1
Unordered Samples without Replacement
Consider population of n elements a1 , a 2 ,...., an .
Any unordered arrangement of r elements is called an unordered sample of size r. Two unordered samples
are different only if one contains an element not contained in the other.
Consider unordered sample of size r. This sample can be used to make r! Ordered samples (r!
permutations).
Example:
Combinations unordered sample:
(a,b,c)
Permutation ordered samples:
(a,b,c); (a,c,b); (b,a,c) (b,c,a); (c,a,b); (c,b,a)
The total number of ordered samples (n r ) =6
Therefore,
Number of unordered samples=
Number of ordered samples
Number of ordered samples per
unordered sample
=
=
(n )r
r1 !
=
n (n − 1 )... (n − r + 1 )
1 * 2 * ... * (r − 1 )r
n!
=
(n − r )! r !



n
r


Example
A box contains 75 good IC chips and 25 defective chips, and 12 chips are selected at
random. Let:
A= ”at least one chip is defective”
Find P(A)
 100 

 12 
S =
A = ”no chip of sample is defective”
 75 

 12 
A =
2
()
PA =
A
S
=
 75 


 12 
 100 


 12 
=
75! 12! 88!
75! 88!
=
= 0.025
12! 63! 100! 63! 100!
Hence,
()
P ( A) = 1 − P A = 0.975
Example:
 52 


13 
= 635,013,559,600 different bridge hands.
 52 


5 
= 2,598,960 different poker hands.
Bridge 
Poker 
For example,
Let A= “hand of poker contains five different face values”
These face values can be chosen in
13

 

5 
ways, and corresponding to each card we are free
to choose one of four suits.
Thus, on each card:
P ( A) =
13 


5 
4 5 
 52 




5 
=
A
= 0.5071
S
Example:
Consider r distinguishable balls in n cells. Define,
Ak = “specified cell contains exactly K balls”
r 
Can place r balls in n cells n r different ways. S = n r . K balls can be chosen in   different
k
 
ways. The remaining (r-k) balls can be placed into the remaining n-1 cells in (n − 1)
r −k
P( Ak ) =
r 
 
 
k
ways.
(n − 1)r −k
nr
Example:
How many lines connect 5 points with no 3 co-linear. To answer this we must select 2 points
(which define a line) from the 5 given points, i.e.
3
5
 
 
 2
Multinomial Coefficients
Now, consider the following situation: A set of distinct items is to be divided into r distinct groups of
∑ n . How many different divisions are possible? To answer this, we
r
respective sizes n1,n2,…nr, where
i
i =1
n
 
n 
 1
note that there are
 n − n1 


 n


2 
n



possible choices for the second group; for each choice of the first two groups there are
− n1 − n2 
n3
possible choices for the first group; for each choice of the first group there are



possible choices for the third group; and so on. Hence it follows from the generalized
version of the basic counting principle that there are
So we can state:
FG n IJ FG n − n IJ FG n − n − n −
n
Hn KH n K H
1
1
1
2
2
nr −1
r
IJ
K
n!
n − n1
(n − n1 − n2 − n3 − nr −1 )!
0!nr !
(n − n1 )!n1 ! (n − n1 − n2 )! n2 !
n!
=
possible divisions.
n1 ! n2 ! nr !
=
If n1 + n 2 +
n


 n ,n ,
 1 2
…

=
n r 
Thus,

+ n r = n, we define 
n
…n
 n1 ,n 2 ,
r




by
n!
n1! n 2 ! n r !
n


 n ,n ,
 1 2
…


n r 
represents the number of possible divisions on n distinct objects into
r distinct groups of respective sizes n1, n2 …nr
Example:
A police department in a small city consists of 10 officers. If the department policy is to
have 5 of the officers patrolling the streets, 2 of the officers working full time at the station,
and 3 of the officers on reserve at the station, how many different divisions of the 10b
officers into 3 groups are possible?
4
Solution: There are
10!
= 2520 divisions.
5!2!3!
Example:
There are 10 boys who are to be divided into an A team and a B team of 5 boys each. The A
team will play in one league and the B team in another. How many different divisions are
possible?
Solution: there are
10!
possible divisions.
5!5!
Example:
In order to play a game of basketball, 10 boys at a playground divide themselves into two
teams of 5 each. How many different divisions are possible?
Solution: Note that this example is different from the previous one because now the order of
the two of the two teams is irrelevant. That is, there is no A and B teams but just a division
consisting of 2 groups of 5 boys each. That is, there are twice (2!) as many divisions in the
first case as in this case. If there were three teams, then there would be 3! As many
divisions than if there were teams A, B and C. Hence the desired answer is
10! / 5!5!
=126
2!
Summary: This is a good place to summarize.
Permutation:
Definition: The number of distinct arrangements that can be made from the n elements of
S, using r of them at a time, is denoted by (n )r and called the number of permutations of
n things taken r at a time. Note that r ≤ n . Order is important.
Combination:
Definition: The number of distinct subsets of size r that can be formed from the n
n
elements if S is denoted by   , and is called number of combinations of n things taken
r
 
