Bulletin of the Marathwada Mathematical Society Vol. 11, No. 1, June 2010, Pages 11–18.

Problems on Probability involving Infinite Sample Spaces S.R.Joshi 8, Karmayog, Tarak Colony,Opp to Ram-krishna Ashram, Beed bye pass Road, Aurangabad - 431517 and B.B. Kulkarni, Department of mathematics, N.S.B. College, Nanded ,431601 Abstract The main purpose of this article is to show how the concepts and results in calculus of a real variable are used in solving some problems on probability for which the corresponding sample spaces are countable or uncountable.

1

INTRODUCTION

For any numerical problem on probability ,it is observed that the corresponding sample space is finite or infinite. Of course there are more problems on probability for which the corresponding sample space is finite. If the sample space happens to be infinite , it may be countable or uncountable. If it is countable , then the corresponding problem can be solved either by reducing the sample space to a finite sample space or by using the concept of limit in Calculus. If the sample space is uncountable, then, some problems may be solved by using concepts of Calculus of a real variable , such as continuity and monotonic nature of a function, maxima and minima, integral of a function etc. along with some known results in the theory of numbers and combinatorics. In this article, we first discuss some problems on probability, for which events are countable sets.After this we illustrate some problems for which the corresponding sample space is uncountable and solve them by applying the results in Calculus without considering the aspects related to a continuous random variable.For this purpose we also make a suitable definition of the probability of an event in R2 or R3 . 11

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2

Problems on Probability involving an Infinite Sample Space PRELIMINARIES

It is assumed that the readers are familiar with the terms such as sample space, event and its probability, countable and uncountable sets, limit and continuity of a function increasing and decreasing functions, bounded functions, maximum and minimum values of a function, integral of a real valued function, etc. For these concepts the readers may refer to [1, 2, 3, 4].We shall use the symbols N, Q, and R1 to represent the set of positive integers, the set of rational numbers and the set of real numbers respectively.If E is an event with respect to a finite sample space , then we denote the number of elements in E by N(E) and the probability of E by P(E). We also require some results in Calculus. These are given below. Result 2.1 :If a, b, c and d are four real numbers such that a 6= 0 and c 6= 0 then lim (an + b)/(cn + d) = a/c .

n→∞

Result 2.2 : lim

n→∞

n/(2n ) = 0 .

Result 2.3 : If a region R is bounded by a curve y = f (x), a ≤ x ≤ b,the lines x= a , x = b and the X-axis, where f(x) is continuous on the interval a ≤ x ≤ b , then the area, A(R), of the region R is given by Z b A(R) = f (x)dx . (2.1) a

Result 2.4 : Let R be the region as given in Formula(2.3) and let V(R) be the volume of the solid , which is obtained by revolving the region R about the x-axis. Then, Z b V (R) = π f 2 (x)dx . a

Wa also state the intermediate value theorem for a real valued function f(x) defined on a closed interval.This will be used in solving the Problem 4.3 . Theorem 2.1 Let f(x) be a real valued continuous function defined on the closed interval a ≤ x ≤ b.Let m and M be the minimum and maximum values of f(x). If d is any real number between m and M , then there exists a real number c satisfying a ≤ c ≤ b, such that f(c) = d . Lastly we make the following definition for the probability of an event E in a sample space S, where S is a subset of R2 or R3 .

Problems on Probability involving Infinite Sample Spaces

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Definition 2.1 If E is an event in S where S is a subset of R2 or R3 . then we define the probability of E by the relation P (E) =

m(E) m(S)

where m(A) is the area or volume of the set A depending on whether A is a subset of R2 or R3 and m(S) is non-zero. Remark 2.1 The real valued set function m defined above is a well defined function on a suitable set of subsets of R2 or R3 and it is particular case of the Lebesgue measure on Rn . For details about the Lebesgue measure, see [4, 229–236].

