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anonymous

  • one year ago

Probability: A balanced die is tossed six times, and the number on the uppermost face is recorded each time. What is the probability that the numbers recorded are 1, 2, 3, 4, 5, and 6 in any order?

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  1. anonymous
    • one year ago
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    I just need help setting it up. I've got the computational power.

  2. Vocaloid
    • one year ago
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    intuitively I'd say 6!/(6^6)

  3. Vocaloid
    • one year ago
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    there are 6^6 total possibilities for one die being tossed 6 times since we want 1,2,3,4,5,6 in any order: for the first die, our possibilities are 1,2,3,4,5,6 for the second die, we've already used up one of the 6 options, so we're left with 5 for the third die, 4 choices, etc.

  4. anonymous
    • one year ago
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    Right, I do agree there's no replacement. Let me think about what you've written for a moment. Thank you

  5. anonymous
    • one year ago
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    In my textbook, the formula is n!/r!(n-r)!. It warns against making it n!/n^r because that doesn't work when there's no order.

  6. anonymous
    • one year ago
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    n being the sample and r being the selection from it

  7. Vocaloid
    • one year ago
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    hm..

  8. anonymous
    • one year ago
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    But if I do that, it's 6!/1!5! and that just turns out to be 6...which seems really small and an odd result

  9. Vocaloid
    • one year ago
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    that's what I was thinking, too. the formula should yield the same results

  10. anonymous
    • one year ago
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    What does 6 MEAN, then?

  11. anonymous
    • one year ago
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    What is that telling me?

  12. Vocaloid
    • one year ago
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    @satellite73 I think I'm missing something obvious here ;-;

  13. dan815
    • one year ago
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    hi okay so

  14. anonymous
    • one year ago
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    I really appreciate the effort here btw, this is the last question.

  15. dan815
    • one year ago
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    we have 6 numbers and 6 spots we want no repetition there are 6! of getting 1 to 6 in some order there are 6^6 possibilities for anything to happen therefore 6!/6^6 = chance of 1 to 6 in any order

  16. anonymous
    • one year ago
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    okay, let me look again at what my book says

  17. anonymous
    • one year ago
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    I guess my question is why doesn't the formula with no replacement and no order work here...

  18. anonymous
    • one year ago
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    Why do I have to make something up?

  19. dan815
    • one year ago
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    no replacements and no order eh

  20. dan815
    • one year ago
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    ya sure it works

  21. anonymous
    • one year ago
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    Okay, so how does n!/r!(n-r)! work here?

  22. dan815
    • one year ago
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    I am not fully sure what you mean by that formula though, i dont really know these formulas like that

  23. dan815
    • one year ago
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    okay so what u have there is (N choose R)

  24. anonymous
    • one year ago
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    Well I've got four of them for the variety of situations with order/without order and with/without replacement. I promise they're right.

  25. anonymous
    • one year ago
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    yep!

  26. dan815
    • one year ago
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    okay so for your formula, would you do 6 choose 6 then

  27. anonymous
    • one year ago
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    ohh maybe! Let me try

  28. anonymous
    • one year ago
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    No, then the formula gives me 6!/6!0! which is 1

  29. anonymous
    • one year ago
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    which kind of makes sense for the logic of "six choose six" I guess...I feel like I'm missing something big. I'm going to write down your improvised formula and then ask her before class tomorrow.

  30. dan815
    • one year ago
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    yep but wait there is nothing wrong with this

  31. anonymous
    • one year ago
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    Okay, go ahead.

  32. dan815
    • one year ago
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    you see how we didnt care about the order, now what are the total possibilitites where we dont care about the order again

  33. anonymous
    • one year ago
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    I don't know exactly what you're referring to, or saying. Please elaborate?

  34. misty1212
    • one year ago
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    HI!!

  35. misty1212
    • one year ago
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    there is replacement since you are tossing six different dice you do not use up a number once it is rolled

  36. anonymous
    • one year ago
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    oh right.

  37. dan815
    • one year ago
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    6^6 is how many total possibilties there are where arrangement matters so 6^6/6! so u see how if u do 1/6^6/6! = 6!/6^6 we are abck the first solution again

  38. anonymous
    • one year ago
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    The formula for no order and replacement is (n+r-1)!/r!(n-1)!

  39. anonymous
    • one year ago
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    I guess maybe I'm not clear on what n and r are. I think n is 6, but is r 1 or 6?

  40. anonymous
    • one year ago
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    ah misty left. nevermind

  41. anonymous
    • one year ago
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    I don't want to logic this out, dude. I want the forumlas to work. That's the point of having them. I won't get credit for my work if I pull something out of the blue. I will ask her tomorrow.

  42. dan815
    • one year ago
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    okay

  43. anonymous
    • one year ago
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    Ah it doesn't even matter if there's replacement, you get the same answer.

  44. carlyleukhardt
    • one year ago
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    im not im middle school......XD tf im in college

  45. carlyleukhardt
    • one year ago
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    bye now

  46. carlyleukhardt
    • one year ago
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    Kinda yeah XD

  47. misty1212
    • one year ago
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    my suggestion in doing probability is not to get married to any formula use whatever you need when you need it i wouldn't even use \[\frac{n!}{k!(n-k)!}\] in that form ever

  48. anonymous
    • one year ago
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    Well you and my ph.d. professor would disagree, but I guess thanks.

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