/sci/ - Science & Math

Hint: The answer is 800 supposedly. Been trying to find out how to find it for the past two hours with no luck...

depends.
do we care about the order?

Doesn't say.

I came up with only 250.

>>4906688
>>4906695

Order Doesn't Matter: C(5,3) or 10.
Order Does Matter: P(5,3) or 60.

Or something like that. Correct me if I am wrong.

So in the required roll there should be two dice giving any number from 1-5 and three 6's. That would be 25 possible combinations.
From there I need to know if the dice "position" matters or a 2-4-6-6-6 would count as a different one if it was 4-2-6-6-6 or even if one of the 6's exchanged places with one another.

It's shit like this that makes me hate probability.

If order matters:
66655, 66654, 66653,...., 66645, 66644,...., 66611
There are 25 possible rolls in the case where you roll 6's for your first three rolls. Each other possible placing of sixes in the five slots _ _ _ _ _ has 25 more possible rolls.

>>4906703
The answer is 800. What I need to know is how to find the answer.

The closest ive gotten to an explanation is number 9 in this worksheet: http://www.madandmoonly.com/doctormatt/mathematics/dice1.pdf
And the explanation of "exactly" on this website: http://mathbits.com/mathbits/tisection/statistics2/binomialatmost.htm

666xx
66x6x
66xx6
6x66x
6x6x6
6xx66
x666x
x6x66
x66x6
xx666

My thinking, with order not mattering, is that this the are the only ten outcomes you can get in the five tosses of the die, with exactly three of them being heads.

Order matters because this is the case where order doesn't matter. First, choose two of the five die which are not sixes: <span class="math">\binom{2}{5}=10[/spoiler]. Then, of those two die, if they're the same there's five different options; different there's 10 different options; order doesn't matter. Therefore, we get 150 different ways.

Let's say order does matter. In the case that the other two die are the same, we have five different "other pairings," and the bookkeeping rule gives us <span class="math">\frac{5!}{3!2!}=10[/spoiler] different orderings for a total of 50 for a "full house" situation. The other case has <span class="math">\frac{5!}{3!}=20[/spoiler] different orderings, with 20 possible pairs for the non-six die, giving us 400, for a total of 450 different ways. I can't combinatorics....

>>4906708
>>4906714
I think you guys are on the right track. How does one get 800 though?

there are 25 combinations of which numbers can be used, eg 66611, 66612 and so on

let a be a number other than 6, you can have order of
666aa 66a6a 6a66a a666a 66aa6 6a6a6 a66a6 6aa66 a6a66 aa666
10 different ways to order, each with 25 possibilities, so the answer is 250

there is a faster way to calculate it with combinatorics, but i cant remember how to and cba to look up

>>4906721

Didn't even think about the other two dice being rolled as important. Than yes, that changes the entire question.

First, consider how many different arrangements of "xx" there can be along with the three sixes. Six opportunities for one, six for the other, independent of one another so you can multiply them, ending with 36 different positions for the "xx" value.

Then, multiply this with the 10 different arrangements of sixes, and you 360. That's the only conclusion I can come up with this at this time.

OP here. The thing that pisses me off the most is that theyre giving this type of stuff to us in beginning statistics.. For the summer term...

>>4906721
Find a way to multiply it by 32.

5^6 - 3^5

>>4906735

Then fuck me, you can't have any more sixes, so that makes my intial number smaller.

>>4906736
And that's what happens when a /sci/entist gets a job as a teacher and decides to troll the whole lot of students all summer long.

>>4906735
Five opportunities. If a six comes up it fucks everything.

>>4906738
That's 15625 - 243 wtf

>>4906727

C(5,3)*5*5

Some of the answers here are confusing the hell out of me. Here is the exact question I was given, just so you have extra assurance.

>>4906765
Question isn't even grammatically correct, lol.

5 rolls need 3 6's, so 5^2 others. have 5! total orderings with 6 repeating 3 times, so 25*5!/3! = 500? How is this different to the classic "how many ways to arrange letters of the word" problem?

800 is wrong. Using a binomial distribution, we can calculate the probability of getting exactly three sixes in five rolls of the dice:

choose(6,3) * (1/6)^3 * (5/6)^2 = 0.03215021

(Not used to /sci/, so I don't know the code)

Now, there are 6^5 possible outcomes to the five rolls of the dice. Guess what happens when we plug 800 in?

800/6^5 = 0.1028807 (wrong)
250/6^5 = 0.03215021 (right)

So the answer to the problem is wrong.

I think that was just a troll question. Its a ten question quiz and we have three tries -- every try has different questions. So far This question has appeared once on both tries ive taken so far. So yeah, I guess its just to see who uses up all their tries...

>>4906836

Typo, should be:

choose(5,3) * (1/6)^3 * (5/6)^2 = 0.03215021

>>4906756
maybe this time i remember that part c(5,3), it makes some sense but i always seem to forget about it, thanks

someone post more questions like this

>>4906852

If I roll 5 fair dice, how many different ways can one get all sixes?

>>4906854

>>4906852
lol do you want me to post all ten questions? im on my third try with an hour and a half to spare...

hehehe

What the fuck happened to muh old "GTFO WITH UR HOMEWORK UNDERAGE" /sci/?
Faggots.

>>4906875

For #1, use De Morgan's laws

>>4906883
seems simple enough

>>4906854
That's unsolvable.