probability! (1 Viewer)

[lookoutastroboy]

just one question that i came across in my mind when making a tree diagram for some questions. Say that there is a tree diagram with like 4 options at first then 5 options out of each one and out of those 5 options, there are 6 options for each of those 5 options and so on... to get the results, you obviously have to look carefully and link up the options in succession. I was just wonderin if there was any way to find the outcomes easily and efficientlly rather than trace through so many links (can be thousands..) to get the outcomes. Secondly, if for example you have a subset such as this P(HD,HD,HD, HD,HD,D,D), is it possible to find out the total number of combinations from it? And if so, could you show me in detail?

 

[kurt.physics]

just one question that i came across in my mind when making a tree diagram for some questions. Say that there is a tree diagram with like 4 options at first then 5 options out of each one and out of those 5 options, there are 6 options for each of those 5 options and so on... to get the results, you obviously have to look carefully and link up the options in succession. I was just wonderin if there was any way to find the outcomes easily and efficientlly rather than trace through so many links (can be thousands..) to get the outcomes. Secondly, if for example you have a subset such as this P(HD,HD,HD, HD,HD,D,D), is it possible to find out the total number of combinations from it? And if so, could you show me in detail?

Now this, i know, is combinatorics.

For the first question is linked with the multiplication principal. I think in the past i didnt really explain it, so i will use the example that made me understand it. (I have also attached a diagram to help).

There are 3 roads from town A to town B and 4 roads from B to C. How many different ways are there to travel from A to B to C?

Consider one particular road from A to B; by traveling along this road, there are 3 ways to travel from A to B to C. Similarly, by taking another road from A to B, there are another 3 ways to travel from A to B to C. Hence there are 4 x 3 ways to travel from A to B to C.

(Just note this example was from fitzpatick)

This tells use how many outcomes there are. It doesnt tell us each individual outcome.

 

[kurt.physics]

I noticed you had some trouble with the product rule (or what ever it is called). I will use an example to show you.

Example: Lets say i rolled a die, the flipped a coin, and then flipped a coin again. What is the probability i will get a 6, a head, and a tale?

Firstly, just to show you how you would use the multiplication principal, the number of outcomes is 6 x 2 x 2 or 24. Do you see this. You have 6 numbers of the die, 2 choices with the coin and two for the coin again.

Anyway. What we want is to find P(6, H, T). Just note here that 6, H, T is not the same as 6, T, H!

So we will be using the formula P(A,B) = P(A) x P(B). This is for compound events, like the example, its where there are two or more things going on.

Also note that the probability of something goes next the the branch in the tree diagram. And the probabilities are without respect to everything else. So you would have like 10 dies rolling one after the other, and then a coin flipped, and the probability of a head is still going to be 1/2.

So first we must list the probability of getting a 6. There are six numbers and 6 is one of them, so the probability of getting a 6 is 1/6.

P(6) = 1/6

The probability of a Head is

P(H) = 1/2

and a tale is

P(T) = 1/2

So for a tree diagram, you would follow the 6 branch, then the Head branch that comes off of the 6 branch, then the Tale branch coming off of our head branch. You would multiply these probabilities together, this would give you the probability of that who event happening.

So

P(A,B) = P(A) x P(B) (note this can be done any times ie P(A,B,C,D) = P(A) x
P(B) x P(C) x P(D).)

So P(A,B,C) = P(A) x P(B) x P(C)

This is the same as P(6, H, T), so

P(6,H,T) = P(6) x P(H) x P(T)

= 1/6 x 1/2 x 1/2
= 1/24

 

[lookoutastroboy]

extra thanks kurt, ur a legend for hepn me ous so much, just with the subset thing e.g. (HD,HD,D,HD,HD,D,D,D) where HD = High Distinction and D = Distinction, is there a formula or way to calculate the number of combinations from it (order doesnt matter). Just say an easier one (HD,HD,HD,D) where there are 3 HDS and 1 D u can arrange it in 4 ways.
1st way: (HD,HD,HD,D)
2nd way: (HD,HD,D,HD)
3rd way: (HD,D,HD,HD)
4th way: (D,HD,HD,HD)
That is what i am trying to get at for the above subset, how would you show the total number of combinations for the easy one applied to the harder one above with the 4 HDs and 4 Ds?





thanks in advance, lookoutastroboy

 

[kurt.physics]

Im just typing it up now lookoutastroboy, so if you wait online for a second i can post it up

 

[kurt.physics]

extra thanks kurt, ur a legend for hepn me ous so much, just with the subset thing e.g. (HD,HD,D,HD,HD,D,D,D) where HD = High Distinction and D = Distinction, is there a formula or way to calculate the number of combinations from it (order doesnt matter). Just say an easier one (HD,HD,HD,D) where there are 3 HDS and 1 D u can arrange it in 4 ways.
1st way: (HD,HD,HD,D)
2nd way: (HD,HD,D,HD)
3rd way: (HD,D,HD,HD)
4th way: (D,HD,HD,HD)
That is what i am trying to get at for the above subset, how would you show the total number of combinations for the easy one applied to the harder one above with the 4 HDs and 4 Ds?

thanks in advance, lookoutastroboy

Now, i would just like to clarify again that i have only been doing combinatorics for 2 days, so dont crucify me if im wrong, but i will try to convince you though logic and we will see what you think.

