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Mathematics Dictionary
Dr. K. G. Shih
Probability
Subjects


  • AL 17 00 | - Outlines
  • AL 17 01 | - Permutation
  • AL 17 02 | - Arrangement
  • AL 17 03 | - Combination
  • AL 17 04 | - Arrangement and sample
  • AL 17 05 | - Arrangement of n elements in circle pattern
  • AL 17 06 | - Probability
  • AL 17 07 | - Arrangement of n elements in a line. One special element always at first
  • AL 17 08 | - Binomial probability
  • AL 17 09 | - Hypergeometric probablity
  • AL 17 10 | - Hypergeometric probablity
  • AL 17 11 | - Normal distribution
  • AL 17 12 | - Examples : C(n,r)
  • AL 17 13 | - Examples : Dice and sample spaces
  • AL 17 14 | - Examples : Color balls
  • AL 17 15 | - Examples : Color number cards
  • AL 17 16 | - Examples : Color balls
    • Answers


    AE 17 01. Permutation

    Defintion and examples
    • Definition : arrangement of symbols without repetition
    • Formula
      • P(n,r) = n*(n-1)*(n-2)*....*(n-r+1)
      • P(n,n) = n! = n*(n-1)*(n-2)*.......*3*2*1
      • n! is called factorial
    • Examples
      • P(1,1) = 1
      • P(2,2) = 2*1 = 2
      • P(3,3) = 3*2*1 = 6
      • P(4,4) = 4*3*2*1 = 24
      • P(5,5) = 5*4*3*2*1 = 120
    • Each time take 2 symbols from A,B,C. How many difference arrangement ?
      • AB and BA
      • AC and CA
      • BC and CB
      • Hence there are six different arrangement
      • The computation is P(3,2) = 3*2 = 6
    • Each time take 2 symbols from A,B,C. Find arrangement if symbol can be same
      • AB and BA
      • AC and CA
      • BC and CB
      • AA, BB, CC.
      • Hence there are nine different arrangement
      • The computation is 3^2 = 9
    Prove that P(n,r) = n!/(n-r)!
    • P(n,r) = n*(n-1)*.....*(n-r+1)
    • P(n,r) = n*(n-1)*.....*(n-r+1)*(n-r)!/(n-r)!
    • P(n,r) = n!/(n-r)!
    Prove that P(n,0) = 1
    • P(n,0) = n!/(n-0)! = 1
    Exercise : Find P(10,10)
    Go to Begin

    AE 17 02. Arrangement

    Formula
    • 1. Take r from n elements without duplication : P(n,r)
    • 2. Take r from n elements with duplication : n^r
    Example
    • A set a,b,c. Take all three element for arrangement with duplication
    • The answer is 3^3 = 27
    • List of all arrangement
      • aaa, bbb, ccc
      • aab, aba, baa, aac, aca, caa
      • bbc, bcb, cbb, bba, bab, abb
      • cca, cac, acc, ccb, cbc, bcc
      • abc, acb, bca, bac, cab, cba
      • Hence there are 27 diiferent arrangments

    Go to Begin

    AL 17 03. Combination

    • Definition : A subset of r elements chosen from a set of n elements
    • Formula
      • C(n,r) = n*(n-1)*(n-2)*....*(n-r+1)/(r!)
      • C(n,n) = 1
    • Examples
      • C(1,1) = 1
      • C(2,2) = 1
      • C(3,3) = 1
    • Each time take 2 symbols from A,B,C. Find arrangement without cosidering position
      • AB and BA as one count
      • AC and CA as one count
      • BC and CB as one count
      • Hence there are three different subsets
      • The computation is c(3,2) = 3*2/(2!) = 3
    Exercise : Find P(20,10)
    Go to Begin

    AL 17 04. Arrangement and sample

    Arrangement
    • It is permutation. The sub-set will consider the position of the element
    • Example
      • Sub-set elements A,B is different from sub-set B,A
      • They are two different sub-set
    Sample
    • It is combination. The sub-set will not consider the position of the element
    • Example
      • Sub-set elements A,B is same as sub-set B,A
      • There is only one sub-set

