Probability query. (1 Viewer)

[CHUDYMASTER]

If you get a question that goes something like, "If a die is thrown twice, what is the probability that the sum of the outcomes will be 7?" Is there ANY WAY to possibly do this without drawing up a table of possible outcomes? I mean a MATHEMATICAL WAY!! No tree diagrams, no diagrams period. Anyone?

 

[Jellymonsta]

uh yes (im almost sure...)
but with dice, why bother when its easier to do diagrams?
if you want to do well in your final exams, you have to be prepard to be ruthless, and exploit everyone and everything to the utmost. this does not include exploiting telegraph poles or small furry animals. you cant be too picky about the methods you employ. nor should you go to exams and mention the history of the donkeys in the armed services.

 

[spice girl]

You can usually do the thing in your head, but for more complicated stuff, you need a box diagram. Depends on how familiar you are with the stuff.

 

[McLake]

You know its /36 right?

Not too many combos to count ...

 

[CHUDYMASTER]

No...

I didn't mean it like that Mclake. I was talking about a problem which involved finding the probability of rolling a total of 7, an odd no., etc..

Doesn't matter though, I guess I'll have to live with silly, non-mathematical methods.

 

[underthesun]

hey is this in the syllabus?

hopes not

 

[McLake]

Originally posted by underthesun
hey is this in the syllabus?

hopes not

Although harder 3U probility isn't specfically mentioned (but none of the harder 3U is really ...) it seems to come up a lot, so ...

Edit: Wait, this is a 3U forum. Um, i dont know if that's 3U ...

 

[underthesun]

never saw this, so i'll take the shortcut and take it that it won't be in this year's hsc

 

[CHUDYMASTER]

Actually...

I wouldn't...this IS in the 3 unit course for the HSC.

The questions based on this aren't that hard. But be prepared.

 

[underthesun]

but it won't ask the sum of blah, if the dice is thrown 3 or more times right? I don't know how to do that

 

[freestyler__54]

isnt it just binomial prob!?!

 

[BlackJack]

edit: there is one...
In these probability stuff... you can also derive the maths from a tree diagram.

however, each of the branches 1 to 6 has one secondary branch to make the sum 7... which is like...
P(sum=7) = 1 * 6 / 6 * 6

For a general method in two dice... these two dice are identical
Let te range of dice values be a to b going up one at a time
It should be
P(sum = x) = [ (b-a) - | x- (b+a)| + 1 ] / (b-a+1)^2

You'll have to wait for a three dice one... AND the explanation for this... 'cause I'm only doing this for public interest...

 

[BlackJack]

Okay... the distribution of the results from the sum of the 3 dice throws forms a series of trianglular numbers from both directions... to make a formula however, there is a problem with my conventional one if you think about it...

The distribution looks like this
Sigma 3 * D6:
3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
1 3 6 10 15 21 28 36 36 28 21 15 10 6 3 1
I think somebody can join the dots...

 

[CHUDYMASTER]

Uhh...

I'm absolutely, positively perplexed....

 

[BlackJack]

Sorry, I wasn't in the best of moods.

There IS a way of representing the probability from a dice throw mathematically.

When you look at all the results, i.e. draw up table and sum the no. of time this sum comes up... it is symmetrical.

For the two dice it goes like:
1,2,3,4,5,6,5,4,3,2,1

Therefore, it can be represented with the formula (which isn't a nCr, but that's not the point).

For a three dice throw it's 1,3,6,10,...,10,6,3,1 BUT there can be 1 or 2 highest numbers (alternates btwn them)

For 4 dice it is 1,4,10....10,4,1

Therefore I thought it is a form of combination.

edit: that formula I worked out, say the dice is faced 1,2,3,4,5,6 then a=1 b=6 and x is the sum.