Distribution of balls in boxes

RANDOM EVENTS AND THEIR PROBABILITIES

1.3. Distribution of balls in boxes

Let there be rballs and n boxes, which are numerated by the numbers i 1,2,...,n. Denote the set of boxes by : ₀ {1,2,..., }n .

Let us first consider the case of distinguishable (i.e., having some differences from each other – number, color, etc.) balls.

Denote by : a sample space, corresponding to a random distribution of r balls into n boxes (here and further «random distribution of balls in boxes» means that any ball can get into any box with the same probability). If we denote byi_j ( 1,2,..., )j r the number of box into which the ball No. j got, then the sample space corresponding to the given experiment can be described as follows:

^ `

r j

r i j r

i i

i₁, ₂,..., : :₀, 1,2,..., :₀u:₀u...u:₀

: (17)

From this we see that the experiment consisting in placing r distinguishable balls into n distinguishable boxes and the experiment corresponding to the choice of a random sample of size r from the general population of size n are described by the same sample space (see the previous paragraph 1.1).

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Remark. Above we used the figurative language of «balls» and «boxes», but the sample space, constructed earlier for this scheme, allows a large number of interpre- tations.

For the convenience of further references, we present now a number of schemes that are visually very different but essentially equivalent to the abstract arrangement of r balls in n boxes in the sense that the corresponding outcomes differ only in their verbal description. In this case, the probabilities attributed to elementary events can be different in different examples.

Example 5. а) Birthdays. The distribution of birthdays of r students corresponds to the distribution of r balls into n 365 boxes (it is assumed that there are 365 days in a year).

b) When firing at targets, the bullet corresponds to the balls, and the targets to the boxes.

c) In experiments with cosmic rays, particles that fall into Geiger counters play the role of balls, and the counters themselves are boxes.

d) The elevator leaves (rises) with r people and stops on n floors. Then the distribution of people into groups, depending on the floor on which they exit, corresponds to the distribution of r balls in n boxes.

e) The experiment consisting in throwing r dice corresponds to the distribution of r balls in n 6 boxes. If the experiment consists in throwing r symmetrical coins, then n 2.

From the above formula (17), according to Theorem 2 of the preceding section, it follows that : n^r. The latter means that r distinguishable balls can be distributed over n distinguishable boxes in n^r ways.

In many cases it is necessary to consider the balls indistinguishable (the balls are the same and they do not differ from each other in color, shape, weight, etc.). For example, when examining the distribution of birthdays by days of the year, only the number of people born on a particular day is of interest (the number of balls that have fallen into a particular box).

To show that depending on whether the balls are distinguishable or indistin- guishable, the number of possible balls distributions in the boxes may be different.

Let us give an example.

Example 6. Placement (distribution) of three balls in three boxes.

а) Let the balls be distinguishable. We denote these three balls by a, b, c, and each of the boxes is represented as an interval between two vertical segments. Then each of the possible distributions is an indecomposable outcome of the experiment (an elementary event). All possible outcomes of the experiment, consisting of placing three distinguishable balls into three distinguishable boxes, are presented in Table 4.

An event A= {there is a box containing at least two balls} occurs when distri- bution is 1 – 21, and we express this by saying that an event is a set of elementary events 1 – 21. Similarly, an event В = {the first box is not empty} can be described as a set of points (elementary events) 1, 4 – 15, 22 – 27. An event С = АВ (A and В

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occurred) is a set consisting of thirteen elementary events 1, 4 -15. It is easy to see that A BB : : = {all the 27 outcomes from Table 4} is a certain event. An event

A (A didn’t occur) consists of points 22 -27 and can be described as A= {there are no empty boxes}.

Table 4

№ 1 2 3 № 1 2 3 № 1 2 3

1 аbс - - 10 а bс 19 - а bс

2 - аbс - 11 b ас - 20 - b ас

3 - - аbс 12 с аb - 21 - с аb

4 аb с - 13 а - bс 22 а b с

5 ас b - 14 b - ас 23 а с b

6 bс а - 15 с - аb 24 b а с

7 аb - с 16 - аb с 25 b с а

8 ас - b 17 - aс b 26 с а b

9 bс - а 18 - bс а 27 с b а

Finally, an event {The first box is empty and there is no box containing more than one ball}ൌ ‡(an impossible event, i.е. it cannot occur).

b) Balls are indistinguishable. All three balls are the same, which means that further we do not make a difference between such placements as 4, 5, 6; 7, 8, 9; 10, 11, 12; 13, 14, 15; 16, 17, 18; 19, 20, 21; 22 - 27.