    Go to Begin

    AL 17 05. Arrangement of n elements in circle pattern

    Formula
    • It is P(n,n)/n = (n-1)!
    Test
    Go to Begin

    AL 17 10. Hypergeometric probability (2)

    Definition
    • Sample sapce N
    • Defective samples is n
    • Choosing samples s
    • Hence getting x defective
    • Hence H(x) = C(n,x)*C((N-n),(s-x))/C(N,s)
    Example : 10000 samples have 500 defective. Now take 10 samples randomly     Example : Use binomial distribution to solve above question
    • Use program 01 03
      • Probability to get defective sample p = 500/10000 and q = 1 - p
      • B(x) = C(n,x)*(p^x)*(q^(n-x))
    • Input data
      • 1. Sample n = 10, expect defective sample x = 0 and p = 0.05
      • 2. Sample n = 10, expect defective sample x = 1 and p = 0.05
      • 3. Sample n = 10, expect defective sample x = 2 and p = 0.05
      • 4. Sample n = 10, expect defective sample x = 3 and p = 0.05

    Go to Begin

    AL 17 11. Normal distribution

    Standard normal distribution on computer
    • Computer program
    • Facts
      • N(z gt +a) = 1 - N(z le +a)
      • N(z lt -a) = 1 - N(z le +a) since distribution f(z) is symmetrical
      • From table we know that N(z le 1) = 0.8413
    • Example : From table we know that N(z le 1) = 0.8413
      • Find N(z gt +1)
      • Find N(z lt -1)
    Normal distribution with mean u and standard deviation d
    • Computer program
    • Convert to standard : Let z = (x - u)/d
    • Example : A student grade has mean 80 with d = 10
      • Find probability his amrk is between 70 and 90
      • Find probability his amrk is greater than 90
      • Find probability his amrk is is less than 70
    • Solution : Between 70 and 90
      • N(70 - u) GE N(x - u) LT N(90 - u)
      • N((70 - u)/d) GE N((x - u)/d) LE N((90 - u)/d)
      • Let z = (x - u)/d
      • Since u = 80 and d = 10
      • Hence N(-1) GE N(z) LE N(1)
      • Hence N(z) = N(z lt 1) - N(z lt -1)
      • Hence N(z) = 0.8413 - (1 - N(1))
      • Hence N(z) = 0.8413 - (1 - 0.8413)
      • Hence N(z) = 0.8413 - 0.1587 = 0.6826 = 68.26%
    • Solution : GT 90
      • N((x - 80)/d) LT N((90 - 80)/d)
      • Hence N(z GT +1) = 1 - N(z LE 1)
      • Hence N(z GT +1) = 1 - 0.8413 = 0.1587 = 15.87%
    • Solution : LT 70
      • N((x - 80)/d) GT N((70 - 80)/d)
      • Hence N(z LT -1) = 1 - N(z LE 1)
      • Hence N(z LT -1) = 1 - 0.8413 = 0.1587 = 15.87%

    Go to Begin

    AL 17 12. Examples : C(n,r)

    • 1. Solve C(n,1) = 16.
      • C(n,1) = n
      • Hence n = 16
    • 2. Solve C(n,2) = 28.
      • C(n,2) = n*(n-1)/(2!) = 28
      • Hence n*(n-1) = 56 or n^2 - n - 56 = 0
      • (n - 8)*(n + 7) = 0
      • Since n is positive, hence n = 8
    Binomial distribution
    Go to Begin