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PROBLEMS RELATED TO COUNTABLE EVENTS

In this section we illustrate some problems on probability for which the corresponding sample space is countable or uncountable and the events are countable sets. Problem 3.1 A positive integer is selected at random. Find the probability that (i) it is even, (ii) it is divisible by 4 or 6, (iii) it is of the form 2n . Solution (i) The problem can be solved in two ways. In the first way the sample space N is reduced to a finite set depending on the digit in the unit place. This digit can be any one of the ten digits 0 to 9 . Hence we can take the sample space as the set S = {0, 1, 2, 3, · · · , 9}. Further if E1 is the required event related to (i) , then E1 with respect to S will be E1 = {0, 2, 4, 6, 8}. Hence P (E1 ) is given by P (E1 ) =

N (E1 ) = 5/10 = 1/2. N (S)

In the second way consider first 2n or (2n+1) positive integers, where n ∈ N . In each of these two cases there are n even integers. Hence the required probability P (E1 ) is obtained by taking the limit of n/(2n) or n/(2n + 1) as n tends to ∞. Using Result 2.1 , we get P (E1 ) = 1/2 in each case. Remark 3.1 The problem (i) can also be viewed as follows: Among any two consecutive numbers in N , one is always even and hence the answer to this problem is 1/2. (ii) Let A , B, C be three events given by A = {x ∈ N | x is divisible by 4},

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Problems on Probability involving an Infinite Sample Space B = {x ∈ N | x is divisible by 6}, C = {x ∈ N | x is divisible by 12}.

We want to find the value of P (A ∪ B),which is obtained by using the result P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = P (A) + P (B) − P (C). Note that C is the same as A ∩ B

(3.1)

since 12 is the l.c.m. of 4 and 6.

Now P (A) is obtained as in (i) by considering the first 4n , or 4n + 1, or 4n + 2, or 4n + 3 positive integers. Representing each of these integers by m, we observe by Result 2.1 that the limit of n/m in each of these four cases is 1/4 since there are n integers among the first m integers which are divisible by 4 . Similarly we can show that P (B) = 1/6 and P (C) = 1/12. Hence using (3.1) we get P (A ∪ B) = 1/4 + 1/6 − 1/12 = 1/3. (iii) To reduce the sample space to a finite set , we consider the first 2n integers in N . Let Sn = {1, 2, . . . , 2n }. If E3 is the required event, then we observe that E3 = {21 , 22 , . . . , 2n }.Clearly E3 contains n elements. Hence by Result (2.2)we get P (E3 ) = lim

n→∞

n/(2n ) = 0

Compared to the part(i) and part(ii), the answer to this part is not a positive number. This is some what startling. If we put n = 1, 2, 3, 4, . . . in Sn , we get S1 = {1, 2}, S2 = {1, 2, 3, 4}, S3 = {1, 2, 3, . . . , 8} etc. We observe that the probabilities of the event E3 are 1/2, 2/4, 3/8, 4/16, etc. These values form a decreasing sequence tending to 0. Problem 3.2 A number x ∈ R1 is selected at random. Find the probability that it is a member of Q . Solution We know that the set R1 , which is also the sample space for the given problem is uncountable.If E is any set , then the number of elements in E is known as its cardinality, [1] . If E is countable its cardinality is denoted by ℵ0 . Here the required event is Q which is known to be countable.The cardinality of R1 is given by c = 2ℵ0 . Hence using the Result (2.2), we get P (Q) = 0.

Remark 3.2 By using the result P (E 0 ) = 1 − P (E), where E 0 is the complement of E with respect the corresponding sample space ( here our sample space is R1 ), we observe that the probability of a real number being an irrational number is 1 i.e. P (Q0 ) = 1. In fact if E is any countable set in R1 then its probability is 0 and the probability of its complement is 1 .

Problems on Probability involving Infinite Sample Spaces

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PROBLEMS RELATED TO UNCOUNTABLE EVENTS

In this section we solve a few problems on probability, for which the corresponding sample space is uncountable ,by using Results 2.3 , 2.4 and several concepts in Calculus along with Definition 2.1 . √ Problem 4.1 Let R be a plane region bounded by the curve y = x , the lines x = 0, x = 1 and y = 0. A point P is selected at random from the region R.Find the probability that it lies below the line y = x . Solution :Here the sample space is R which is the set of all points belonging to the dotted portion shown in Fig.4.1 .Clearly R is uncountable. If E is the required event,then E consists of all points in the triangular region OAB shown in the figure.To find P (E), we have to find the areas of R and 4OAB.By Result 2.1 we have

Z

A(R) =

1√

x = 2/3 .