Thus, in this case, Table 4 is reduced to the following Table 5 (in Table 5, indistinguishable balls are indicated by asterisks (*)).

Table 5

№ 1 2 3 № 1 2 3

1 *** - - 6 * ** -

2 - *** - 7 * - **

3 - - *** 8 - ** *

4 ** * - 9 - * **

5 ** - * 10 * * *

As we see, in this case the sample space consists of only 10 points.

If the balls are indistinguishable, but the boxes are distinguishable (for example, numbered), then each placement (r₁,r₂,...,r_n), where r_j is the number of balls in the j- th box, is completely described by the relations given below:

1 2 ... _n , _i 0, 1,2,...,

r r r r r t i n, (17) or «filling numbers», and nonnegative integers r₁,r₂,...,r_n completely describe all pos- sible combinations of filling numbers.

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We now prove the following lemma.

Lemma. At random placement of r indistinguishable balls in ndistinguishable boxes:

а) The number of distinguishable placements (that is, the number of different solutions of equation (17)) is equal to

А_r,n C_n^r_r1 C_nⁿ¹_r1 . (18) b) The number of distinguishable placements, when there will not be a single empty box, is equal to C_rⁿ₁¹.

Proof. а) We denote the balls by asterisks (*) and draw n boxes in the form of n spaces between n+1 vertical dashes. For example, the placement

| | | |

Is used to indicate the placement of r 6 balls into n 5 boxes so that these boxes contain 2, 1, 0, 0, 3 balls, respectively. With the introduction of such a notation, dashes necessarily stand at the beginning and at the end, but the remaining n1 dashes and r asterisks can be arranged in random order. Hence it follows that the number of distinguishable placements is equal to the number of ways to choose r places for asterisks from the available n r 1 places (or the number of ways to select n1 places for dashes from available n r 1 places), i.e. the formula (18) takes place.

b) The condition that there is no empty box means that there are no two dashes nearby. Between r asterisks there are r1 intervals, in n1of which there is one dash, so we have C_rⁿ₁¹ choices.

Example 7. а) Simultaneous throwing of r indistinguishable dice gives C_r⁵₅ distinguishable outcomes (in formula (18), the number of all outcomes n = 6).

b) Partial derivatives. It is known that partial derivatives of order r of an analytical function f x x( , ,..., )_{1 2} x_n of n variables do not depend on the differentiation order, but depend on how many times the function is differentiated for each variable. Thus, each variable plays the role of a box, and, consequently, in this case there are C_n^r_r1 different partial derivatives (see the lemma).

Thus, the function of three variables has fifteen derivatives of the fourth order and 21 derivatives of the fifth order.

c) r indistinguishable balls are randomly placed into n boxes. If we consider all indistinguishable distributions to be equally likely, then the probability that there will not be any empty boxes (of course, in this case r nt )) is equal to (see the lemma)

1 1 1

1 n r n n r

p C C

. (19)

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1.3.1. Statistics of Maxwell-Boltzmann, Bose-Einstein and Fermi-Dirac We consider several important distribution problems arising in the study of certain particle systems in physics and statistical mechanics.

In statistical mechanics, phase space is usually divided into a large number n of small regions or cells so that each particle is assigned to one cell. As a result, the state of the whole system is described as a random arrangement of r particles (balls) in n cells (boxes).

The Maxwell-Boltzmann system is characterized as a system of rdistingui- shable (different) particles, each of which can be in one of n cells (states), regardless of where the remaining particles are located. In such a system it is possible to have n^r different arrangements of r particles into n cells. If, in doing so, all such arrangements (states of the system) are considered equally probable, then we speak of Maxwell- Boltzmann statistics. Thus, in the Maxwell-Boltzmann system (statistics), the proba- bility of each state (elementary event) is n^r.

The Bose-Einstein system is defined as a system of r indistinguishable particles, each of which independently of the others can be in one of n cells. Since the particles are indistinguishable, each state of this system is given by «filling numbers»

rn

r

r₁, ₂,..., , where r_j is the number of particles in the cell No. j. If in this case all states of the system are considered equiprobable, then we speak of Bose-Einstein statistics.

Thus, the probability of each state (elementary event) in the Bose-Einstein system is

¹ 1

¹

n

r

Cn (see formula (18)).

Note that if in the Bose-Einstein system we additionally require that no cells remain empty in each state of the system (clearly, this should be r nt ), then the num- ber of possible states of the system will be reduced to C_rⁿ₁¹ (this we proved above, in part b) of the last lemma).

