Binomial Randomization 3/13/02

Transcription

1 Binomial Randomization 3/13/02 Define to be the product space 0,1, á, m â 0,1, á, m and let be the set of all vectors Ðs, á,s Ñ in such that s á s t. Suppose that \, á,\ is a multivariate hypergeometric random vector. That is, for all s, á,s, T \ =, á,\ = where Q m á m and = á = Þ = â = Q For the purposes of Theorems 1 and 2 which follow we we will suppose that ], á,] are independent binomial random variables parameterized such that for #á,,, C C ) ) T ] C 1 C ß ß á, Þ C ) ) It is well known that in this case parameters Q and ). Therefore T ] =, á,] = Y3 t ] á ] T ] =, á,] = I = 3 TY 3 t = 3 3 = ) 3 ) 3 I = Q 3 ) Q ) 3 3 will follow a binomial distribution with 1

2 â = = = 3 Q 3 Q ) ) I = Q 3 T \ =, á,\ = I = 3 ) Q ) Q 3 3 Theorem 1.. I 1 \ á \ Q,, ) I1 ] á ] x.,, ) where the in I is used to denote that \ á \. Proof. Q Q I1 ], á,] I1 ], á,] Y 3 t T Y 3 t =, á, = T ] =, á,] = Y3 tt Y 3 t ) Q 3 3 Q 1=, á=, T\ =, á\, = T Y 3 t 3 Q I 1 \, á,\ T Y t 3 3 Q Q I 1 \, á,\ ) ) Q and Q ) Q I1 ], á,] I 1 \, á,\ ) Q Therefore, 2

3 . Q. Q ) I 1 ] á ] I 1 \ á \.,,.,, ) ) ) ) Q Q Q. I 1 \, á,\.) ) Q Q I 1 \, á,\ x I I 1\ á \ Q,, x IßßáßQ Hence, ) ). I 1 \ á \ Q,, ) I1 ] á ] x.,, ) Q ) Theorem 2. Let T and define T T. Then for, T \ á \. T ] á ] x. Q,, T ),, T Q ) ) Proof. Apply Theorem 1 with 1= á= =, á, = T,, else so that and I 1 \, á,\ T \, á,\ T I1], á], T ], á], T. 3

4 For the purposes of Theorem 3 which follows we will suppose that ], á,] are independent binomial random variables parameterized such that for #á,,, T ] C C C : : Cßßá, Þ C Suppose an urn contains objects of Type, á, objects of Type and that objects are drawn from this urn without replacement. Let Q á. When ; 3 objects of Type 3 have been drawn we will say Type 3 has reached its quota. Let [ :; ßáß; á ; [ : U represent the waiting time until exactly different,, types have reached their quota. 5 2 U : U : Let I [ represent the 5 ascending moment of [. That is, 5 U : U : U : U : I [ I [ [ â [ 5 Theorem 3. and where 5 x 5 5Q 5 I [ U : : T ], á,] U : d: Qx 5 x 5 5 5Q 5 I [ U : [ U : : T ], á,] U : d: Qx (1 Ñ : is the event that at least of the (independent) events T, á, T occur U : is the event that exactly of the (independent) events T, á, T occur U (2) T is the event that ] ;. Proof Define R ;ßáß; R U to be the number of Types that have not reached their quota after balls have been drawn out.

5 Define R as the number of events amongst T, á, T that occur. It follows that U F [ U : Í R U However, it follows from that Thus, Q I1 ], á,] I1 ], á,] Y 3 t T Y 3 t 3 3 Q T Q RU F T R U : : : TR d: 5 F U 5 Q Q : TR U : : d: Q 5 Q U Q : T [ : : : d: Q Q T [ Q 5 Q U : : : d: Q Q T [ U : Q Q x 5 Q Qx 5 T [ Q5x x U : x Q. 5

