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Pulsified333

  • one year ago

Assume that the box contains 6 balls: 1 white, 2 yellow, and 3 green. Balls are drawn in succession without replacement, and their colors are noted until a white ball is drawn. How many outcomes are there in the sample space?

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  1. Pulsified333
    • one year ago
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    I thought there should be 33 in the space but its wrong

  2. anonymous
    • one year ago
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    12C3 = 12! / (9! 3!) = 12 * 11 * 10 / (3 * 2 * 1) = 220 different combinations of three balls among the 12 balls in the box. This is the number of different possible 3-ball draws. There are 6C3 = 6! (3! 3!) = 20 combinations of 3 balls that are all red. There are 4C3 = 4 combinations of 3 balls that are all blue. There are no combinations of 3 balls that are all green. So the probability of drawing 3 balls the same color, without replacement, is (20 + 4) / 220 = 24/220 = 6/55 = about 10.9%

  3. Pulsified333
    • one year ago
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    there are no red balls in this problem

  4. anonymous
    • one year ago
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    ?? Let me re-do. This was my problem, I've had this before.

  5. Pulsified333
    • one year ago
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    ok

  6. anonymous
    • one year ago
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    I did this awhile ago, mine was just with red instead of white, so I'm not sure.

  7. dan815
    • one year ago
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    |dw:1441343388675:dw|

  8. Pulsified333
    • one year ago
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    so there are 6?

  9. dan815
    • one year ago
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    nono so w can be drawn either 1s 2nd 3rd... or 6th

  10. dan815
    • one year ago
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    for each of these 6 cases u can have different combinations of balls before

  11. Pulsified333
    • one year ago
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    right so how do i find out how many outcomes there are

  12. Pulsified333
    • one year ago
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    I don't know ;(

  13. dan815
    • one year ago
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    okay so take a random case see if we can figure out a pattern because i dont like this method it looks too long

  14. dan815
    • one year ago
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    lets say for w showing up on the 4th try _ _ _ W that means we have 3 places for ,2 red and 3 green

  15. Pulsified333
    • one year ago
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    no red just yellow

  16. dan815
    • one year ago
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    that means we either have 3 green,2, green or 1 green in there

  17. Pulsified333
    • one year ago
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    true

  18. dan815
    • one year ago
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    |dw:1441344028052:dw|

  19. dan815
    • one year ago
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    just 3 possiblites

  20. dan815
    • one year ago
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    but i dont wanna count like this, theres gonna be a permutation adn combination way of getting 3

  21. dan815
    • one year ago
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    we have 3 spots, 3 Green,2 Yellow* to fill those 3 spots

  22. Pulsified333
    • one year ago
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    ok

  23. dan815
    • one year ago
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    hmmm

  24. Pulsified333
    • one year ago
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    would it be 35?

  25. dan815
    • one year ago
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    how did u get that

  26. Pulsified333
    • one year ago
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    well I originally made a tree chart and came up with 33

  27. dan815
    • one year ago
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    hmmm

  28. dan815
    • one year ago
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    |dw:1441344262286:dw|

  29. dan815
    • one year ago
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    ugh thats too annoying

  30. Pulsified333
    • one year ago
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    Ik, is there a faster way to do it?

  31. dan815
    • one year ago
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    thats what im trying to think about

  32. Pulsified333
    • one year ago
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    I think i missed a W on my chart so I think it 34

  33. Pulsified333
    • one year ago
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    Got it :D

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