r at a time. Note that r ≤ n . Order is not important, (binomial coefficient)
-When the sample contains several sets of identical elements, we have permutation with
repetition, or undistinguishable sets. The number of permutations of n objects of which n1
are alike, n2 are alike, ... nr are alike, etc
Hence (n )n1 , n2 , ... nr =
n!
− multi-nominal
n1!n2 !...nr !
If we have n objects, n!- permutations exist. If n1 and n2 are alike then
distinguishable permutations.
n!
= # of
n1! n2 !
5
Example
How many different signals, each consisting of 8 flags hung in a vertical line, can be
formed from a set of 4 indistinguishable red flags, three indistinguishable white flags, and
a blue flag? We seek the number of permutations of 8 objects of which 4 are alike (the
red flags) and 3 are alike (the white flags). By the above theorem, there are,
8! 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
=
= 280 different signals.
4!3!
4 * 3 * 2 *1* 3 * 2 *1
Example:
How many ways can 12 people be divided into 3 rows of 4 each? (Order is not important)
12  8  4 
 12 

3! =
 

 

 
3! = 
 4,4,4 
 4  4  4 
7000
inter-changing rows once selected
Example:
Two sets of times are included in a group of eight people. How many ways can six
distinguishable people from this group be arranged in a row?
 2  2  4 

 

 

 
6! =
1 1  4 
2880
taking care of permutations in the row
Example
A dance class consists of 22 students, 10 women qnd 12 men . if 5 men and 5 wiomen are
to be chosen and then paired off, how many results are possible?
 10  12 
   possible
 5  5 
Solution: There are 
choices of the 5 men and 5 women. They can then
be paired up in 5! Ways, su=ince if we arbitrarily order the en then the first man can be
paired with any of the 5 women. Thenwxt with any of the remaining 4, and so on. Hence
 10  12 
 
 5  5 
possible results.
there are 5! 
Example
In howmany ways can n identical balls be distributed into r urns so that the ith urn contains
at least mi balls, for each i=1,2,…r? Assume that n ≥
∑m .
r
i =1
… + x = n,
Solution: The number of integer solutions of
x1 +
r
xi ≥ mi
Is the same as t=he number of nonnegative solutions of
i
6
y1 +
…+ y = n − ∑ m ,
r
r
i
yi ≥ 0.
1
Proposition 6.2 in our text gives

n−
the result 



∑ m + r −1 .
r





i
1
r −1
Example
A pair of six sided dice is thrown until an 8 or 11 appear. What is the probability that an 8
appears first?
5/36
2/36
other 29/36
8
8
8
5/36
11
5/36
11
2/36
2/36
Other 29/36
other
A= “8 appears first”
B= “11 appears first”
2
∑
5
5 29 5  29 
5 ∞  29 
P ( A) =
+
+
+   + ... =


36 36 36 36  36 
36 k =0  36 
=



1 
 5 


 36 
 1 − 29 

36 


2 
P(B ) =

36 
=
5
← most prob.
7

1  2
=
29  7
1− 
36 

k
11
29/36
7
Now, Let us look at distributing balls into sum when all the balls are undistinguishable. We know n
distinguishable balls can be distributed to r possible urns in rn possible outcomes. When n
indistinguishable balls are used, then the n balls put into r urns is described by the outcome;
(x1 , x2 ,...x r ), Where xi →# balls in ι th sum
So we have to find the number of distinct, non-negative, integer-valued vectors, (x1,x2,…xr), such that;
X1+x2+ . . .+xr = n
In order to find this, suppose we first have n indistinguishable objects lined-up, and we want to divide
them into r non-empty groups.
0 x 0 x 0 x.... x 0 x 0
n object, 0, with x denoting a space
We need to select r-1 spaces of the n-1 spaces, x, available between adjacent objects.
i.e. If n=8, r=3, then the two dividers are 000/000/00
and the vector is ( x1 = 3, x 2 = 3, x 3 = 2) = (3,3,2 )
− 1

