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)Similarly, it can be shown that
P(Ā ∩ B) = P(A)  P(A ∩ B)Theorem : (Addition of Probabilities):
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 :
Corollaries:
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 ∩ B∩ C)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(A_{i})ΣΣP(A_{i}∩ A_{j})+ΣΣΣ P(A_{i}∩ A_{j}∩ A_{k})+(1)^{n}P(A_{1}∩ A_{2}∩ ...∩ A_{n})(i ≠j≠ k,) P(A1U A2....U An)=1 P(A_{1}∩A_{2}∩....∩A_{n})(if the events are mutually exclusive)
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)= 650/1000150/1000= 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)= 400/1000150/1000=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+400300/1000=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+400150/1000= 9/10Alternative Method: The above probabilities can easily be computed by the following ninesquare table :
From the above table, we can write
(i) p(A ∩ B) = 500/1000=1/2This 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+42/52=7/13Example : 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
(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 ninesquare table can be written as follows :
(ii) Here n(C) = 6, n(D) = 10, n(C∩D) = 2 and n(S) = 36. Thus, we have
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 A_{1} and A_{2} mutually exclusive? Obtain P(A_{1}∩A_{2}) and P(A_{1}∪A_{2}). Further, let A_{1} be the event that both coins show heads and A_{2} be the event that both show tails. Are A_{1} and A_{2} mutually exclusive? Find P(A_{1}∩A_{2}) and P(A_{1}∪A_{2}). .
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 A_{1}∩A_{2} = {(T, H)}, Since A_{1}∩ A_{2}≠ Ø , A_{1} and A_{2} are not mutually exclusive. Further, the coins are given to be unbiased, therefore, all the elementary events are equally likely.
(ii) When both the coins show heads; A_{1} = {(H, H)},When both the coins show tails; A_{2} = {(T, T)}
Here A_{1}∩ A_{2}= Ø ,.....A_{1} and A_{2} are mutually exclusive. Thus, P (A_{1}∪ A_{2})=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,
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)
If we multiply the numerator and denominator of the above expression by n(S), we get
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).Thus, the above equation can be written as
Substituting this value in the formula of conditional probability theorem, we get P(A∩B) = P(A).P(B).
Corollaries:
Pairwise 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 pairwise independent. From the above we note that mutually independent events will always be pairwise independent but not viceversa.
Example: Among 1,000 applicants for admission to M.A. economics course in a University, 600 were economics graduates and 400 were noneconomics graduates; 30% of economics graduate applicants and 5% of noneconomics 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
Alternative Method: Writing the given information in a ninesquare table, we have
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 A_{2} be the event that second is black. We want to find P (A_{2}/A_{1}).
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(A_{1}) = 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 (A_{1}∩A_{2}) = 3 x 2 = 6
(b) Here we have to find P (A_{1}/A_{2}). The second black ball can be drawn in the following two mutually exclusive ways:
Thus, n(A_{2}) = 3 x 2 + 2 x 1 = 8,
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
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 18_{C2} equally likely ways.
(a) Two red balls can be drawn from 10 red balls in 10_{C2} ways.
(b) One red ball can be drawn in 10_{C1}ways and one black ball can be drawn in 8_{C1}ways
Example: Five cards are drawn in succession and without replacement from an ordinary deck of 52 well shuffled cards :
Solution:
(a)
(b)
(c)
(d)
Example : Two cards are drawn in succession from a pack of 52 wellshuffled cards. Find the probability that
Solution:
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
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.
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.
We have to find P(A∪B) . Note that A and B are independent.
Example : If A and B are two events such that
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 :
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 A_{i} and B_{i} be the respective events that A and B throw a six in I^{th} toss, i = 1, 2, .... . B will win the game if any one of the following mutually exclusive events occur:
Example: A bag contains 5 red and 3 black balls and second bag contains 4 red and 5 black balls.
Solution:
(a) Required Probability = [Probability that ball from both bags are red] + [Probability that balls from both bags are black]


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