    AL 17 13. Examples : Dice and samples space

    Two dice have sample space 6*6 = 36
    • (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
    • (2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
    • (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
    • (4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
    • (5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
    • (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)
    Example : Throw 3 dice, the probability to get number 6
    • Dice A : 1, 1, 1, 1, 2, 2, 2, 3, 3, 4
    • Dice B : 1, 4, 2, 3, 3, 1, 2, 2, 1, 1
    • Dice C : 4, 1, 3, 2, 1, 3, 2, 1, 2, 1
    • Since sample space of 3 dice is 6*6*6
    • Hence probablity is 10/(6*6*6)
    Example : Throw 3 dice, the probablity to get two same number
    • Dice : A, B, C
    • A1, B1, C2, C3, C4, C5 and C6 : Sample space = 5
    • A2, B2, C1, C3, C4, C5 and C6 : Sample space = 5
    • A3, B3, C1, C2, C4, C5 and C6 : Sample space = 5
    • A4, B4, C1, C2, C3, C5 and C6 : Sample space = 5
    • A5, B5, C1, C2, C3, C4 and C6 : Sample space = 5
    • A6, B6, C1, C2, C3, C4 and C5 : Sample space = 5
    • Hnace Sample space S2 = C(6,1)*C(5,1) = 6*5 = 30
    • For A, B, C dice, the sample space : (A,B,C), (B,C,A) and (C,A,B)
    • Sample sapce S3 = C(3,1) = 3
    • Hence probabilitye is C(6,1)*(C(5,1)*C(3,1)/(6*6*6) = 90/216

    Go to Begin

    AL 17 14. Examples : Color balls

    Example 1 : 3 red balls and 2 white balls in a bag. Each time take two
  • Sample space = C(6,2)*C(4,2)*C(2,2)/(3*2*1) = (15*6*1)/(3*2*1) = 15
  • Each box same color : only one case, hence P = 1/15
  • 1st box same color
    • 3*C(4,2)*C(2,2)/(2*1)
    • 2nd box may have same color
    • 3rd box may have same color
    • It include all bax same color. Hence 3 cases of same color
  • 1st and 2nd boxes same color
    • 3*C(2,2)/1
    • It include all bax same color. Hence 3 cases of same color
  • All box different color = 15 - 3*C(4,2)*C(2,2)/(2*1) + 3*C(2,2)/1 - 1 Example 3 : 2 red balls, 2 green balls, 2 white and 2 blue balls. Pick 2 balls each time and put into 4 boxes
  • Sample space = C(8,2)*C(6,2)*C(4,2)*C(2,2)/(4*3*2*1) = (28*15*6*1)/(4*3*2*1) = 105
  • All box same color P = 1/105
  • 1st box same color : S1 = C(6,2)*C(4,2)*C(2,2)/(3*2*1)
  • 1st 2 boxes same color : S2 = C(4,2)*C(2,2)/(2*1)
  • 1st 3 boxes same color : S3 = C(2,2)/(1)
  • All box diff color P = 105 - 4*S1 + 6*S2 - 4*S3 + 1
    Go to Begin

    AL 17 00. Outline

    Properties of P(n,r)
    • P(n,0) = 1
    • P(n,1) = n
    • P(n,n) = n!
    Properties of C(n,r)
    • C(n,0) = 1
    • C(n,1) = n
    • C(n,n) = 1
    • C(n,n-1) = n
    Definition of distributions
    • 1. Permutation formula : P(n,r)
    • 2. Combination formula : C(n,r)
    • 3. Binomial distribution : B(x) = C(n,x)*(p^x)*(q^(n-x))
      • There are n trials, wish to get x events
      • The probablisty to an event is p and q = 1 - p
    • 4. Hypergeometric formula : H(x) = C(n,x)*C((N-n,n-x)/C(N,n)
      • Sample space is N, take n samples from N
      • Expect to have x samples
    • 5. Hypergeometric formula : H(x) = C(n,x)*C((N-n,s-x)/C(N,s)
      • Sample space is N, defetive samples is n
      • Take s sample and expect x defective samples
    Arrangement
    • Take r different things from n different things without repeating : P(n,r)
    • Arrange n different things in circle pattern : P(n,n)/n = (n-1)!
    • Arrange n different things in line and special one at 1st position : (n-1)!
    • Arrange n different things with repeating : n^n

    Go to Begin
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