 possible selections, we can conclude.
r − 1
n
Since there are 
Theorem I
− 1

 distinct positive integer-valued vectors ( x1 , x 2 ,... x r ) satisfying
r − 1
n
There are 
x1 + x 2 + ...,+ x r = n x > 0 i = 1, r
8
To obtain the number of non-negative solutions (that is, to allow empty cells), the number of nonnegative solutions of x1 + x 2 + ... + x r = n is identical to the number of positive integer solutions of
y1 + y 2 + ... + y r = n + r (Which we can see by letting yi = xi + 1, i = 1, ...r )
From Theorem I, above, we get
 N − 1




r
−
1


=
(n + r − 1)! , with N = n + r
(n!)(r − 1)!
And, Theorem II is
“There are (n+r-1) distinct non-negative integer valued vectors
(x1 , x2 + ...xr ) satisfying x1 + x2 + ... + xr
=h
Example:
How many ways can we distribute 3 black, indistinguishable balls into two sum as alike the #
of distinct non-negative integer valued solutions of x1 + x2 = 3 are possible.
∴ n = 3, r = 2
∴
 3 + 2 − 1 = 4 ⇒ (0,3)(1,2)(2,1)(3.0)
 3 
Example:
Consider an investor has $20,000 to invest among 4 possible investments.
How many different investment strategies are possible?
If: 1) all money is to be invested
2) not all money is to be invested.
Solution:
Let xι ; i = 1, 2, 3, 4 , be the number of 1000’s of dollars invested in investment i.
a) Then when all money is to be invested,
x1 + x2 + x3 + x 4 = 20 xι ≥ 0
Hence, when
9
n = 20, r = 4 ,
 20 + 4 − 1



20 

 23 


 20 
= 
=
23!
23 * 22 * 21 23 * 22 * 7
=
=
70!−3!
3* 2
2
= 1771 possible stategies
b) If not all money is to be invested, we let x5 denote the amount kept in reserve.
In this case, a strategy is a non-negative integer-valued vector ( x1 , x 2 , x 3 , x4 , x5 ) = 20
∴ n = 20, r = 5
n



+ r − 1  24   24 
24!



= 10626
=

=

=
n
4!20!
  20   4 
Example
Consider a set of n antennas, of which m are defective and n-m are functional. Assume that the
defective and all the functioning antennas are indistinguishable. How many linear ordering are
there in which no two defective antennas are consecutive, n − m < m .
Solution:
Imagine n-m functional antennas lined up.
0-functional antennas
ϖ-defective antennas
n-m
0 0 0 0…0
Now, if no two defections are to be consecutive, then the spaces between the functional
antennas must contain at most one defective antenna.
∧
0 ∧ 0 ∧ 0... ∧ 0 ∧ 0
That is in the n-m+1 possible positions between the n-m functional antennas, we must select m
of these to put the defective antennas.
Hence:
 n − m + 1




 m

m < n2
These are the possible orderings in which there is at least one functional antenna between two
defective ones.
10
A Useful combination identity is;
n

 

r 
− 1  n − 1


+

 =1≤ r ≤ n
r − 1  r 
n
= 
− 1
combination of size r


r − 1
n
i.e. Consider n object and focus on object #1. Now, there are 
that contains object 1.
n
Also, there are 

− 1
 combinations of size r that do not contain object 1.
r 
n
Since then are   combination of size r.
r
 
 n
 
 
r 
 n − 1  n − 1
+

 

 r − 1  r 
= 
QED
Proof:
n

 

r 
(n − 1)!
(n − 1)!
− 1   n − 1
=
+

+


(n − 1) − (r + 1)! (r − 1)! r! (n − r − 1)!
 r −1  r 
n
=
(n − r ) = (n − 1)!  r + n − r 


r (n − r )! 
 r! (n − r )! 
r +
= (n − 1)! 

