Some math help...

[carnageX]

Ok, I'm stuck. Section my class is doing is currently dealing with Permutations/Combinations/Probability. Been a while since I've done any of that (and I've always hated doing 'em). I basically have forgotten how to do it; and it all went over my head when my teacher was explaining it...so yeah xD.

Ok, here's a few examples of problems (with the answer provided, b/c it was in the back of my book):

A six sided die is rolled twice.
Q: Find the probability if the sum is less than 11.
A: 11/12

Q: Find the probability if the sum is odd and no more than 7.
A: 1/3


Ok, I'm quite confused on how to do these...if somebody could explain I'd love that.

 

[veg1992]

this is what we're doing in our Alg II class right now

first you consider all the ways you get to the sum of 11

pretty much every way except 6 and 6 (since theres 6 sides each its out of 12) so you get 11 for the numerator
___________________________________________________________________

ok for the second one you consider the ways to get no more than 7

aka 1,2 2,1 1,3 3,1 4,1 1,4 1,5 5,1 6,1 1,6 5,2 2,5

ok so thats 12 of them right there, out of 36, is 12/36, or 1/3

sorry I suck at explaining

 

[Luke]

Dice - Wikipedia, the free encyclopedia

that might help, i didn't read it all though. I would try to explain some of it but i really suck at explaining stuff.

 

[carnageX]

THanks for trying Vern, but I still don't understand it >.<

How'd you get 11 for the first problem? When I typed out all the ways i could get 11, I got 35 possibilities . apparently I did something wrong...

Edit:
Ok, I kinda figured that part out now: it can't be equal to 11, and there were 2 possibilities that equaled 11, so I took those out, and I got 33/36 possibilities, and that reduces to 11/12.

 

[veg1992]

well I didn't list all the ways to get less than 11, cuz I felt kind of lazy

but I can tell you its all the ways except for 6,6 5,6 and 6,5 (I believe it said less than 11)

 

[Merkwürdigeliebe]

ok for the second one you consider the ways to get no more than 7

aka 1,2 2,1 1,3 3,1 4,1 1,4 1,5 5,1 6,1 1,6 5,2 2,5

ok so thats 12 of them right there, out of 36, is 12/36, or 1/3

sorry I suck at explaining

the sum of the combination can't be even though, so the ones I highlighted in your post will not work

here's how I'd do it:

1: 1,2,3,4,5,6
2: 1,2,3,4,5
3: 1,2,3,4
4: 1,2,3
5: 1,2
6: 1

There are 21 combos there that are < or = 7

Take out the ones where the sum is even (bolded). There's 9 of them, 21-9 = 12

12/36 = 1/3



That doesn't sound entirely right to me though. I always thought that calculating the probability of getting two or more numbers in a row was an exponential relationship

Like the probability of getting seven '6's in a row would be (1/6)^7. Isn't that right?


ah wait a minute, interesting, taking out the combinations that have an even sum means taking out any combinations that have the same number turning up in a row (obviously), so the exponential relationship isn't an issue..


as for the first question, vernong explained it perfectly fine, there are only 3 combinations that don't work with that parameter (being less than 11), 6/6, 5/6 or 6/5
so from your 36 possible combinations, take out the 3, and you'll you 33/36 = 11/12

 

[peterhuang913]

well I didn't list all the ways to get less than 11, cuz I felt kind of lazy

but I can tell you its all the ways except for 6,6 5,6 and 6,5 (I believe it said less than 11)

Correct. That's leaves 3 out of the possible 36. So the probability is 33/36 or 11/12.

 

[carnageX]

Ok, I figured out the 11/12 problem.

There's an easier way to find out how many different ways isn't there? Other than counting them all out, or is that about the best way to do it?

 

[Merkwürdigeliebe]

I don't see how it can get easier than the way we've mentioned it here dude.

You've got numbers 1,2,3,4,5 and 6.

Combinations of two numbers that are > or = 11: 6/6, 5/6, 6/5 (3 combinations)

36 - 3 = 33

33/36 = 11/12

It's a quick mental math problem.

 

[peterhuang913]

You don't have to list all of it, just the ones that don't work.