6 Q Qx T [ 5 Q5x : Q5 U 5 Qx T [ : 5 t Q5x 5 Qx I [ 5 Q5x 5 U 5 ÀU 5 But 5 ÀU 5 ÀU [ 5 [ hence Qx : TR d : I [ Q5x 5 5 F 5 U ÀU and 5 5 Q5x 5 I [ : T F R ÀU : Qx U d where TR U F Tat least types do not obtain their quotabinomial model Therefore, where Q5x 5 I [ [ : T F R ÀU ÀU : Qx U d TR U F Texactly types do not obtain their quotabinomial modelþ Theorem. Suppose we draw balls without replacement from an urn initially containing balls of color, ßáß. 6

7 Again letting G equal the number of times color is selected, now let H 5 equal the number of G's which equal 5, 5 ß ß á ß. If á, then.. ) - EGH, H ßá, H ßßßá / EG^ ß^, á (0.0.1) x.-.) - ) where ^ß^ßáare independent and ^ µ Poisson )-. Models: Multivariate Hypergeometric Grouped Fermi Dirac Distribution, Urns with cells, at most one ball per cell, balls identical.

8 Applications Problem 1. Suppose we draw balls without replacement from an urn containing copies of each of different colored balls. Let Z equal the number of colors which are selected exactly times. Find @ Proof Define T +, á, + I and + á + T +, á, + I It follows from Theorem 2 that T \, á,\ T. ) T ] á ] x.,, T ) ) Let F represent the event that ]. Then by the generalized inclusion-exclusion principle we have

9 T ], á,] T T of the events F occur?@ 1 ) ) 1 1 ) x.?@?@ ) 1 ) ) x.?@?@ 1 ) @ @ ) 1 )?@?.)?@ x I?@? 9

10 @?? x 1 @? @? 1 Alternative proof using Theorem. T ) - / T x.. - ) - ) Let ^ µ Poisson )-. Then.. / x.-.) / ) - x@x.-.) ) / / - ) - ) / )- - x x

11 EXTRA Using Theorem, Let ^ µ Poisson )-. Then T H ßH ßáßH.. ) - / T Z ßZ ßáßZ x.. - ) - ).. ) - / TZ ßZ ßáßZ x.-.) - ).. x.-.) ) - )- / / - ).. x.-.) / ) ). x x = x G ß=ß - 5 = - 11

12 5 5 where G ß=ß â â 5 ßáß5 5 á5 = 5 ßßáß = x 5 G ß=ß = G ß=ß = 5 = = G ß=ß 5 = Is there an expression for the G??? Problem 2. Suppose we draw balls without replacement from an urn containing copies of each of different colored balls. Find the probability that every color is drawn at least once. Answer x Cß ß where C ß ß are the G-numbers defined by Cß ß Þ x and where we adopt the standard falling factorial notation â + x Þ 12

13 Proof The event that every color is drawn at least once is the event that notation of Problem 1. Therefore, from Problem 1, Z using the 3 3 TZ x C ßßÞ letting 3 3 Continuing with this notation we can identify new numbers which we will name the extended G -numbers and @ C ß ß ß 1 with the properties that Cß ß ß Cß ß and T x C ßßßß@. Problem 3. Suppose we continue to draw balls without replacement from an urn containing copies of each of different colored balls until all colors have been selected at least 5 times each. Let [ 5 equal the required waiting time (number of draws). Find the probability that it takes Adraws to get secure one of every color, i.e. T [ A. Then find the probability that it takes Adraws to secure two of every color, i.e. T [ A.+ Answer # 13

14 Proof Continuing with the notation leading up to Theorem 3, we let [ 5 [ : 5ßáß5á,,; and similarly R R 5 5ßáß5. It follows that [ 5 Ÿ A Í R5A and T [ 5 A T [ 5 ŸAT [ 5 ŸA TR A TR A 5 5 Now consider first the case 5. By Problem 2, x x T [ A CAßß CAßß A A x Aßß A Aßß C C A x CAßß A This last equality can be verified directly. We see that 1