=
n 
n(n − 1)!
n!
=
=  
r! (n − r )! r! (n − r )!  r 
We are now ready to give the formal axiomatic definition of probability.
“Let S be the set of all outcomes ε i of an experiment ε . A is a set (or sub set of S of event points.
Hence we have a probability space. (S,F,P)”
If we can assign P(A) = Probability Measure of event A in F such that the following axioms will be
satisfied:
i) 0 ≤ P( A) ≤ 1
11
ii) P (S ) = 1 - Probability of certain event is 1
iii) P(0) = 0 - Probability of impossible event
is 0
iv) If AB = 0 -no points in common, then
P ( A + B ) = P ( A) + P ( B )
if Aβ ≠ 0
P( A + B ) = P( A) + P(B ) − P( AB )
Example:
We had two events A and B.
A = A'+C 
 A + B = A'+ B'+C '
B = B'+C '
∴ P( A) = P( A'+C ') = P( A') + P(C ')
P(B ) = P(B'+C ') = P( B') + P(C ')
∴ P( A + B ) = P( A'+ B '+C ')
= P( A') + P( B') + P(C ')
But,
P( A') = P( A) − P(C ')
P(B') = P(B ) − P(C ')
∴ P (a + B ) = P (a ) + P (B ) − P(c' )
∴ P (a + B ) = P (a ) + P (B ) − P(aB ) ← QED
We can extend to three and four events as follows,
P( A + B + C ) = P(a ) + P (B ) + P (c ) − P ( AB ) − P( AC ) = P (BC ) + P ( ABC )
P( A + B + C + D ) = P( A) + P(B ) + P(C ) + P(D ) − P( AB ) − P( BC ) − P(BC )
− P( AC ) − P( AD ) − P(BD ) − P(CD ) + P( ABC ) + P( ABD ) + P( BCD )
+ P( ACD ) − P( ABCD )
Generalizing:

P
∪ Ε = ∑ P(Ε ) − ∑ P(Ε Ε
n
i
 L=1
+ (− 1)




r +1
+ (− 1)
n +1
n
i
L=1
∑ P(Ε
L1 < L2 < Lr
L1
i1 <i2
i1
Ε L2 ...Ε Lr ) + ...
P(Ε1Ε 2 ...Ε n )
i2
) + ...
12
The Summation:
∑ P(Ε Ε ...Ε ) is taken over all
i1
i2
ir
n

 

r 
possible subsets of size r of the set,
{1,2,…,n}.
Example
Consider tossing three coins and observing how they land. The Space, S, can be described
as follows;
A1
A2
A3
A4
A5
A6
A7
A8
=
=
=
=
=
=
=
=
1st
H
H
H
T
H
T
T
T
2nd
H
H
T
T
T
T
H
H
3rd
H
T
T
T
H
H
H
T
Let A = 2 heads occur
B = First coin is a head
A = {A2 , A5 , A7 }
B = {A1 , A2 , A3 , A5 }
AB = {A2 , A5 }
Assuming equal probability for each even
P( A1 ) = P ( A2 ) = ... = 1
8
P( A) = 3 8 , P(B ) = 4 8 , P ( AB ) = 2 8
∴ P ( A + B ) = P ( A) + P(B ) − P ( AB ) = 3 8 + 4 8 − 2 8
P( A + B ) = 5 8
13
Example
An urn contains n balls, of which one is special. If k of these balls are withdrawn one at a
time, with each selection being equally likely to be any of the balls that rem,ain at the time,
what is the probability that the special ball is chosen?
Solution: Since all the balls are identical, then the set of k balls can be one of the
 n
  sets of k balls. So
k
P{special ball is selected} =
 1  n − 1 
 

 1  k − 1 
 n
 
k
=
k
n
This could also have been obtained by letting Ai denote the event that the special
ball is the ith ball to be chosen, i=1,…k. Then, since each one of the n balls is
equally likely to be the ith ball chosen, it follows that P(Ai)=1/n. Hence, since these
events are clearly mutually exlusive, then
k
k


k
P{special ball is selected} = P  Ai  = P( Ai ) =
n
i =1
 i =1

∪
∑
It could have been argued that P(Ai)=1/n, by noting that there are
n(n − 1) (n − k + 1) = n !/(n − k )! equally likely outcomes of the experiment, of
which (n − 1)(n − 2) (n − i + 1)(1)(n − i ) (n − k + 1) = (n − 1)!/(n − k )! result in the
special ball being the ith one chosen. From this it follows that
(n − 1)! 1
P( Ai ) =
=
n!
n
Example
A football team consists of 20 offensive and 20 defensive players. The players are
to be paired in groups of 2 for the purpose of determining roommates. If the pairing
is done at random, what is the probability that there are no offensive-defensive
roommate pairs? What id =s the probability that there are 2i offensive-defensive
pairs, I=1,2,…10?
Solution ‘ We have 40 player that must be paired onto groups of 2, Hence there are
40  (40)!


