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

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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