Posing Probability Problems Mathematically

The principle that lies at the foundation of the mathematical theory of probability is the following: to speak of the probability of a random event , a probability space on which the event is defined must first be set up. In this section we show how several problems, which arise frequently in applied probability theory, may be formulated so as to be mathematically well posed. The examples discussed also illustrate the use of combinatorial analysis to solve probability problems that are posed in the context of finite sample description spaces with equally likely descriptions.

Example 2A . An urn problem. Two balls are drawn with replacement (without replacement) from an urn containing six balls, of which four are white and two are red. Find the probability that (i) both balls will be white, (ii) both balls will be the same color, (iii) at least one of the balls will be white.

Solution

To set up a mathematical model for the experiment described, assume that the balls in the urn are distinguishable; in particular, assume that they are numbered 1 to 6. Let the white balls bear numbers 1 to 4, and let the red balls be numbered 5 and 6.

Let us first consider that the balls are drawn without replacement. The sample description space of the experiment is then given by (3.1) of Chapter 1; more compactly we write

In words, one may read (2.1) as follows: is the set of all 2-tuples whose components are any numbers, 1 to 6, subject to the restriction that no two components of a 2-tuple are equal. The th component of a description represents the number of the ball drawn on the th draw. Now let be the event that both balls drawn are white, let be the event that both balls drawn are red, and let be the event that at least one of the balls drawn is white. The problem at hand can then be stated as one of finding (i) , (ii) , (iii) . It should be noted that , so that . Further, and are mutually exclusive, so that . Now \begin{align} A &= \{(1,2),(1,3),(1,4),(2,1),(2,3),(2,4), \\ &\quad\quad\quad (3,1),(3,2),(3,4),(4,1),(4,2),(4,3) \} \\ &= \{(z_1, z_2) \mid z_1, z_2 \in {1,2,3,4}, z_1 \neq z_2 \}, \tag{2.2} \end{align} whereas . Let us assume that all descriptions in are equally likely. Then

The answers to the questions posed in example are given, in the case of sampling without replacement, by (i) , (ii) , (iii) . These probabilities have been obtained under the assumption that the balls in the urn may be regarded as numbered (distinguishable) and that all descriptions in the sample description space given in (2.1) are equally likely. In the case of sampling with replacement, a similar analysis may be carried out; one obtains the answers

It is interesting to compare the values obtained by the foregoing model with values obtained by two other possible models. One might adopt as a sample description space . This space corresponds to recording the outcome of each draw as or , depending on whether the outcome of the draw is white or red. If one were to assume that all descriptions in were equally likely, then , . Note that the answers given by this model do not depend on whether the sampling is done with or without replacement. One arrives at a similar conclusion if one lets , in which 0 signifies that no white balls were drawn, 1 signifies that exactly 1 white ball was drawn, and 2 signifies that exactly two white balls were drawn.

Under the assumption that all descriptions in are equally likely, one would conclude that .

The next example illustrates the treatment of problems concerning urns of arbitrary composition. It also leads to a conclusion that the reader may find startling if he considers the following formulation of it. Suppose that at a certain time the milk section of a self-service market is known to contain 150 quart bottles, of which 100 are fresh. If one assumes that each bottle is equally likely to be drawn, then the probability is that a bottle drawn from the section will be fresh. However, suppose that one selects one bottle after each of fifty other persons have selected a bottle. Is one’s probability of drawing a fresh bottle changed from what it would have been had one been the first to draw? By the reasoning employed in example 2B it can be shown that the probability that the fifty-first bottle drawn will be fresh is the same as the probability that the first bottle drawn will be fresh.

Example 2B . An urn of arbitrary composition. An urn contains balls, of which are white and are red. A sample of size 2 is drawn with replacement (without replacement). What is the probability that (i) the first ball drawn will be white, (ii) the second ball drawn will be white, (iii) both balls drawn will be white?

Solution

Let denote the event that the first ball drawn is white, denote the event that the second ball drawn is white, and denote the event that both balls drawn are white. It should be noted that . Let the balls in the urn be numbered to , the white balls bearing numbers to , and the red balls bearing numbers to .

