Mathematics Dictionary

Mathematics Dictionary
Dr. K. G. Shih

Probability

Subjects

Symbol Defintion

Example : Sqr(x) = square root of x

Q01 | - 3 pairs Color balls

Q02 | - 3 pairs Color balls

Q03 | - 3 pairs Color balls

Q04 | - 3 pairs Color balls : Proposed formula

Q05 | - 3 pairs Color balls : Verify by sample space

Q06 | - 4 pairs Color balls

Q07 | - 4 pairs Color balls

Q01. 3 pairs Color balls

Question

2 red, 2 green and 2 blue balls
Pick 2 balls and put into 3 boxes
Find probability for same color in each box

Solution

Total sample space n(S) = C(6,2)*C(4,2)*C(2,2) = (15*6*1) = 90
Sample space in box 1 : 6 balls take 2
- n(S) = C(6,2) = 15
- n(same 1) = 3
- n(diff 1) = 12
Sample space in box 2 : 4 balls take 2 with balls in box 1 same color
- n(S) = C(4,2) = 6
- n(same 2) = 2
- n(diff 2) = 4
Sample space in box 2 : 4 balls take 2 with balls in box 1 diff color
- n(S) = C(4,2) = 6
- n(same 3) = 1
- n(diff 3) = 5
Hence sample space for all same
- Box 1 must be same and box 2 must be same
- Hence sample space = n(all) = n(same 1)*n(same 2) = 3*2 = 6
Hence probability p = n(all same)/n(S) = 6/90 = 1/15

Go to Begin

Q02. 3 pairs Color balls

Question

2 red, 2 green and 2 blue balls
Pick 2 balls and put into 3 boxes
Find probability for same color in only one box

Solution

Total sample space n(S) = C(6,2)*C(4,2)*C(2,2) = (15*6*1) = 90
Sample space in box 1 : 6 balls take 2
- n(S) = C(6,2) = 15
- n(same 1) = 3
- n(diff 1) = 12
Sample space in box 2 : 4 balls take 2 with balls in box 1 same color
- n(S) = C(4,2) = 6
- n(same 2) = 2
- n(diff 2) = 4
Sample space in box 2 : 4 balls take 2 with balls in box 1 diff color
- n(S) = C(4,2) = 6
- n(same 3) = 1
- n(diff 3) = 5
Sample space for one same
- Box 1 same and box 2 diff
- Box 1 diff and box 2 same
- n(1) = n(box 1, box 2)
- n(1) = n(same 1)*n(diff 2) + n(diff 1)*n(same 3)
- n(1) = 3*4 + 12*1 = 24
Hence probability p = n(1)/n(S) = 24/90 = 4/15

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Q03. 3 pairs Color balls

Question

2 red, 2 green and 2 blue balls
Pick 2 balls and put into 3 boxes
Find probability for diff color in each box

Solution

Total sample space n(S) = C(6,2)*C(4,2)*C(2,2) = (15*6*1) = 90
Sample space in box 1 : 6 balls take 2
- n(S) = C(6,2) = 15
- n(same 1) = 3
- n(diff 1) = 12
Sample space in box 2 : 4 balls take 2 with balls in box 1 same color
- n(S) = C(4,2) = 6
- n(same 2) = 2
- n(diff 2) = 4
Sample space in box 2 : 4 balls take 2 with balls in box 1 diff color
- n(S) = C(4,2) = 6
- n(same 3) = 1
- n(diff 3) = 5
Sample space in box 3 : 2 balls take 2 with balls in box 1 and box 2 diff color
- n(S) = C(2,2) = 1
- n(same 3) = 1
- n(diff 3) = 5
Sample space for all diff color in each box
- Box 1 has 12 diff sample spaces
- Box 2 and box 3 have 2 same and 4 diff sample spaces
- n(0) = n(box 1, box 2, box 3)
- n(0) = 12*4 = 48
Hence probability p = n(0)/n(S) = 48/90 = 48/15

Go to Begin

Q04. 3 pairs Color balls : Proposed formula

Question

2 red, 2 green and 2 blue balls
Pick 2 balls and put into 3 boxes
Find probability for diff color in each box

Using formula :

Sample space n(S) = C(6,2)*C(4,2)*C(2,2)/(3*2*1) = (15*6*1)/(3*2*1) = 15
Sample for all diff color in each box
- n(0) = n(S) - 3*C(4,2)*C(2,2)/(2*1) + 3*C(2,2)/1 - 1
- n(0) = 15 - 3*3 + 3*1 - 1 = 8
Hence p = n(0)/n(S) = 8/15
This agrees with the answer of sample structure in question 5

