?
2. If a bag is selected at random and two balls are drawn from it, what is the probability that they are of (i) same colour, (ii) different colours?

Solution:

(a) Required Probability = [Probability that ball from both bags are red] + [Probability that balls from both bags are black]

=
5
8
x
4
Quantitative Techniques for management Topics