15 CAß ß A CA ß ß A A A x x A A x A A x x CA ß ß Now consider first the case 5#Þ A T [ A TRA TRA # # # 2# Aßß 2# Aßß A A where we define 25Aßß â â. B B B ába B3 5ßáßß 3 ßáß Now take any set B ßáßBlB3 ß ß á ß 3 ß á ß and define Z â B B Then with ƒ # 2 Aßß Zƒ # # B ßáßB lb á B Aß B3 #ß á ß ß 3 ß á ß Þ Now define sets 15

16 U B ßáßB lb á B Aß B ß B3 #ß á ß ß 3 ß á ß ß 3 Á as subsets of universal set ƒ Þ Then, and w w ƒ2 U â U 2 Aßß Zƒ # # ZU w w â U Zƒ ZU â U Zƒ ZU â U 2 Aßß 2 Aßß 2 A ß ß Therefore, 16

17 T [ A # 2# Aßß 2# Aßß A 2 Aßß A A 2 Aßß A CA ß ß x Ax A CA ß ß x A x A Problem. Suppose we continue to draw balls without replacement from an urn containing copies of each of different colored balls until all colors have been selected at least 5 times each. Let [ 5 equal the required waiting time (number of draws). Find I [ and I[ #. Answer 5 5x E[ : ßáß x 5 5 Proof We have from Theorem 3 that where 5 x 5 5Q 5 I [ U : : T ], á,] U : d: Qx 1

18 (1 Ñ : is the event that at least of the (independent) events T, á, T occur U (2) T is the event that ] ;. We consider the two special cases, ß; and ß;#. In this first case we will find the 5 2 ascending moment while in the later case we will consider only the first ascending moment. If we take ; á ; ;, then by the general inclusion-exclusion principle, T ], á,] Tat least of the events T, á, T occur U : Tat least of the ] ; T ] ; ; C C : : C C C : : C ; C C Therefore, : T ], á,] U : d: 5 5 C C : : : d: C ; C Now suppose ;. Then 1

19 5 5Q 5x I [ 5 : ßáß Qx : T ], á,] U : d: 5Q 5x 5 Qx : : : d : 5Q 5x 5 Qx : : d: 5Q 5x 5 x x Qx 5 x 5x 5x x x 5 x 5x x 5 5 5x Þ x 5 5 Now suppose ß ; #ß 5. Then x I [ : #ßáß# x : T ], á,] U : d: : : : : d: : : : d: 19

20 : : d: : : d: x x x Now suppose ß ; #. 5 5Q 5x I [ 5 : #ßáß# Qx : T ], á,] U : d: 5 5x 5 C C x : C : : d: C x d 5 5x 5 : : : : : : 5 x d 5 5x : : : : : 20

21 5 5x 5 : : : d: x 5 5x 5 x : : d: 5 5x 5x x x 5x 5 5x 5x x x 5x 5 5x x 5 5 Now suppose ;# x E[ 5 : #ßáß# x : T ], á,] U : d: 5 5x 5 : C C x C : : d: C x 5 5x 5 : : : : d: 5 5x x 5 5x x : : : : d: 5 21

22 5 : : : d: 5 5x x : : d: 3 5 5x x 3 53 x x x 5 5x x x x 53x 3 5 5x x Now suppose ; x E[ 5 : ßáß x : T ], á,] U : d: 5 5x 5 x : T ], á,] U : d: 22

23 5 5x 5 x : C C C : : d: C 5 5x 5 x : : d: 5 5x 35 3 x : 3 d: 3 5 5x 3 35 x 3 : d: 3 3 x x 5 5x 3 x x x x x x x 35 5x x x x x Problem 5. 23

24 Suppose we draw balls without replacement from an urn containing copies of each of different colored balls. Let Z equal the number of colors which are selected times. Find IZÞ Answer Proof IZ I I \ I I \ T \ Problem 5. Show that and Proof By the law of total probability 2

25 3 3 3 TZ from which the first identity follows immediately. To establish the second identity, we note from Problems 1 and 2 taken together that, I Z 3T Z It follows that Problem 6a. Suppose we draw balls without replacement from an urn containing copies of each of different colored balls. Let ^ equal the number of different colors that are selected in this sample. where T ^ 5 5 G ß5ß Gß5ß 5x 25