We consider first the case of sampling with replacement . The sample description space of the experiment consists of ordered 2-tuples , in which is the number of the ball drawn on the first draw and is the number of the ball drawn on the second draw. Clearly, . To compute , we use the fact that a description is in if and only if its first component is a number 1 to (meaning a white ball was drawn on the first draw) and its second component is a number 1 to (due to the sampling with replacement the color of the ball drawn on the second draw is not affected by the fact that the first ball drawn was white). Thus there are possibilities for the first component, and for each of these possibilities for the second component of a description in . Consequently, by (1.1) , the size of is . Similarly, , since there are possibilities for the first component and possibilities for the second component of a description in . The reader may verify by a similar argument that the event , (a white ball is drawn on both draws), has size . Thus in the case of sampling with replacement one obtains the result, if all descriptions are equally likely, that

We next consider the case of sampling without replacement . The sample description space of the experiment again consists of ordered 2-tuples , in which (for ) denotes the number of the ball drawn on the th draw. As in the case of sampling with replacement, each is a number 1 to . However, in sampling without replacement a description must satisfy the requirement that its components are not the same. Clearly, . Next, , since there are possibilities for the first component of a description in and possibilities for the second component of a description in ; the urn from which the second ball is drawn contains only balls. To compute , we first concentrate our attention on the second component of a description in . Since is the event that the ball drawn on the second draw is white, there are possibilities for the second component of a description in . To each of these possibilities, there are only possibilities for the first component, since the ball which is to be drawn on the second draw is known to us and cannot be drawn on the first draw. Thus by (1.1) . The reader may verify that the event has size . Consequently, in sampling without replacement one obtains the result, if all descriptions are equally likely, that

Another way of computing , which the reader may find more convincing on first acquaintance with the theory of probability, is as follows. Let denote the event that the first ball drawn is white and the second ball drawn is white. Let denote the event that the first ball drawn is red and the second ball drawn is white. Clearly, . Since , we have

To illustrate the use of (2.5) and (2.6) , let us consider an urn containing balls, of which are white. Then and in sampling with replacement, whereas and in sampling without replacement.

The reader may find (2.6) startling. It is natural, in the case of sampling with replacement, in which , that the probability of drawing a white ball is the same on the second draw as it is on the first draw, since the composition of the urn is the same in both draws. However, it seems very unnatural, if not unbelievable, that in sampling without replacement . The following remarks may clarify the meaning of (2.6) .

Suppose that one desired to regard the event that a white ball is drawn on the second draw as an event defined on the sample description space, denoted by , which consists of all possible outcomes of the second draw. To begin with, one might write . However, how is a probability function to be defined on the subsets of in the case in which the sample is drawn without replacement. If one knows nothing about the outcome of the first draw, perhaps one might regard all descriptions in as being equally likely; then, . However, suppose one knows that a white ball was drawn on the first draw. Then the descriptions in are no longer equally likely; rather, it seems plausible to assign probability 0 to the description corresponding to the (white) ball, which is not available on the second draw, and assume the remaining descriptions to be equally likely. One then computes that the probability of the event (that a white ball will be drawn on the second draw), given that the event (that a white ball was drawn on the first draw) has occurred, is equal to . Thus represents a conditional probability of the event (and, in particular, the conditional probability of , given that the event has occurred), whereas represents the unconditional probability of the event . The distinction between unconditional and conditional probability is made precise in section 4 .

The next example we shall consider is a generalization of the celebrated problem of repeated birthdays . Suppose that one is present in a room in which there are people. What is the probability that no two persons in the room have the same birthday? Let it be assumed that each person in the room can have as his birthday any one of the 365 days in the year (ignoring the existence of leap years) and that each day of the year is equally likely to be the person’s birthday. Then selecting a birthday for each person is the same as selecting a number randomly from an urn containing balls, numbered 1 to 365. It is shown in example that the probability that no two persons in a room containing persons will have the same birthday is given by

The value of (2.7) for various values of appears in Table 2A .

n	P _n	Q _n
4	0.984	0.016
8	0.926	0.074
12	0.833	0.167
16	0.716	0.284
20	0.589	0.411
22	0.524	0.476
23	0.493	0.507
24	0.462	0.538
28	0.346	0.654
32	0.247	0.753
40	0.109	0.891
48	0.039	0.961
56	0.012	0.988
64	0.003	0.997

TABLE 2A . In a room containing

persons let

be the probability that there are not two or more persons in the room with the same birthday and let

be the probability that there are two or more persons with the same birthday.