Discussion

Coefficients 1, -3, +3 and -1 are similar as binomial coeffcients

Sample space of box 1 : C(6,2)/3 = 1 same and 4 diff

Sample space of box 2 : C(4,2)/2 = 1 same and 2 diff

Sample space of box 3 : C(2,2)/1 = 1 same or 1 diff

Hence total sample space = n(S) = C(6,2)*C(4,2)*C(2,2)/(3*2*1)

C(4,2)*C(2,2)/(2*1) = 1 same and 2 diff color samples

Box 1, 2, 3 give C(3,1) = 3

Remove 1 same color box : C(4,2)*C(2,2)/(2*1)
- 4th term : we remove 3 boxes all same color
  - All box different color = 15 - 3*C(4,2)*C(2,2)/(2*1) + 3*C(2,2)/1 - 1
  - = 15 - 3*3 + 3 - 1 = 8
  Go to Begin
  Q05. 3 pairs Color balls : Sample structure
  
  Verify
  
  Sample space n(S) = C(6,2)*C(4,2)*C(2,2) = 90
  Six balls pick 2 and put into 1st box
  
  (a1,a2) (a1,b1) (a1,b2) (a1,c1) (a1,c2)
  (a2,b1) (a2,b2) (a2,c1) (a2,c2)
  (b1,b2) (b1,c1) (b1,c2)
  (b2,c1) (b2,c2)
  (c1,c2)
  Total 3 same color samples : (a1,a2), (b1,b2), (c1,c2)
  Total 12 different color samples
  
  Box 1 same color (a1,a2) combine with other 4 balls b1,b2,c1,c2
  
  B1 .... B2 .... B3
  (a1,a2) (b1,b2) (c1,c2) : box 1,2,3 same color boxes
  (a1,a2) (b1 c1) (c2,b2) : box 1 same color box
  (a1,a2) (b1,c2) (b2,c1) : box 1 same color box
  (a1,a2) (b2 c1) (c2,b1) : box 1 same color box
  (a1,a2) (b2,c2) (b1,c1) : box 1 same color box
  (a1,a2) (c1,c2) (b1,b2) : box 1,2,3 same color boxes
  Hence (a1,a2) combine with 4 different color samples 2 same color samples
  One same color in (a1,a2) = 4
  One same color in (b1,b2) = 4
  One same color in (c1,c2) = 4
  Hence box 1 same and box 2 same color samples = 3*2 = 6
  Hence box 1 same and box 2 diff color samples = 3*4 = 12
  Different color box samples = none
  
  One different color (a1,b1) combine with other 4 balls (a2,b2,c1,c2)
  
  B1..... B2 .... B3
  (a1,b1) (a2,b2) (c1,c2)
  (a1,b1) (a2,c1) (b1,c2)
  (a1,b1) (a2,c2) (b1,c1)
  (a1,b1) (b2,c1) (b2,c2)
  (a1,b1) (b2,c2) (c1,a2)
  (a1,b1) (c1,c2) (b2,a2)
  Box 1 is diff color and box 2 has 1 same 5 diff color samples
  Box 1 is diff color and box 2 has 1 same 5 diff color samples
  Hence box 1 diff color with box 2 or box 1 same color = 12*2 = 24
  Hence one same color box = 24
  Hence all different color box samples = 12*4 = 48
  All box different color : P = 48/90 = 8/15
  
  This agrees with the proposed formula with coeff 1, -3, +3, -1 in Q4
  Summary
  
  all same color samples = 3*2*1 = 6
  One same color samples = 12 + 24 = 36
  Two same color samples = none
  all diff color samples = 48
  Go to Begin
  Q06. 4 pairs Color balls
  
  Quuestion
  
  2 red, 2 green, 2 white and 2 blue balls
  Pick 2 balls each time and put into 4 boxes
  Solution
  
  Balls : A, A, B, B, C, C, D, D
  Sample space
  
  n(S) = C(8,2)*C(6,2)*C(4,2)*C(2,2)/(4*3*2*1)
  n(S) = (28*15*6*1)/(4*3*2*1)
  N(S) = 105
  
  All box same color P = 1/105
  One box and two boxes same color : S1 = C(6,2)*C(4,2)*C(2,2)/(3*2*1)
  Two boxes same color : S2 = C(4,2)*C(2,2)/(2*1)
  Three boxes same color : none
  four boxes same color = C(2,2)/(1)
  All box diff color n(0) = 1*n(S) - 4*S1 + 6*S2 - 4*S3 + 1
  n(0) = 105 - 4*15 + 6*3 - 4*1 + 1 = 60
  Probability for all boxes in diff color = P = n(0)/n(S) = 60/105
  Coefficients 1, -4, +6, -4, + 1 are same binomial coefficients
  
  Go to Begin
  
  Q7. 4 pairs Color balls
  
  Quuestion :
  
  2 red, 2 green, 2 white and 2 blue balls
  Balls : A, A, B, B, C, C, D, D
  Pick 2 balls each time and put into 4 boxes
  Find one box same color
  Box 1 : C(8,2) = 28 samples.
  