26 E^ = = = References: Walton Korwar Charalambides, On a Restricted Occupancy Model and its Applications, Biom. Journal, Vol. 23, no. 6, 191, pages Problem 6b. Suppose we draw balls without replacement from an urn containing copies of each of different solid colored balls and = identical striped balls. Let Y equal the number of different solid colors that are selected in this sample and let Z be the number of different striped balls that are selected in this sample. where = TY 5ßZ Kß5ßß= = Kß5ßß= EY Z / / = / / = / = EY / = / / = 26

27 References: Charalambides, On a Restricted Occupancy Model and its Applications, Biom. Journal, Vol. 23, no. 6, 191, pages Problem. We have distinguishable urns, each with = distinguishable cells. A cell cannot hold more than ball. Identical balls are randomly distributed (all empty cells equally likely at each turn) until 5 urns, among the specified urns, are occupied by at least one ball. Let Q equal the number of turns required. for 5ß 5 ß á where T Q =5 G ß 5 ß =ß = = = 5 Gß5ß=ß= == 5x 5 5 is the non-central G-number (Charalambides, Koutras, On the Differences of the Generalized Factorials at an Arbitrary Point and Their Combinatorial Applications, Discrete Mathematics, Vol., 193, pages ) Charalambides, Ch. A., A Unified Derivation of Occupancy and Sequential Occupancy Distributions, Advances in Combinatorial Methods and Applications to Probability and Statistics, N. Balakrishnan (editor), 199, pages EQ 5 5 x= = = 5 3 == Problem. 2

28 An urn consists of # balls of each of = different colors. Balls are drawn without replacement until both balls of some color have been drawn out. Let R equal the number of draws required. ER #= xx #= xx ERR # #= Reference: Blom, Gunnar and Holst, Lars, Embedding Procedures for Discrete Problems in Probability, Mathematical Scientist, 16, 29-0, Problem 9. Suppose we draw balls without replacement from an urn containing copies of each of different colored balls. Let H equal the number of colors which are selected exactly times. The joint descending factorial moment EH âh is given by EH âh V V W where V á and W á. The special case EH Ð i.e. á Ñsimplifies to EH in agreement with C. Charalambides, On a Restricted Occupancy Model and its Applications, Biom. Journal 23 (191), no. 6,

29 Proof.. ) - EH âh / E^ â^ x.-.) - ).. ) - / x.. E - ) ^ - ).. ) - / x.. )- - ) - ).. ) - V W / ) - x.-.) where V á - ) and W á x. V W - - x Vx. - x V x x Vx W - V V W. 29

30 Method 2 Blom, Gunnar and Holst, Lars, Embedding Procedures for Discrete Problems in Probability, Mathematical Scientist, 16, 29-0, Let ÐY ß, á,y ß Ñ, á, ÐY 3ß, á,y 3ß Ñ, á, ÐY ß, á,y ß Ñ be a random sample of size M m 3 á m from the continuous UniformÐ0,1 Ñ distribution. Let Y Y á Y be the ordered values of Y. ÐÑ Ð#Ñ ÐMÑ 3ß We will say that Y is from the i group provided Y ÐY,,Y Ñ. ÐÑ 2 ÐÑ á 3 3ß 3ß For 1 Ÿ t Ÿ M, let N 3ß equal the number of values among Y ÐÑ á Y Ð Ñ which are from the group. i 2 Now consider drawing objects one by one at random and without replacement from an urn initially containing m3 objects of color i, i 1, á, n. For 1 Ÿ t Ÿ M, let C equal the number of objects of color i drawn in the first t draws from this urn. 3ß Clearly for all 1 Ÿ t Ÿ M, N, á,n d C, á,c. ß ß ß ß When q objects of color i have been drawn or equivalently when q values from the i group of uniform 3 3 variates ÐY 3ß, á,y 3ß Ñ have appeared in the list YÐÑ Y Ð#Ñ á Y Ð Ñ we will say this color (or 3 M group) has reached its quota. Let Ðk;q, á,q Ñ 5 represent the draw (index) when exactly k non-specified colors (groups) reach their quota. Our goal is to find a formula for E. 5 2 Define Y ÐÑ 0. We note that Y D Ð Ñ 5 5 where D Y ÐÑ Y ÐÑ. Furthermore it is well know that D, á,d are exchangeable random variables following a Beta distribution with parameters 1 and M m á m. That is the density function f( y) of D is f( yñ MÐ1 y Ñ M. 30