0

Also the area of 4OAB = (1/2)OA × AB = 1/2 since OA = AB = 1. Hence using Definition 2.1 we get P (E) = 1/2 2/3 = 3/4. Note that there are uncountable points in R and in 4OAB, but the areas of R and 4OAB are finite positive numbers. Further 4OAB ⊂ R and hence A(R) ≥ Area of 4OAB. The next problem is related to the volume of a solid of revolution. Problem 4.2 A point is selected at random from a solid sphere of radius 1. Find the probability that it belongs to the solid segment BCAD, for which the height AB is 1/3 units.(see Fig.4.2) Solution : In this problem the sample space S is the set of all points inside and on the sphere with O as the center and radius 1 . Further the requited event E is a subset of S consisting of all points in the solid segment BCAD, for which the height AB is 1/3 units.Both S and E are uncountable sets. To obtain the value of P (E) we shall again use Definition 2.1 after finding the volumes of S and E The values of V (E) and V (S) are obtained by using the Result 2.4 given in Section 2.

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Problems on Probability involving an Infinite Sample Space

It is known that V (S) = 4π 3 , since the radius is 1. To obtain the volume of E consider the plane region R given by R = {(x, y) | x2 + y 2 ≤ 1 , 2/3 ≤ x ≤ 1 , y = 0}. √ In other words R is the plane region bounded by the curve y = (1 − x2 ) , the lines x = 2/3, x = 1 and the X-axis. Note that OB = 1 and AB = 2/3. Hence using Result 2.4 , we get Z 1 √ V (E) = π (1 − x2 )dx 2 3

= 8 π/81. Hence P (E) = V (E)/V (S) = 8π / 4π = 2/27 . √81 3 Problem 4.3 Let g(x) = [ ( 2 x + 1)6 ] , 0 ≤ x ≤ 1 be a real valued function defined on the iinterval I = {x ∈ R1 | 0 ≤ x ≤ 1 }, where [t] represents the integer part of the real number t. If x ∈ I is selected at random and g(x) is considered as a member of the related sample space , find the probability that g(x) is a prime number, or is divisible by 6 or 8 . Solution Firstly note that for the given problem the domain of the function g(x) i.e. the interval I is not the sample space. In facr the sample space S is the set of all values of g(x),where x ∈ I . Since for each x ∈ I , g(x) is an integer. S consists of different integers depending on the values of x ∈ I. We shall now show that S is a finte set. For this purpose define f (x) on I as follows. √ f (x) = ( 2 x + 1)6 , x ∈ I.

(4.1)

Since f (x) is a polynomial of 6th degree in x, f (x) is continuous. Further I is a closed √ √ interval. Hence f (x) ia bouded. Since f 0 (x) = 6 2( 2 x + 1)5 , and because f 0 (x) is positive for all x ∈ I, f (x) is an increasing function on I . This further implies that the minimum and maximum values of f (x) are f (0) and f (1) respectively. But √ f (0) = 1 . Let m = f (0) = 1 and M = f (1) = ( 2 x + 1)6 . Note that M is an irrational number, and the range of f is the closeed interval [m, M ]. This range of f is an uncountable set. But this is not the range of the function g , though the domain of each of f and g is I . The range of g is the closed intervcal [g(0), g(1)], √ where g(0) = [1] = 1 and g(1) = [f (1)] = [ ( 2 + 1)6 ].This is so because g(x) is also increasing on I . Note that g(x) is a continuous function on I , because f (x) is continuous, [x] is continuous , and continuous function of a continuous function is a continuoous function. Since g is increasing and continuoous, g is bounded on I. Further each value of g(x) is a positive integer, and hence the range of g is a fintite set. Hence by Theorem 2.1, for each integer d ∈ [g(0), g(1)] , there exists c ∈ I , such that g(c) = d. Thus the range of g, and hence the sample space for the given

Problems on Probability involving Infinite Sample Spaces

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problem is the set of all integers k satisfying g(0) ≤ k ≤ g(1) . Clearly g(0) = 1. To determine the integer g(1) , we use the following result related to binomial theorem.