  4 same color and 24 different color
  AA, AB, AB, AC, AC, AD, AD
  AB, AB, AC, AC, AD, AD
  BB, BC, BC, BD, BD
  BC, BC, BD, BD
  CC, CD, CD
  CD, CD
  DD
  4 box same color : AA, BB, CC, DD
  Box 1 same color AA and other 3 boxes BB, CC, DD
  
  AA combine with 6 all three boxes same color : 6 all 4 boxes same color
  AA combine with 36 one box same color : 36 two boxes same color
  AA combine with 48 all boxes different color : 48 one box same color
  Total samples
  
  1 box same color 4*48 = 192
  2 boxes same color 4*36 = 144
  3 boxes same color = 0
  4 boxes same color = 4*6 = 24
  
  Box 1 different color AB and box 2 different color AB and 2 boxes C, C, D, D
  
  AB, AB, CC, DD
  AB, AB, CD, DC
  AB, AB, CD, DC
  AB, AB, CD, DC
  AB, AB, CD, DC
  AB, AB, DD, CC
  Samples
  
  1 box same color = 0
  2 boxes same = 2
  4 boxes different color = 4
  
  Box 1 = AB box 2 = AB : 2 (2 boxes same) and 4 (all boxes diff)
  Box 1 = AB box 2 = AC : 2 (1 box same) and 4 (all boxes diff)
  Box 1 = AB box 2 = AC : 2 (1 box same) and 4 (all boxes diff)
  Box 1 = AB box 2 = AD : 2 (1 box same) and 4 (all boxes diff)
  Box 1 = AB box 2 = AD : 2 (1 box same) and 4 (all boxes diff)
  Box 1 = AB box 2 = BC : 2 (1 box same) and 4 (all boxes diff)
  Box 1 = AB box 2 = BC : 2 (1 box same) and 4 (all boxes diff)
  Box 1 = AB box 2 = BD : 2 (1 box same) and 4 (all boxes diff)
  Box 1 = AB box 2 = BD : 2 (1 box same) and 4 (all boxes diff)
  Total
  
  1 boxes same color = 8*2 = 18
  All different color boxes = 8*4 = 32
  
  Box 1 different color AB and box 2 same color CC and 2 boxes A, B, D, D
  
  AB, CC, AB, DD
  AB, CC, AD, DA
  AB, CC, AD, DA
  AB, CC, BD, DB
  AB, CC, BD, DB
  AB, CC, DD, CC
  Box 1 = AB Box 2 = CC : 2 (2 same color boxes) and 4 (1 same color box)
  Box 1 = AB Box 2 = DD : 2 (2 same color boxes) and 4 (1 same color box)
  Total
  
  2 boxes same color = 2*2 = 4
  1 box same color = 2*4 = 8
  
  Box 1 different color AB and box 2 different color CD and 2 boxes A, B, C, D
  
  AB, CD, AB, CD
  AB, CD, AC, BD
  AB, CD, AD, BC
  AB, CD, BC, AD
  AB, CD, BD, AC
  AB, CD, CD, AB
  Box 1 = AB box 2 = CD : All different color samples = 6
  Box 1 = AB box 2 = CD : All different color samples = 6
  Box 1 = AB box 2 = CD : All different color samples = 6
  Box 1 = AB box 2 = CD : All different color samples = 6
  Total
  
  All boxes diff color = 4*6 = 24
  
  Sumnary
  
  Total sample spaces n(S) = 28*15*6*1 = 2520
  All boxes diff color n(0) = 24*(4*6 + 1*4 + 8*4) = 24*60
  Two boxes same color n(2) = 144 + 24*2 + 24*4 = 288
  One box same color n(1)= 192 + 24*16 + 24*8 = 768
  All boxes same color n(4) = 24
  Probability
  
  All boxes are in diff color : P = n(0)/n(S) = (24*60)/2520 = 60/105
  One box in same color : P = n(1) = 768/2520 = 32/105
  Two boxes in same color : P = n(2)/n(0) = 288/2520 = 12/105
  All boxes in same color : P = n(4)/n(0) = 024/2520 = 01/105
  
  Go to Begin
  
  AL 17 00. Outline
  
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