31 Now let YÐ3ßÑ Y Ð3ß#Ñ á Y Ð3ß Ñ be the ordered values of the iid UniformÐ0,1) variates 3 ÐY,,Y Ñ. á 3 3ß 3ß It follows that 2 Ð Ñ Ðß; Ñ Ð#ß; Ñ Ðß; Ñ Y k largest value in the set of independent variates Y, Y, á, Y. 5 # 2 Ñ However it is well know that the j largest value from a set of n iid Uniform(0,1 variates follows a Beta distribution with parameters q and m q 1. That is the density function f( y) of Y is Ðß; Ñ m Ðq 1 Ñ Ðm q Ñ f( yñ y ; ; Ð1 yñ. Finally we can show that 5 is a stopping time. Hence EY Ð 5 Ñ 5 E D E ED 5 E 1 5 M1 Therefore, E 5 ÐM 1 Ñ EY Ð 5Ñ where 2 Ð Ñ Ðß; Ñ Ð#ß; Ñ Ðß; Ñ Y is the k largest value in the set of independent Beta variates Y, Y, á, Y 5 # where Y Ð3ß; Ñ µ Beta distribution Ðq, m q 1 Ñ. 3 In our particular problem 31

32 m á m m and q á q 1. 32

33 For the purposes of Theorem 3 which follows we will suppose that ], á,] are independent binomial random variables parameterized such that for #á,,, T ] C C C : : Cßßá, Þ C Suppose an urn contains objects of Type, á, objects of Type and that objects are drawn from this urn without replacement. Let Q á. When ; 3 objects of Type 3 have been drawn we will say Type 3 has reached its quota. Let [ :; ßáß; á ; [ : U represent the waiting time until exactly different,, types have reached their quota. 5 2 U : U : Let I [ represent the 5 ascending moment of [. That is, 5 U : U : U : U : I [ I [ [ â [ 5 Theorem 3. where 5 x 5 5Q 5 I [ U : : T ], á,] U : d: Qx (1 Ñ : is the event that at least of the (independent) events T, á, T occur U (2) T is the event that ] ;. T ], á,] Tat least of the events T, á, T occur U : Tat least of the ] ; 33

34 T ] ; ; C C C : : C ; C C C : : C 5 : T ], á,] d: U : 5 : C C C : : d: ; C Suppose. ; 5 C C : C : : d: C If ; 5 : : : d: : : : 5 d : 5 d: 5 3

35 5 5 using Gould's identity (1.1) If ; # 5 C C : C : : d: C 5 : : : : : d: 5 : : : : d: 5 : : : d: 5 : : d: 5xx 5 x 5 For the special case 5, this simplifies to However, it is well known that for nonnegative integers + and,, 35

36 : : : +, d +x,x +, x C T ] ; : : ; 5 C C T ] ; ŸŸ T Suppose E, E, á, E are sets within a universal set. Define : # H B H ± Bis an element of at least of the sets E, E, á, E # Define TE â E 5 Ÿ5Ÿ Ð ß á, Ñ& where 5 is the set of all samples of size 5 drawn without replacement from ß #ß á ß ß where the order of sampling is not considered important. Then, T 36

