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Mathematics Dictionary
Dr. K. G. Shih
Probability

Question
Three pairs color balls and Each two balls put into a box


  • Q01 | - Find probability for all 3 boxes having same color
  • Q02 | - Find probability for only one box having same color
  • Q03 | - Find probability for all 3 boxes having different color
  • Q04 | - 3 pairs Color balls : Proposed formula
  • Q05 | - 3 pairs Color balls : Verify by sample space

    • Q01. Find probability for all 3 boxes having same color

    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

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    Q02. Find probability for only one box having same color

    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. Find probability for all 3 boxes having different color

    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

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    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)
      • 1 same in box 1 and 2 diff in box 2
      • 1 same in box 1 and 1 same in box 2
      • Hence 2nd term : remove 3 same and 6 diff
      • Hence 3rd term : we put 3 same back
      • 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

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    Q05. 3 pairs Color balls : Sample structure

    Sample spaces
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