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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?
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I just need help setting it up. I've got the computational power.

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.0intuitively I'd say 6!/(6^6)

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.0there 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, I do agree there's no replacement. Let me think about what you've written for a moment. Thank you

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In my textbook, the formula is n!/r!(nr)!. It warns against making it n!/n^r because that doesn't work when there's no order.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0n being the sample and r being the selection from it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But 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
 one year ago
Best ResponseYou've already chosen the best response.0that's what I was thinking, too. the formula should yield the same results

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What does 6 MEAN, then?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What is that telling me?

Vocaloid
 one year ago
Best ResponseYou've already chosen the best response.0@satellite73 I think I'm missing something obvious here ;;

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I really appreciate the effort here btw, this is the last question.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1we 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
 one year ago
Best ResponseYou've already chosen the best response.0okay, let me look again at what my book says

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I guess my question is why doesn't the formula with no replacement and no order work here...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Why do I have to make something up?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1no replacements and no order eh

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so how does n!/r!(nr)! work here?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1I am not fully sure what you mean by that formula though, i dont really know these formulas like that

dan815
 one year ago
Best ResponseYou've already chosen the best response.1okay so what u have there is (N choose R)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well I've got four of them for the variety of situations with order/without order and with/without replacement. I promise they're right.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1okay so for your formula, would you do 6 choose 6 then

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh maybe! Let me try

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, then the formula gives me 6!/6!0! which is 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which 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
 one year ago
Best ResponseYou've already chosen the best response.1yep but wait there is nothing wrong with this

dan815
 one year ago
Best ResponseYou've already chosen the best response.1you see how we didnt care about the order, now what are the total possibilitites where we dont care about the order again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't know exactly what you're referring to, or saying. Please elaborate?

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0there is replacement since you are tossing six different dice you do not use up a number once it is rolled

dan815
 one year ago
Best ResponseYou've already chosen the best response.16^6 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
 one year ago
Best ResponseYou've already chosen the best response.0The formula for no order and replacement is (n+r1)!/r!(n1)!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I guess maybe I'm not clear on what n and r are. I think n is 6, but is r 1 or 6?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah misty left. nevermind

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ah it doesn't even matter if there's replacement, you get the same answer.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im not im middle school......XD tf im in college

misty1212
 one year ago
Best ResponseYou've already chosen the best response.0my 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 \[\frac{n!}{k!(nk)!}\] in that form ever

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well you and my ph.d. professor would disagree, but I guess thanks.
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