Now the first thing we have to do is to make a distinction between a combination and a permutation. Permutations are arrangements, position matters, ie AB is not the same as BA.

Combinations is selections, AB is the same as BA. So if A and B were two people, then AB and BA are two different arrangements but only one selection of two people.

Now in this question, order doesnt matter, ie it is a combination problem.

And just for some more background information on notation. 5! (reading 5 factorial) is 5 x 4 x 3 x 2 x 1. In general n! is all numbers from (and including) n to 1 multiplied together. If we wanted to know how many ways we can arrange the letters in the word DOG is would be 3!. This is just an extension on the multiplication principal. For the first position, any of the 3 letters can occupy that position, say O. Then for the second position, because we have already used the word O, we only can choose from D or G ie 2 letters. So lets say we choose G, then for the last position we only have 1 letter left, namely D. So the new word is OGD, but we can use the multiplication principal and say that there are 3 x 2 x 1 or 6 ways to arrange those letters.

(Im going to brake these long post up, so i will post this one now and continue instantly, this way you can read one, then i will do another as you read it etc)

 

[lookoutastroboy]

soz kurt, that subset with the 4 HDs and 4 Ds is supposed to be a permutation problem, not a combination problem, order does matter, so could you show me any formula to calculate all the results out of that? (lol i wouldnt insult or offend anyone for any matter based on maths)

 

[kurt.physics]

So, to continue,

Say we have four people A, B, C, and D. In how many ways can they be arranged, taking two at a time? How many selections of two can be made?

We will list all the possibilities.

AB, AC, AD, BC, BD, CD,
BA, CA, DA, CB, DB, DC

(notice that the second line is just the opposite of the one above it.)

So there are, in total, 12 arrangements.

Now the selections remember is for combinations, ie AB is same as BA, there for the number of selections (or combinations) is 6, namely

AB, AC, AD, BC, BD, CD.

So lets try a way to get these combinations into permutations.

each of AB, AC, AD, BC, BD and CD can have each letter arranged 2! ways, eg for AB, it can be either AB or BA. We can consider the set (A,B) and ask how many different ways they can be arranged and conclude the first spot can be taked by either 2 and the last by the leftover 1. So there are 2 x 1 or 2! ways to arrange them.

So if we have 6 groups each of which (in their group) be arranged 2! ways, then the permutations is

6 x 2! or 12

This agrees with out previous example

Simple math tells us that if we have to multiply by the arrangements in each group of the combination to get the permutation. Then we can divide by the number of ways a particular group can be arranged to get the combination.

ie 12 / 2! = 6

This is the amount of combinations.

 

[kurt.physics]

soz kurt, that subset with the 4 HDs and 4 Ds is supposed to be a permutation problem, not a combination problem, order does matter, so could you show me any formula to calculate all the results out of that? (lol i wouldnt insult or offend anyone for any matter based on maths)

are you sure that its a permutation problem? Just out of curiosity, where is this problem from?

 

[kurt.physics]

In your example with the small subset, (HD,HD,HD,D), im fairly sure thats a combination problem?

 

[kurt.physics]

In your example with the small subset, (HD,HD,HD,D), im fairly sure thats a combination problem?

The reason why i think it is is because if it is a permutation problem then the arrangements of them would be 4! or 24, which contradicts your logic.

I will show you (just if your curious) as to how you got 4 combinations. (or at least how i think.

 

[shaon0]

just one question that i came across in my mind when making a tree diagram for some questions. Say that there is a tree diagram with like 4 options at first then 5 options out of each one and out of those 5 options, there are 6 options for each of those 5 options and so on... to get the results, you obviously have to look carefully and link up the options in succession. I was just wonderin if there was any way to find the outcomes easily and efficientlly rather than trace through so many links (can be thousands..) to get the outcomes. Secondly, if for example you have a subset such as this P(HD,HD,HD, HD,HD,D,D), is it possible to find out the total number of combinations from it? And if so, could you show me in detail?

This doesn't look like Harder 3unit.

 

[kurt.physics]

Just to continue from my previous maths post, we concluded that to get from a permutation to a combination then you divide by the amount of permutations of a particular group (like AB is a group of A and B and the different ways to arrange them is 2!).

With you small example (HD, HD, HD, D), the permutations is 4! or 24, but there is a group of three HDs (and just to clarify, even though HD is H and D, we dont consider it as a group of H and D because it is really High Distinction, so High Distinction could be represented by B or 2 or whatever). So in the group of 3 HDs, there can be 3! or 6 arrangements amoung themself, so the combination is

4!/3! or 4.

So if we apply this to you bigger example 4 HDs and 4Ds, ie
(HD, HD, HD, HD, D, D, D, D)

So there are 8! arrangements all together, but the 4 HDs can be arranged amoungst themself 4! times and the 4 Ds can be arranged amounst themselves 4! times, so the total combinations is

8! / (4! x 4!)

which using the definition of a factorial is

8 x 7 x 6 x 5 x (4 x 3 x 2 x 1)
------------------------------------
(4 x 3 x 2 x 1) x (4 x 3 x 2 x 1)

note this can be written as

8 x 7 x 6 x 5 x 4!
------------------------
4! x 4!

Crossing out the two four factoricals gives

(8 x 7 x 6 x 5) / 4!

= 1680 / 24
= 70

So there are 70 different combinations.

What do you think? : )