Consider
Theorem. Let
The definition of
Then, for any integer
In particular, for
The probability that at least
Before giving the proof of this theorem, we shall discuss various applications of it.
Example 6A . The matching problem (case of sampling without replacement) . Suppose that we have
(ii) exactly
\begin{align} \frac{1}{m !}\left\{1-1+\frac{1}{2 !}-\frac{1}{3 !}+\cdots \pm \frac{1}{(M-m) !}\right\}=\frac{1}{m !} \sum_{k=0}^{M-m}(-1)^{k} \frac{1}{k !} \tag{6.6}\\ \doteq \frac{1}{m !} e^{-1} \quad \text { for } M-m \text { large }. \end{align}
The matching problem may be formulated in a variety of ways. First variation : if
There is a considerable literature on the matching problem that has particularly interested psychologists. The reader may consult papers by D. E. Barton, Journal of the Royal Statistical Society , Vol. 20 (1958), pp. 73–92, and P. E. Vernon, Psychological Bulletin , Vol. 33 (1936), pp. 149–77, which give many references. Other references may be found in an editorial note in the American Mathematical Monthly , Vol. 53 (1946), p. 107. The matching problem was stated and solved by the earliest writers on probability theory. It may be of value to reproduce here the statement of the matching problem given by De Moivre ( Doctrine of Chances , 1714, Problem 35): “Any number of letters a, b, c, d, e, f, etc., all of them different, being taken promiscuously as it happens; to find the Probability that some of them shall be found in their places according to the rank they obtain in the alphabet and that others of them shall at the same time be displaced”.
Solution
To describe the distribution of the balls among the urns, write an
It then follows that the sum
Equations (6.5) and (6.6) now follow immediately from (6.8) , (6.3) , and (6.2) .
Example 6B . Coupon collecting (case of sampling with replacement) . Suppose that a manufacturer gives away in packages of his product certain items (which we take to be coupons, each bearing one of the integers 1 to
The symbol
A table of
The problem of coupon collecting has many variations and practical applications. First variation (the occupancy problem): if
Solution
To describe the coupons found in the
It is easy to obtain the probability of the intersection of any number of the events
The quantities
Let
which coincides with (6.9) .
Other applications of the theorem stated at the beginning of this section may be found in a paper by J. O. Irwin, “A Unified Derivation of Some Well-Known Frequency Distributions of Interest in Biometry and Statistics”, Journal of the Royal Statistical Society , Series A, Vol. 118 (1955), pp. 389–404 (including discussion).
The remainder of this section 1 is concerned with the proof of the theorem stated at the beginning of the section. Our proof is based on the method of indicator functions and is the work of
The method of indicator functions proceeds by interpreting operations on events in terms of arithmetic operations. Given an event
The two basic properties of indicator functions, which enable us to operate with them, are the following.
First, a product of indicator functions can always be replaced by a single indicator function; more precisely, for any events
To prove (6.15) , one need note only that
Second, a sum of indicator functions can in certain circumstances be replaced by a single indicator function; more precisely, if the events
The proof of (6.16) is left to the reader. One case, in which
Our ability to write expressions in the manner of (6.18) for the indicator functions of compound events, in terms of the indicator functions of the events of which they are composed, derives its importance from the following fact: an equation involving only sums and differences (but not products) of indicator functions leads immediately to a corresponding equation involving probabilities; this relation is obtained by replacing
The principle just enunciated is a special case of the additivity property of the operation of taking the expected value of a function defined on a probability space. This is discussed in Chapter 8 , but we shall sketch a proof here of the principle stated. We prove a somewhat more general assertion. Let
In words,
The sum
Finally, from (6.23) , and the fact that
We now apply the foregoing considerations to derive (6.2) . The event

Finally, from (6.24) to (6.28) and (6.1) we obtain
Theoretical Exercises
6.1. Matching problem (case of sampling without replacement) . Show that for
6.2. Matching (case of sampling with replacement) . Consider the matching problem under the assumption that, for
6.3. A man addresses
(i) If the
(ii) If the
(iii) In part (ii) the probability that each bill and each check will be in a wrong envelope is equal to the square of the answer to part (i).
6.4. A sampling (or coupon collecting) problem . Consider an urn that contains
Hint :
6.5. Verify the formulas in row I of Table 6A .
6.6. Verify the formulas in row II of Table 6A.
6.7. Verify the formulas in row III of Table 6A.
6.8. Verify the formulas in row IV of Table 6A.
Exercises
6.1. If 10 indistinguishable balls are distributed among 7 urns in such a way that all arrangements are equally likely, what is the probability that (i) a specified urn will contain 3 balls, (ii) all urns will be occupied, (iii) exactly 5 urns will be empty?
Answer
(i)
6.2. If 7 indistinguishable balls are distributed among 10 urns in such a way that not more than 1 ball may be put in any urn and all such arrangements are equally likely, what is the probability that (i) a specified urn will contain 1 ball (ii) exactly 3 of the first 4 urns will be empty?
6.3. If 10 distinguishable balls are distributed among 4 urns in such a way that all arrangements are equally likely, what is the probability that (i) a specified urn will contain 6 balls, (ii) the first urn will contain 4 balls, the second urn will contain 3 balls, the third urn will contain 2 balls, and the fourth urn will contain 1 ball, (iii) all urns will be occupied?
Answer
(i)
6.4. Consider 5 families, each consisting of 4 persons. If it is reported that 6 of the 20 individuals in these families have a contagious disease, what is the probability that (i) exactly 2, (ii) at least 3 of the families will be quarantined?
6.5. Write out (6.2) and (6.4) for (i)
Answer
(i)
(ii)
(iii)
- The remainder of this section may be omitted in a first reading of the book. ↩︎