=
20
 2, 2,..., 2  (2!)
ways od sorting the players into ordered pairs.(i.e. there is a first pair, second pair,
etc) To get unordered pairs we simply divide by 20!.
Now, a division will result in no offensive-defensive pairs if the offensive (and
consequently, the defensive) players are paired among themselves. It should be
 (20)! 
obvious that there are  10
 such division for both offensive and defensive
 2 (10)!
players. Hence, the probabiolity of no offensive-defensive pairing, PO , is
14
PO =
 (20)! 
 10

 2 (10)! 
2
=
[(20)!]3
2
(40)!
(10)!] (40)!
[
220 (20)!
To determine P2i , the probability that there are 2i offensive-defensive pairs, we
note there are
 20 


 2i 
2
ways of selecting the 4i players involved (2i offensive and 2i
defensive). These 4i players can then be paired up (2i)! Pairs. The remaining 202i (offensive and defensive) players are then paired among themselves, it follows
that there are
 20 


 2i 
2
 (20 − 2i)! 
(2i)  10−i
(10 − i )!
2
2
divisions which lead to 2i offensive-defensive pairs. Hence
P2i =
 20 


 2i 
2
 (20 − 2i )! 
(2i)!  10

 2 (10 − i)!
2
, i=0,1,...,10
(40)!
220 (20)!
another example which illustrates the Axioms of probability and the mathematical dexterity
needed for solving problems is as follows.
Example:
Suppose that each of N men at a party throws his hat into the center of th e room,
The hats are first mixed up and then each man randomly selects a hat. What is the
probability that
!a) none of the men selects his own hat;
(b) exactly k of the men select their own hat?
Solution:
(a) We first go in “the back door” and calculate the probability of the
complementary event of at least one man’s selecting his own hat. Let Ei, i=1,2,…N,
be the event that the ith man selects his own hat. Now by the general expression for
the union of N non-mutually exclusive events, the probability that at least one of the
men selects his ownhat is
(−1) n +1
∑
i1 < i2

P
∪
N
 i =1

P( Ei1 Ei2
Ei  =

Ein )
in
∑ P( E ) − ∑ P( E E ) +
N
i
i =1
+
i1 <i2
i1
i2
+ (−1) N +1 P ( E1 E2
+ (−1) n +1
∑
i1 < i2
EN )
P ( Ei1 Ei2
in
Ein )
15
Consider this experiment resulting in a vector whose elements represent the number of the hat
taken by the ith man. For example, the vector (E1E2 …En) is the vector for the event that each
man selects his own hat.. Then there are N! possible such vectors
Now, the event that each of n men, i1,i2,…,in, selects his own hat can occur in any of
(N-n)x(N-n-1)x…x3x2x1=(N-n)! possible ways, since there is only one way for the n men to
select their own hat, then, of the N-n men remaining, the first can select any of (N-n) hats, the
second can select (N-n-1), and so on. So, assuming all N! possibilities equally likely,
Continuing, there are
∑
i1 < i2
<in
P( Ei1 Ei2
N
 
n
Ein ) =
terms in P{Ei1 Ei2
Ein ) =
( N − n)!
, so
N!
N !( N − n)!
1
=
( N − n)!n ! N ! n !
and thus,

P

∪E
N
i
i =1

 = 1−

1 1
+ −
2! 3!
(−1)
N +1
1
N!
So, the probability that none of the men selects his own hat =s
1 1
1
1 − 1 + − + (−1) N
2! 3!
N!
−1
And for large values of N, this can be approximated by e ≈ .36788 . In other words, for large N,
the probability that none of the men selects his own hat is approximately .37. Note, it does not go
to infinity, as might be expected.
(b) Consider the event that exactly k of the men select their own hats, The number of ways that
only these k men can select their own hats is the sam as the number of ways in which the other
N-n men can select among the remaing hats such that none of them select their own hats. But
1 1
1
1 − 1 + − + (−1) N −k
is the probability we just found for not
2! 3!
( N − k )!
one of the N-k men selecting their own hat.. Then the number of
ways in which the set of men selecting their own hats is


1 1
1
( N − k )! 1 − 1 + − + (−1) N −k
2! 3!
( N − k )!!

16
Since there are
N
 
n
ways to select k men from N,

P{Exactly k men select their own hats} =
( N − k )! 1 − 1 +

1−1 +
=
e −1
For large N, this can be approximated by
.
k!
1 1
− +
2! 3!
1 1
− +
2! 3!
N!
+ (−1) N −k
k!
(−1) N −k

1
( N − k )!!
1
( N − k )!