The Fermi-Dirac system is defined as a Bose-Einstein system, which in addition to the Pauli exclusion principle requires that no more than one particle is in each cell.

Since in this case the particles are also indistinguishable, the state of the system is characterized by the numbers r₁,r₂,...,r_n, where rj = 0 or rj = 1 (because in each cell there can be no more than one particle), j = 1,2,…,n, and mandatory r nd .

You can specify the system status by specifying the filled cells. The latter can be selected by C_n^r ways (r cells can be selected from ncells for r particles in C_n^r ways), the Fermi-Dirac system has the same number of states. If all states are equi- probable, then we speak of Fermi-Dirac statistics. Thus, the probability of each state (elementary event) in Fermi-Dirac statistics is C_n^{r 1}, r nd .

Example 8. r distinguishable (for example, numbered) particles are arranged in n cells according to the Maxwell-Boltzmann system.

Find the probabilities of the following events:

a) Exactly k (0d dk r) particles fell in a certain cell (say, in cell No. 1).

b) Exactly k (0d dk r) particles fell in some cell.

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c) Particle No. 1 fell into cell No. 1, particle No. 2 fell into cell No. 2.

d) At least one cell is empty.

Solution. а) A sample space can be described by the formula (17). If k particles fell in a certain cell (for example, in the cell No. 1), then the remaining r k particles must be placed in the remaining n1cells (except for the selected ones). For the cell No. 1 we can select k particles from r particles in C_r^k ways, and the remaining r k particleswe can distribute into the remaining n1 cells in (n1)^rk ways. Then, by the classical definition of probability, the required probability is

1

( 1) 1 1

1 .

k r k

r k

k k

r r r

p C n C

n n n

§ · §¨ ¸ ¨© ¹ © ·¸¹

b) We can select one cell from all n cells in n ways. So, the required probability p2 is n times greater than the probability p1, i.е. p₂ np₁.

c) It is not difficult to guess that we need to find the probability of an event

^ `

3 (1,2, ,..., )3 _r : _j 0, 3,..., ,

A i i : i : j n :₀

^

1,2,...,n

`

. Then the required probability is

2 3

3 2

r 1

r

A n

p n n

: .

d) Let’s introduce the events:

А = {at least one cell is empty},

Аk= {cell No. k is empty}, k = 1,2,…, n.

So

1

,

( 1) 1

P( ) 1 , 1 .

n k k r r

k r

A A

A n k n

n n

§¨© ·¸¹ d d

k, A_kk

An event A A_{k l} (k lz ) means that the cells k and l are empty, therefore

2 2

1 1

r r

k l r

(n )

Ρ(A A ) , k,l n,k l.

n n

§¨© ·¸¹ d d z

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Similarly, for different , ,k l m:

3 3

1

r r

k l m r

(n )

Ρ(A A A ) .

n n

§¨© ·¸¹ etc.

Finally, by the addition of the probability formula, the required probability is

2 2 1

4

1 2 1

1 1 1 1

r r r

n n

n n

p Ρ(A) n C ... ( ) C n

n n n

§ · § · § ·

¨ ¸ ¨ ¸ ¨ ¸

© ¹ © ¹ © ¹

1 1 1

1 1

( 1) 1 ( 1) 1 .

r r

n n

j j j j

n n

j j

j j

C C

n n

§¨© ·¸¹ §¨© ·¸¹

¦ ¦

Example 9. r particles are distributed into n cells, 0൑ r nd .

Assuming that the «particle-cell» system obeys the Maxwell-Boltzmann, Bose- Einstein, and Fermi-Dirac statistics, find the probabilities of the following events:

а) There is one particle in each of predetermined r cells.

b) There is one particle in some r cells.

Solution. For Maxwell-Boltzmann statistics, the number of all possible distribu- tions is n^r, and the number of all possible distributions of r distinguishable particles into r cells is r! In addition, in the case b) r cells (for the distribution of r particles) from n cells can be selected in C_n^r ways. So, the wanted probabilities are

а) !

r

p r

n ;

b) ^!

^!

^!

r n

r r

C r n

p n n r n

˜

˜ .

In Bose-Einstein statistics, particles are indistinguishable, the number of all pos- sible distributions is C_nⁿ¹_r1, in addition, the replacement of particles in places in pre- determined cells does not give a new distribution. Therefore

а)

1

1

r n r

p C ; b)

1

!( 1)!

( )!( 1)!

r n r n r

C n n

p C n r n r

.

It is easy to see that in the case of Fermi-Dirac statistics, the required proba- bilities are:

а) p C_n^{r 1}; b) p 1.

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