(a + b)n + (a − b)n = 2[ an + C2 an−2 b2 + C4 an−4 b4 + ... + Cr an−r br ] where a, b ∈ R1 , n ∈ N , Ck = is odd.Now let

n! k!(n−k)!

(4.2)

and r = n if n is even and r = n − 1 if n

√ f (1) = ( 2 + 1)6 = g(1) + f0

(4.3)

where g(1) and f0 are the integer part and the fractional part of f (1) respectively. √ Putting x0 = ( 2 − 1)6 and using (4.2) we have √ √ f (1) + x0 = ( 2 + 1)6 + ( 2 − 1)6 √ √ √ = 2 [( 2)6 + 6C 2 ( 2)4 + 6C 4 ( 2)2 + 6C 6 ]

= 198.

Thus f (1) + f0 + x0 = 198 = an integer.

(4.4)

√ √ But x0 = ( 2 − 1)6 = a positive fraction since 2 − 1 < 1.Also f0 is a positive fraction. Further g(1) is a positive integer. Hence from (4.4), we conclude that f0 + x0 = 1 and hence g(1) = 197. This shows that the range of the fuction, g(x) is the set S given by S = {1, 2, 3, . . . , 197}. This S is the sample space for the given Problem. Let E be the required event, whose probability is to be determined. E can be divided into two exclusive events E1 and E2 ,where E1 is the set of all prime numbers in S , and E2 is the set of all integers in S , which are divisible by 6 or 8 . By actual counting we observe that E1 contains 48 prime numbers. Let A be the set of all those integers in S which are divisible T by 6 and B be the set of all those integers in S which are divisible by 8. Then (A B) will be the set of all integers in S, which are divisible by 6 and 8 both, i.e. divisible by 24 (l.c.m. of 6 and 8). Now N (A) = [197/6] = 32, N (B) = [197/8] = 24 and N (A

T

B ) = [197/24] = 8 ,

where by [t] we mean the integer part of the real number t. Hence N (E2 ) = N (A) + N (B) − N (A‘ ∩ B ) = 32 + 24 − 8 = 48.This implies N (E) = N (E1 ∪ E2 ) = N (E1 ) + N (E2 ) = 48 + 48 = 96. Hence P (E) =

N (E) N (S)

= 96/197, and the solution is complete.

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Problems on Probability involving an Infinite Sample Space REMARKS AND CONCLUSIONS

From the problems that we discussed in this article, it is clear that , some problems on probability for which the corresponding sample space is an infinite set can be solved with the help of Calculus of a real variable. There can be some problems which may be solved without using any concept from calculus. As far as the problems for which corresponding sample spaces are uncountable are concerned, the authors think that the problems can not be solved without the help of Calculus and the definition 2.1. Further the authors are not aware of those practical problems on probability, for which the concepts in Calculus are actually used. ACKNOWLEDGEMENT The authors are very much thankful to the referee and Dr.D.Y.Kasture for making useful suggestons in bringing out this paper.

References [1] S.G.Deo, D.Y.Kasture, H.V. Kumbhojkar and V.G.Tikeker, Popular Lectures on Mathematics, Universities Press(India), Pvt. Ltd.(2009). [2] Meyer P.L., Introductory Probability and Statistical Applications, Oxford and IBH Publishing Co. , New delhi, (1969) [3] S Narayan and T.K.M. Pillay, Calculus Vol.I and II, S. Vishwanathan Printers and Publishers Pvt.Ltd. (1995). [4] Walter Rudin. , Principles of Mathematical Analysis, (2nd Ed.) McGraw Hill Book Co., New York (1964)