Can anybody help me out with a maths problem?

[Conelton]

Hey guys.

I know this isn't exactly Yahoo Answers but the community here has always been pretty intelligent and helpful so I'm turning to you guys for help.

The simple question is: why is (1/26)^3 not equal to 1/(26P3) ?

The longer question is about the probability of choosing a particular permutation of 3 randomly-selected letters from the alphabet.

I.e.

3 letters are chosen randomly from the alphabet. What is the probability that they are, let's say, ABC, in that order.

The correct answer is apparently (1/26)^3. I understand that. But why does 1/(26P3) give a different answer, or why isn't it also correct? The number of permutations of 3 from 26 is 26P3, and ABC is one such permutation.

Any help would be appreciated. Alternatively I can just ask my stats teacher, but I thought I'd give you guys something to do

[NInety]

No, but I can help you out with your grammars.

JK I know people from England say 'Maths' instead of Math.

 

Answer: Order of operations is why.

[Conelton]

What is this 'math' you speak of?

I can't see why the order of operations is causing a problem here, could you please explain?

[Nineshadow]

How is (1/26)³ not equal to 1/(26³)?It fucking is.

(1/26) = 1³ / 26³ ,which is equal to 1/26³ ,because ,obviously, 1³=1

I mean,we have a proportion : a/b.

(a/b)ⁿ = aⁿ/bⁿ

[MangoJuiceStain]

How is (1/26)³ not equal to 1/(26³)?It fucking is.

(1/26) = 1³ / 26³ ,which is equal to 1/26³ ,because ,obviously, 1³=1

I mean,we have a proportion : a/b.

(a/b)² = a²/b²

Yeah, you're right... 

 

edit: wait if you do bedmas, then you do the brackets first, so 1/26 which is equal to 0.38. Then you do 0.38³ which is 0.055. so is the answer 0.055? then for the second one 1/(26³) is equal to 1/17576 which is 0.000057.

[_ASSASSIN_]

How is (1/26)³ not equal to 1/(26³)?It fucking is.

(1/26) = 1³ / 26³ ,which is equal to 1/26³ ,because ,obviously, 1³=1

I mean,we have a proportion : a/b.

(a/b)² = a²/b²

however if the numerator is not 0 or 1 it won't work.

[rjshield]

I think it's how you are writing the question to us...

By 1/(26P3) do you mean 1/(26³) ?

[Nineshadow]

however if the numerator is not 0 or 1 it won't work.

But it does with 1.

(1/2)²=1/4=0.25

(1/2)²=0.5²=0.25

Another point :

(1/26)³=1/26 * 1/26 * 1/26 = 1/(26*26*26) = 1/26³

[_ASSASSIN_]

But it does.

(1/2)²=1/4=0.25

(1/2)²=0.5²=0.25

Another point :

(1/26)³=1/26 * 1/26 * 1/26 = 1/(26*26*26) = 1/26³

 

however if the numerator is not 0 or 1 it won't work.

[Fred Castellum]

There ya go OP, is your answer resolved now?

[Nineshadow]

 

Which one is the numerator?a? (a/b)

[_ASSASSIN_]

Which one is the numerator?a? (a/b)

how do you not know what a numerator is? It's the number on the top yo.

Well now that I think of it, it won't matter, numerator or denominator, same thing happens either way.

[Nineshadow]

how do you not know what a numerator is? It's the number on the top yo.

I'm not English.

[MangoJuiceStain]

Fuck it. ABANDON THREAD.

[Nineshadow]

however if the numerator is not 0 or 1 it won't work.

But it works.

(3/5)³=0.6³=0.216

(3/5)³=27/125=0.216

It doesn't have any reason not to.

(a/b)ⁿ=(a/b) * (a/b) * (a/b) * (a/b) .... * (a/b)   -> where we multiply a/b by itself n times.

we know a/b * c/d = (a*c)/(b*d) so :

(a/b)ⁿ=aⁿ/bⁿ

[Conelton]

Rjshield got it (can't quote, on mobile)

Sorry guys. There is some confusion here. Again, because I'm on mobile I couldn't really format properly.

By 26P3 I mean ²⁶P₃.

I realise that also doesn't make much sense but it's the best I can do.

It's a notation used in statistics (probability) meaning permutations. nPr, similar to nCr. Perhaps somebody who knows what I'm getting at can explain it better.

Edit: sorry nineshadow, accidentally posted without finishing. Also, I just found the quote button. Could this have gone much worse? :')

[Nineshadow]

Rjshield got it (can't quote, on mobile)

Sorry guys

What?

[Conelton]

What?

See edited post

[rjshield]

26P3 is 26*25*24  = 15600

26^{3    is 26*26*26 = 17576}

[Nineshadow]

See edited post

Oooooooooooooooooooohhhh!

[TideRower]

With 26P3 you take one letter away after you got it.

In your problem you just choose a random letter and then write it down.

In the second case you can get AAA which isn't possible in the first one. As a consequence there are less options.

 

You should view this problem as choosing three times a random letter which makes it easier:

The possibility of getting A is 1/26.

Doing this three times: (1/26)*(1/26)*(1/26)=1/26^3

You can do this because they don't influence each other (i.e. it doesn't matter which letter you got the first time, the possibilities of the second round don't change)

[Conelton]

26P3 is 26*25*24  = 15600

26^{3    is 26*26*26 = 17576}

 

Yes! That's my point. They give 2 different answers which is no big deal until you put it into the context of the question.

 

The probability of getting a particular permutation of 3 numbers from a set of 26 (e.g. 'ABC':

P('ABC') = P(1st number = A) * P(2nd number = B) * P(3rd number  = C) = 1/26 * 1/26 * 1/26 = 1/17576

 

But, the number of permutations of the 3 selected letters is 26P3. 'ABC' is one permutation. Therefore the probability P('ABC') = 1/26P3.

 

Except I must have understood something wrong here.

 

EDIT: never mind, I've got it now!

[Conelton]

With 26P3 you take one letter away after you got it.

In your problem you just choose a random letter and then write it down.

In the second case you can get AAA which isn't possible in the first one. As a consequence there are less options.

 

You should view this problem as choosing three times a random letter which makes it easier:

The possibility of getting A is 1/26.

Doing this three times: (1/26)*(1/26)*(1/26)=1/26^3

You can do this because they don't influence each other (i.e. it doesn't matter which letter you got the first time, the possibilities of the second round don't change)

 

Thank you! This makes sense now. I never considered that permutations are only useful for calculation the probability of a dependent event. For an independent event like this, it makes much more sense to consider them individually (by definition, in fact).

 

If only I could give you an imaginary internet maths point...