From Table 2A one determines a fact that many students find startling and completely contrary to intuition. How many people must there be in a room in order for the probability to be greater than 0.5 that at least two of them will have the same birthday? Students who have been asked this question have given answers as high as 100, 150, 365, and 730. In fact, the answer is 23!

Example 2C . The probability of a repetition in a sample drawn with replacement . Let a sample of size be drawn with replacement from an urn containing balls, numbered 1 to . Let denote the probability that there are no repetitions in the sample (that is, that all the numbers in the sample occur just once). Let us show that

The sample description space of the experiment of drawing with replacement a sample of size from an urn containing balls, numbered 1 to , is

The th component of a description represents the number of the ball drawn on the th draw. The event that there are no repetitions in the sample is the set of all -tuples in , none of whose components are equal. The size of is given by , since for any description in there are possibilities for its first component, possibilities for its second component, and so on. The size of is . If we assume that all descriptions in are equally likely, then (2.8) follows.

Example 2D . Repeated random digits . Another application of (2.8) is to the problem of repeated random digits . Consider the following experiment. Take any telephone directory and open it to any page. Choose 100 telephone numbers from the page. Count the numbers whose last four digits are all different. If it is assumed that each of the last four digits is chosen (independently) from the numbers 0 to 9 with equal probability, then the probability that the last four digits of a randomly chosen telephone number will be different is given by (2.8) , with and . The probability is .

The next example is concerned with a celebrated problem, which we call here the problem of matches . Suppose you are one of persons, each of whom has put his hat in a box. Each person then chooses a hat randomly from the box. What is the probability that you will choose your own hat? It seems reasonable that the probability of choosing one’s own hat should be , since one could have chosen any one of hats. However, one might prefer to adopt a more detailed model that takes account of the fact that other persons may already have selected hats. A suitable mathematical model is given in example 2E . In section 6 the model given in example 2E is used to find the probability that at least one person will choose his own hat. But whether the number of hats involved is 8,80, or , the rather startling result obtained is that the probability is approximately equal to that no man will choose his own hat and approximately equal to that at least one man will choose his own hat.

Example 2E . Matches (rencontres) . Suppose that we have urns, numbered 1 to , and balls, numbered 1 to . Let one ball be inserted in each urn. If a ball is put into the urn bearing the same number as the ball, a match is said to have occurred. In section 6 formulas are given (for each integer ) for the probability that exactly matches will occur. Here we consider only the problem of obtaining, for the probability of the event that a match will occur in the th urn. The probability corresponds, in the case of the persons selecting their hats randomly from a box, to the probability that the kth person will select his own hat.

To write the sample description space of the experiment of distributing balls in urns, let represent the number of the ball inserted in the th urn (for ). Then is the set of -tuples , in which each component is a number 1 to , but no two components are equal. The event is the set of descriptions in such that ; in symbols, . It is clear that ! and !. If it is assumed that all descriptions in are equally likely, then . Thus we have proved that the probability of a person’s choosing his own hat does not depend on whether he is the first, second, or even the last person to choose a hat.

Sample description spaces in which the descriptions are subsets and partitions rather than -tuples are systematically discussed in section 5 . The following example illustrates the ideas.

Example 2F . How to tell a prediction from a guess . In order to verify the contention of the existence of extrasensory perception, the following experiment is sometimes performed. Eight cards, four red and four black, are shuffled, and then each is looked at successively by the experimenter. In another room the subject of study attempts to guess whether the card looked at by the experimenter is red or black. He is required to say “black” four times and “red” four times. If the subject of the study has no extrasensory perception, what is the probability that the subject will “guess” correctly the colors of exactly six of eight cards? Notice that the problem is unchanged if the subject claimed the gift of “prophecy” and, before the cards were dealt, stated the order in which he expected the cards to appear.

