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hailtothethief
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probability question
#7389743 - 09/09/07 09:04 PM (16 years, 4 months ago) |
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anyone here good with probability?
A pair of fair dice are tossed 3 times. Find the probability that the sum of 7 or 11 appears exactly twice.
i missed class all last week it was at 8:30
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maggotz


Registered: 06/24/06
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find all possible combinations that add up to 7 or 11 and divide that by the total number of possible combinations. then multiply that by itself.
i think that's how it goes but i'm a little stoned.
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California
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Quote:
hailtothethief said: anyone here good with probability?
A pair of fair dice are tossed 3 times. Find the probability that the sum of 7 or 11 appears exactly twice.
i missed class all last week it was at 8:30
are they tossed by a left handed person, right handed person, or game dice cup?
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hailtothethief
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Re: probability question [Re: maggotz]
#7389810 - 09/09/07 09:20 PM (16 years, 4 months ago) |
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na thats not it but math is always hard stoned so its cool
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hailtothethief
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Re: probability question [Re: California]
#7389854 - 09/09/07 09:33 PM (16 years, 4 months ago) |
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they are spat out of a naked 2 foot midgets mouth onto a strippers tabletop ass
of course my fucking asian calculator friend isnt online now
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maggotz


Registered: 06/24/06
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i got a little rusty but here ya go.
the probability of landing a 7 is 1/6 cuz there are 6 combinations out of 36 that give you a sum of 7. ( 1 and 6, 6 and 1, 2 and 5, 5 and 2, 3 and 4, 4 and 3). and the probability of landing an 11 is 1/18 cuz only 2 out of 36 possible combinations give you a sum of 11. ( 5 and 6, 6 and 5).
now, event A is landing a 7 and event B is landing an 11.
A and B are mutually exclusive therefore P(A U B) = P(A) + P(B) = 1/6 + 1/18 = 2/9
or
the total number of points at which the event "A U B" (n)happens divided by the total number of possible points (N). P(A U B) = n/N = 8/36 = 2/9

Edited by maggotz (09/09/07 11:06 PM)
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hailtothethief
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Re: probability question [Re: maggotz]
#7389922 - 09/09/07 09:52 PM (16 years, 4 months ago) |
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i dont think you are factoring in that it can only happen TWICE, not three times. Its multiple choice and these are the answers... 8/729 28/243 28/729 4/27 5/72
fucking damn curveball of the problem is that it has to be exactly twice
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nakors_junk_bag
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Re: probability question [Re: maggotz]
#7389925 - 09/09/07 09:55 PM (16 years, 4 months ago) |
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kjbkjhilkhih
-------------------- Asshole
Edited by nakors_junk_bag (09/09/07 10:45 PM)
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maggotz


Registered: 06/24/06
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ahh i had read you tossed the dice twice. but you got the idea now, right?
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hailtothethief
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You have two dice. You are rolling them three times. You want to roll a 7 or an 11 on two of the rows. The third roll can be any number but 7 or 11.
There are 8 ways you can roll a 7 or 11, so there are 28 ways to not roll a 7 or 11. Fucking this class was called calculus 125... why am i being taught this bullshit.
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nakors_junk_bag
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oh, so you don't have to roll the same number two out of three times.
You can roll a seven and an eleven, or two sevens or two elevens?
-------------------- Asshole
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maggotz


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Quote:
nakors_junk_bag said: a 6 and 1 is exactly the same thing as a 1 and a 6, there are only three possible rolls to land a seven with two dice, 1 and 6, 5 and two and a 3 and 4.
with an eleven there is 1, a five and a 6.
you need to find the percents. If there are 22 posible rolls., the chance of landing a 7 would be the the number of combinations(3) that create 7 into the the total number of possible outcomes 36.
possible dice rolls: 1,1 1,2 1,3 1,4 1,5 1,6 2,2, 2,3 2,4, 2,5, 2,6 3,3, 3,4, 3,5 3,6 4,4 4,5, 4,6 5,5, 5,6 6,6
that would be like one in every seven. will roll you a seven. Now you need to find the odds for rolling it in three right?
please bare in mind its been ten years since my probability classes.
if i have two coins, each has a red side and a black side. how many possible combinations are there for red and black?
one or two? coin a = red, coin b = black or coin a = black, coin b = red
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hailtothethief
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Re: probability question [Re: maggotz]
#7389954 - 09/09/07 10:04 PM (16 years, 4 months ago) |
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not really, but hey man thanks anyways at least you tried..
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maggotz