37 5 5 á á. 5 Problem 3. 3

38 Sampling Without Replacement Model or Grouped Fermi-Dirac Allocation Model Define to be the product space 0,1, á, m â 0,1, á, m and let be the set of all vectors Ðs, á,s Ñ in such that s á s t. Suppose that \, á,\ is a multivariate hypergeometric random vector. That is, for all s, á,s, T \ =, á,\ = where Q m á m and = á = Þ = â = Q For the purposes of Theorems 1 and 2 which follow we we will suppose that ], á,] are independent binomial random variables parameterized such that for #á,,, C C ) ) T ] C 1 C ß ß á, Þ C ) ) Theorem 1.. I 1 \ á \ Q,, ) I1 ] á ] x.,, ) where the in I is used to denote that \ á \. Q ) Theorem 2. Let T and define T T. Then for, T \ á \. T ] á ] x. Q,, T ),, T Q ) ) 3

39 For the purposes of Theorem 3 which follows we will suppose that ], á,] are independent binomial random variables parameterized such that for #á,,, T ] C C C : : Cßßá, Þ C Suppose an urn contains objects of Type, á, objects of Type and that objects are drawn from this urn without replacement. Let Q á. When ; 3 objects of Type 3 have been drawn we will say Type 3 has reached its quota. Let [ :; ßáß; á ; [ : U represent the waiting time until exactly different,, types have reached their quota. 5 2 U : U : Let I [ represent the 5 ascending moment of [. That is, 5 U : U : U : U : I [ I [ [ â [ 5 Theorem 3. and where 5 x 5 5Q 5 I [ U : : T ], á,] U : d: Qx 5 x 5 5 5Q 5 I [ U : [ U : : T ], á,] U : d: Qx (1 Ñ : is the event that at least of the (independent) events T, á, T occur U : is the event that exactly of the (independent) events T, á, T occur U (2) T is the event that ] ;. Problem 1. (a) Suppose we draw balls without replacement from an urn containing copies of 39

40 each of different colored balls. The probability that colors are selected exactly times equals @ References and The special case and is equation 6.1 of Charalambides, A New Kind of Numbers Appearing in the -Fold Convolution of Truncated Binomial Negative Binomial Distributions, SIAM Journal of Applied Mathematics, Vol 33, No. 2, September 19, pages Charalambides's expresses his solution in the form x Cß ß where C ß ß are the G-numbers defined by Cß ß x (b) The probability that at colors are selected exactly times equals @ (c) The expected number of colors that are drawn exactly times equals 0

41 Problem 2. Suppose we continue to draw balls without replacement from an urn containing copies of each of different colored balls. (a) The probability that A draws will be required to secure at least one ball of every color equals References x A CA ß ß This result is equation 6.2 of Charalambides, A New Kind of Numbers Appearing in the -Fold Convolution of Truncated Binomial and Negative Binomial Distributions, SIAM Journal of Applied Mathematics, Vol 33, No. 2, September 19, pages (b) The probability that A draws will be required to secure at least two balls of every color equals # x # A A# GA #ß ß A Note and The result in 2(b) can be derived from 2(a) by an inclusion-exclusion argument the waiting time to secure at least three balls of every color could be derived from 2(b) by the same inclusion-exclusion argument. Unfortunately continuing in this way does not seem to lead to a succinct formula for the waiting time to secure at least 5 balls of every color. The probability that Adraws will be required to secure 5 balls of every color can be expressed in terms of the Generalized C-Numbers ÐEquation 3.13, Charalambides, The Generalized Stirling and G numbers, Sankhya-, Series A, Vol. 36, Pt., 19, pp ), but there is no succinct formula for the generalized G-numbers, even though many properties and applications of these 1

42 numbers are well known. 2 (c) The 5 ascending factorial moment of the number of draws required to secure of the different colored balls equals x 5x 5 5 The special case 5 and simplifies to We can compare this result with its well known formula with replacement analog á # for the expected number of draws required to secure all different colored balls when sampling with replacement from an urn with different colored balls. Problem 3. (a) Suppose we draw balls without replacement from an urn containing copies of each of different solid colored balls and = identical striped balls. Let Y equal the number of different solid colors that are selected in this sample and let Z be the number of different striped balls that are selected in this sample. = TY 5ßZ Kß5ßß= = where Kß5ßß= are the Gould-Hopper numbers. (b) 2