Solution

Let us call the first card looked at by the experimenter card 1; similarly, for , let the th card looked at by the experimenter be called card . To describe the subject’s response during the course of the experiment, we write the subset of the numbers , which consists of the numbers of all the cards the subject said were red. The sample description space then consists of all subsets of size 4 of the set . Therefore, . The event that the subject made exactly six correct guesses may be represented as the set of those subsets , exactly three of whose members are equal to the numbers of cards that were, in fact, red. To compute the size of , we notice that the three numbers in a description in , corresponding to a correct guess, may be chosen in ways, whereas the one number in a description in , corresponding to an incorrect guess, may be chosen in ways. Consequently, , and

Exercises

In solving the following problems, state carefully any assumptions made. In particular, describe the probability space on which the events, whose probabilities are being found, are defined.

2.1 . Two balls are drawn with replacement (without replacement) from an urn containing 8 balls, of which 5 are white and 3 are black. Find the probability that (i) both balls will be white, (ii) both balls will be the same color, (iii) at least 1 of the balls will be white.

Answer

Without replacement, (i) , (ii) , (iii) ;

with replacement, (i) , (ii) , (iii) .

2.2 . An urn contains 3 red balls, 4 white balls, and 5 blue balls. Another urn contains 5 red balls, 6 white balls, and 7 blue balls. One ball is selected from each urn. What is the probability that (i) both will be white, (ii) both will be the same color?

2.3 . An urn contains 6 balls, numbered 1 to 6. Find the probability that 2 balls drawn from the urn with replacement (without replacement), (i) will have a sum equal to 7, (ii) will have a sum equal to , for each integer from 2 to 12.

Answer

2.4 . Two fair dice are tossed. What is the probability that the sum of the dice will be (i) equal to 7, (ii) equal to , for each integer from 2 to 12?

2.5 . An urn contains 10 balls, bearing numbers 0 to 9. A sample of size 3 is drawn with replacement (without replacement). By placing the numbers in a row in the order in which they are drawn, an integer 0 to 999 is formed. What is the probability that the number thus formed is divisible by 39? Note: regard 0 as being divisible by 39.

Answer

2.6 . Four probabilists arrange to meet at the Grand Hotel in Paris. It happens that there are 4 hotels with that name in the city. What is the probability that all the probabilists will choose different hotels?

2.7 . What is the probability that among the 32 persons who were President of the United States in the period 1789–1952 at least 2 were born on the same day of the year.

Answer

2.8 . Given a group of 4 people, find the probability that at least 2 among them have (i) the same birthday, (ii) the same birth month.

2.9 . Suppose that among engineers there are 12 fields of specialization and that there is an equal number of engineers in each field. Given a group of 6 engineers, what is the probability that no 2 among them will have the same field of specialization?

Answer

2.10 . Two telephone numbers are chosen randomly from a telephone book. What is the probability that the last digits of each are (i) the same, (ii) different?

2.11 . Two friends, Irwin and Danny, are members of a group of 6 persons who have placed their hats on a table. Each person selects a hat randomly from the hats on the table. What is the probability that (i) Irwin will get his own hat, (ii) both Irwin and Danny will get their own hats, (iii) at least one, either Irwin or Danny, will get his own hat?

Answer

(i) ; (ii) ; (iii) .

2.12 . Two equivalent decks of 52 different cards are put into random order (shuffled) and matched against each other by successively turning over one card from each deck simultaneously. What is the probability that (i) the first, (ii) the 52nd card turned over from each deck will coincide? What is the probability that both the first and 52nd cards turned over from each deck will coincide?

2.13 . In example 2F what is the probability that the subject will guess correctly the colors of (i) exactly 5 of the 8 cards, (ii) 4 of the 8 cards?

Answer

(i) 0; (ii) .

2.14 . In his paper “Probability Preferences in Gambling”, American Journal of Psychology , Vol. 66 (1953), pp. 349–364, W. Edwards tells of a farmer who came to the psychological laboratory of the University of Washington. The farmer brought a carved whalebone with which he claimed that he could locate hidden sources of water. The following experiment was conducted to test the farmer’s claim. He was taken into a room in which there were 10 covered cans. He was told that 5 of the 10 cans contained water and 5 were empty. The farmer’s task was to divide the cans into 2 equal groups, 1 group containing all the cans with water, the other containing those without water. What is the probability that the farmer correctly put at least 3 cans into the water group just by chance?

Posing Probability Problems Mathematically

Exercises

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