anonymous
  • anonymous
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?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I just need help setting it up. I've got the computational power.
Vocaloid
  • Vocaloid
intuitively I'd say 6!/(6^6)
Vocaloid
  • Vocaloid
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.

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anonymous
  • anonymous
Right, I do agree there's no replacement. Let me think about what you've written for a moment. Thank you
anonymous
  • anonymous
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.
anonymous
  • anonymous
n being the sample and r being the selection from it
Vocaloid
  • Vocaloid
hm..
anonymous
  • anonymous
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
Vocaloid
  • Vocaloid
that's what I was thinking, too. the formula should yield the same results
anonymous
  • anonymous
What does 6 MEAN, then?
anonymous
  • anonymous
What is that telling me?
Vocaloid
  • Vocaloid
@satellite73 I think I'm missing something obvious here ;-;
dan815
  • dan815
hi okay so
anonymous
  • anonymous
I really appreciate the effort here btw, this is the last question.
dan815
  • dan815
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
anonymous
  • anonymous
okay, let me look again at what my book says
anonymous
  • anonymous
I guess my question is why doesn't the formula with no replacement and no order work here...
anonymous
  • anonymous
Why do I have to make something up?
dan815
  • dan815
no replacements and no order eh
dan815
  • dan815
ya sure it works
anonymous
  • anonymous
Okay, so how does n!/r!(n-r)! work here?
dan815
  • dan815
I am not fully sure what you mean by that formula though, i dont really know these formulas like that
dan815
  • dan815
okay so what u have there is (N choose R)
anonymous
  • anonymous
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.
anonymous
  • anonymous
yep!
dan815
  • dan815
okay so for your formula, would you do 6 choose 6 then
anonymous
  • anonymous
ohh maybe! Let me try
anonymous
  • anonymous
No, then the formula gives me 6!/6!0! which is 1
anonymous
  • anonymous
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.
dan815
  • dan815
yep but wait there is nothing wrong with this
anonymous
  • anonymous
Okay, go ahead.
dan815
  • dan815
you see how we didnt care about the order, now what are the total possibilitites where we dont care about the order again
anonymous
  • anonymous
I don't know exactly what you're referring to, or saying. Please elaborate?
misty1212
  • misty1212
HI!!
misty1212
  • misty1212
there is replacement since you are tossing six different dice you do not use up a number once it is rolled
anonymous
  • anonymous
oh right.
dan815
  • dan815
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
anonymous
  • anonymous
The formula for no order and replacement is (n+r-1)!/r!(n-1)!
anonymous
  • anonymous
I guess maybe I'm not clear on what n and r are. I think n is 6, but is r 1 or 6?
anonymous
  • anonymous
ah misty left. nevermind
anonymous
  • anonymous
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.
dan815
  • dan815
okay
anonymous
  • anonymous
Ah it doesn't even matter if there's replacement, you get the same answer.
carlyleukhardt
  • carlyleukhardt
im not im middle school......XD tf im in college
carlyleukhardt
  • carlyleukhardt
bye now
carlyleukhardt
  • carlyleukhardt
Kinda yeah XD
misty1212
  • misty1212
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
anonymous
  • anonymous
Well you and my ph.d. professor would disagree, but I guess thanks.

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