Question

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?

Answer 1

I just need help setting it up. I've got the computational power.

Answer 2

intuitively I'd say 6!/(6^6)

Answer 3

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.

Answer 4

Right, I do agree there's no replacement. Let me think about what you've written for a moment. Thank you

Answer 5

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.

Answer 6

n being the sample and r being the selection from it

Answer 7

hm..

Answer 8

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

Answer 9

that's what I was thinking, too. the formula should yield the same results

Answer 10

What does 6 MEAN, then?

Answer 11

What is that telling me?

Answer 12

@satellite73 I think I'm missing something obvious here ;-;

Answer 13

hi okay so

Answer 14

I really appreciate the effort here btw, this is the last question.

Answer 15

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

Answer 16

okay, let me look again at what my book says

Answer 17

I guess my question is why doesn't the formula with no replacement and no order work here...

Answer 18

Why do I have to make something up?

Answer 19

no replacements and no order eh

Answer 20

ya sure it works

Answer 21

Okay, so how does n!/r!(n-r)! work here?

Answer 22

I am not fully sure what you mean by that formula though, i dont really know these formulas like that

Answer 23

okay so what u have there is (N choose R)

Answer 24

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.

Answer 25

yep!

Answer 26

okay so for your formula, would you do 6 choose 6 then

Answer 27

ohh maybe! Let me try

Answer 28

No, then the formula gives me 6!/6!0! which is 1

Answer 29

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.

Answer 30

yep but wait there is nothing wrong with this

Answer 31

Okay, go ahead.

Answer 32

you see how we didnt care about the order, now what are the total possibilitites where we dont care about the order again

Answer 33

I don't know exactly what you're referring to, or saying. Please elaborate?

Answer 34

HI!!

Answer 35

there is replacement since you are tossing six different dice
you do not use up a number once it is rolled

Answer 36

oh right.

Answer 37

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

Answer 38

The formula for no order and replacement is (n+r-1)!/r!(n-1)!

Answer 39

I guess maybe I'm not clear on what n and r are. I think n is 6, but is r 1 or 6?

Answer 40

ah misty left. nevermind

Answer 41

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.

Answer 42

okay

Answer 43

Ah it doesn't even matter if there's replacement, you get the same answer.

Answer 44

im not im middle school......XD tf im in college

Answer 45

bye now

Answer 46

Kinda yeah XD

Answer 47

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

Answer 48

Well you and my ph.d. professor would disagree, but I guess thanks.