43 EY Z / / = / / = / = (c) EY / = / / = References Charalambides, On a Restricted Occupancy Model and its Applications, Biom. Journal, Vol. 23, no. 6, 191, pages Problem. cells We have distinguishable urns, each with = distinguishable cells. A cell cannot hold more than ball. Identical balls are randomly distributed (all empty equally likely at each turn) until 5 urns, among the specified urns, are occupied by at least one ball. Let Q equal the number of turns required. (a) T Q =5 G ß 5 ß =ß = = = for 5ß 5 ß á where 5 Gß5ß=ß= == 5x 5 5 is the non-central G-number (Charalambides, Koutras, On the Differences of the Generalized Factorials at an Arbitrary Point and Their Combinatorial Applications, Discrete Mathematics, Vol., 193, pages ) References Charalambides, Ch. A., A Unified Derivation of Occupancy and Sequential Occupancy Distributions, Advances in Combinatorial Methods and Applications to Probability and Statistics, N. Balakrishnan (editor), 199, pages

44 (b) 5 5 x= = = EQ 5 3 == References Charalambides, Ch. A., A Unified Derivation of Occupancy and Sequential Occupancy Distributions, Advances in Combinatorial Methods and Applications to Probability and Statistics, N. Balakrishnan (editor), 199, pages Problem 5. The ascending factorial moment of the number of draws required to draw out all copies of any color of ball when sampling without replacement from an urn containing copies of each of different colored balls equals 5 2 5x x 5 References The two special cases #ß5 and #ß5# are given in Blom, Gunnar and Holst, Lars, Embedding Procedures for Discrete Problems in Probability, Mathematical Scientist, 16, 29-0, 1991.

45 Example 0.0. Suppose we draw balls without replacement from an urn containing copies of each of different solid colored balls and = identical striped balls. Let Y equal the number of different solid colors that are selected in this sample and let Z be the number of striped balls that are selected in this sample. = (a) TY 5ßZ Kß5ßß= = where Kß5ßß= are the Gould-Hopper numbers. 5 K ß5ßß= x 5 5 = 5x Proof T Y 5ßZ T G 5ßH H áh 5

46 Suppose an urn containins copies of each of different colored solid balls and = copies of each of different colored striped balls. The number of ways to select balls without replacement from this urn and get a sample with 5 of the solid colors and a total of striped balls equals = 5x 5 x G ß5ß. where the G-numbers where defined earlier by 5 5x Gß5ß. x 3 3 Charalambides [0] considers this problem. There is a misprint where he uses Gß5ß= instead of Gß5ß. Proof In the notation of Theorem with, the problem asks us to find = TG 5ß]. By Theorem TG 5ß].. = = ) - / T [ 5ß^ x.-.) - - where [ µ Poisson and ^ µ Binomial - ) =, - ) = = = x.-.) 5x - - ) 5.. ) - / ) - = - / - ) =.. = ) - 5 / ) - - x5x.-.) ) 6

47 = =.- - 5x- x5x. 5 x - = x. x5x = 5x = 5 = x. 5 - = = = x = x x = 5 3 = Therefore, = = T Y 5ßZ

48 Example Suppose we draw balls without replacement from an urn containing copies of each of different solid colored balls and = identical striped balls. EG ] / = =@ / = Proof EG ] /.. x.-.) = ) - = - / E[ / ^ -.. x.-.) = ) - = - / E[ / E^ -.. = = - / ) = - = ) - / - x.-.) -. =. = - ) - / - / ) = x.- -.) x. x.- / x = - - = x / x = x = x x / x = =@ =@ = / =@ = =@ - - ) ) ) ) = / =@ x = x = / =@ =

49 EG /.. x.-.) = ) - = - / E[ / - ).. = = ) - / - / ) - x.-.) ). =. = ) - / - / ) - x.-.) ). = / x x.- / x = - - x / x = - x. x.- / x = x x =@ =@ - - x / x = = =@ E].. x.-.) = ) - = - / E^ - ).. = = / = x.-.) - ) ). = - = - = - x.- - = x.-. = = - - = = = x x - - 9

50 = = = 50