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sorry if i'm not explaining myself clearly but, like i said, i'm a little stoned.
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hailtothethief
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you could roll 2 7s, a 7 and 11, an 11 and 7, two 11s, etc. You just have to roll either one of them twice, but only twice out of the three rolls
so like... 7, 11, (any number but 7 or 11)
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nakors_junk_bag
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28/729?
-------------------- Asshole
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hailtothethief
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now thats what i thought, but a girl across the hall submitted that answer and it was wrong.
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maggotz


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Re: probability question [Re: maggotz]
#7389973 - 09/09/07 10:09 PM (16 years, 4 months ago) |
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nooo wait. 2/9 is the probability of landing either a 7 or a 11 trowing the dice once. i didn't really understand what the original problem was. my bad.
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nakors_junk_bag
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Re: probability question [Re: maggotz]
#7389978 - 09/09/07 10:11 PM (16 years, 4 months ago) |
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well we know the probablility of it being one of the answers left just became greater.
-------------------- Asshole
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Iolaa
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ok make a square with 36 spaces, 1-6 on each side. then add the numbers and put them in the corresponding square...
blah blah
5 ways to get 7, 2 ways to get 11. 7+2=9 9/36 = 1/3.
GO TO CLASS NEXT TIME.
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nakors_junk_bag
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let me fininsh working this it the way I had started and let me see what i get.
-------------------- Asshole
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nakors_junk_bag
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Re: probability question [Re: Iolaa]
#7389992 - 09/09/07 10:13 PM (16 years, 4 months ago) |
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yeah, like I said it was ten years ago.
you seem to know what you are doing, carry on.
I recant all my mathematical ineptness.
-------------------- Asshole
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Iolaa
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Re: probability question [Re: Iolaa]
#7389993 - 09/09/07 10:13 PM (16 years, 4 months ago) |
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wait. then square 1/3 to get either a 7 or 11 twice in a row, then add a third for the extra throw. or not. it's been a while. or i'm a dumbass.
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VoidOfsPg
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I'm in statistics right now and I'm totally lost here.
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hailtothethief
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Re: probability question [Re: Iolaa]
#7389999 - 09/09/07 10:15 PM (16 years, 4 months ago) |
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You have to roll the two dice THREE times and get 7 or 11 on two of those rolls. 8:30 is so damn early in the morning, i even went Wednesday but i just passed out 10 minutes in
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nakors_junk_bag
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1/4 according to your calculations Iolaa.
9/36 equal 1/4
-------------------- Asshole
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nakors_junk_bag
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might it be that every 27 times you roll the dice three times, four of those will give you numbers within the desired parameters?
-------------------- Asshole
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hailtothethief
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its one of these. 8/729 28/243 4/27 5/72
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nakors_junk_bag
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shit. lol.
it sems that the odds are going to be against you. becasue of they weren't vegas wouldn't bet on them.
-------------------- Asshole
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nakors_junk_bag
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1 and 6 chance of rolling a seven, and a 1 and 18 chance of rolling an eleven,
1/6 +1/18= 4/ 18 of rolling your numbers.
which is as someone stated 8/36, so don't you need to multiply the fraction times the bumber of times you get to roll? 3
that would be for 108 rolls you get the numbers you need 24 times.
Edited by nakors_junk_bag (09/09/07 10:38 PM)
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nakors_junk_bag
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sorry, I am teh stoopid
-------------------- Asshole
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maggotz


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goddamn, i'm too high for this right now.
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DieCommie


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Quote:
nakors_junk_bag said: 28/729?
Thats what I think too. Not positive though. Is the 'girl down the hall' smart?
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nakors_junk_bag
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so if for every 108 times you get e numbers you need 24 times you need to isolate the incidents when theos numbers happen with one roll of eachother. thus meanins whe ned to devide furhter by the
-------------------- Asshole
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nakors_junk_bag
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hold on give me a moment, whe the dead stops beating my ass
-------------------- Asshole
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nakors_junk_bag
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ah shit, I don't know. Time to break out the dusty ol housemate and his trusty ol notebooks! lol
-------------------- Asshole
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