Help with probability!!

bigslick

I realise this is Leaving Cert stuff but i cant seem to totaly work this problem out.

If a individual is given a choice of 6 boxes (numbered 1,2,3,4,5,6) to choose, one at a time until they have chosen all 6 (once a box is choosen, it is taken away), what is the probability that they will choose them in the order 1,2,3,4,5,6?

I might not have explined it well but is really taxing my mind. Any help would be appreciated.

magma69

bigslick wrote: »

I realise this is Leaving Cert stuff but i cant seem to totaly work this problem out.

If a individual is given a choice of 6 boxes (numbered 1,2,3,4,5,6) to choose, one at a time until they have chosen all 6 (once a box is choosen, it is taken away), what is the probability that they will choose them in the order 1,2,3,4,5,6?

I might not have explined it well but is really taxing my mind. Any help would be appreciated.

Been a long time since I've done any sort of maths (don't know what I'm doing here to be honest!!) so don't take my word for it. I think it would be 6x5x4x3x2x1 = 720
Answer = 720/1

Mark Hamill

bigslick wrote: »

I realise this is Leaving Cert stuff but i cant seem to totaly work this problem out.

If a individual is given a choice of 6 boxes (numbered 1,2,3,4,5,6) to choose, one at a time until they have chosen all 6 (once a box is choosen, it is taken away), what is the probability that they will choose them in the order 1,2,3,4,5,6?

I might not have explined it well but is really taxing my mind. Any help would be appreciated.

Just do it step by step:Whats the odds of picking a particualr box out of 6 boxes? 1/6. Then what is the odds of picking a particular box out of (6-1) boxes? 1/5. Keep doing this and then multiply all the odds together (because you want the odds of picking a particular box from 6 boxes AND the odds of picking a particular box out of 5 boxes AND the odds of ... etc down to picking a particular box out of one box.
This is simply (1/6)(1/5)(1/4)(1/3)(1/2)(1/1) = 1/6x5x4x3x2x1 = 1/720 (wrong way round magma69 )

EDIT: To help get your head around this, just ignore that the boxes are numbered. You are picking a particular box from a sequence of boxes, no particular numbered box is more/less likely than any other, the numbering is arbitrary. Essentially what is done is that you calculate how many possible combinations of boxes there are (720) and say that one particular combination (1,2,3,4,5,6 but it could be any combination) has the odds of 1/720 of occuring

Thomas_S_Hunterson

The odds on any one combination would be 1-719.

Michael Collins

Sean_K wrote: »

The odds on any one combination would be 1-719.

Yep. Just to point out to prevent possible confusion that what Sean has here are the odds on getting the right permutation. Odds are a mathematical concept in their own right which are defined slightly differently to probabilities. The probability of getting the right permutation is 1/720.

The two are closely related:

[latex]\displaystyle odds\ on = \frac{p}{(1-p)}[/latex]

which in this case gives

[latex]\displaystyle odds\ on = \frac{\frac{1}{720}}{1-\frac{1}{720}} = \frac{\frac{1}{720}}{\frac{719}{720}}=\frac{1}{719}[/latex]

or in the usual notation: 1-719 odds on.

Note that a bookies would usually quote this as odds against:

i.e. 719-1 against getting the right permutation. In other words the odds that you won't get it are good.

MathsManiac

It's worth pointing out, by the way, that all of the above analysis is based on an assumption, which was not stated in the original post, that all orderings are equally likely, (or, equivalently, that the person is equally likely at each stage to choose any of the remaining boxes.)

This is a useful assumption for modelling purposes, but is unlikely to be true in reality, unless the person choosing the boxes cannot see the labels, and the labelling is not related to the way the boxes are presented to them.

Mellor

sigh,
it's a leaving cert question. Unless it is stated other wise, then each ball is as likely. No assumptions.

rjt

Mellor wrote: »

sigh,
it's a leaving cert question. Unless it is stated other wise, then each ball is as likely. No assumptions.

The OP said it was "LC stuff" not necessarily actually a leaving cert question. And if this is a real world thing, then each box definitely doesn't have equal probability (as people generally pick 3 more often than 4 for example). There's no harm in pointing this out!

MathsManiac

Mellor wrote: »

sigh,
it's a leaving cert question.

I very much doubt it. Can you give a reference (year, paper, question number)?

Mellor wrote: »

Unless it is stated other wise, then each ball is as likely.

Seems a bizarre premise to work from.
(But then again, since there were no balls in the question, I guess the probability of picking a ball is 0, and so all balls are equally likely. )

Mellor wrote: »

No assumptions.

...apart from the ones being made.

Mellor

rjt wrote: »

The OP said it was "LC stuff" not necessarily actually a leaving cert question. And if this is a real world thing, then each box definitely doesn't have equal probability (as people generally pick 3 more often than 4 for example). There's no harm in pointing this out!

If it's LC stuff, then the general rules of LC questions apply.

also, the numbers only come into it if they are visible. Normally, the words at random is used to get around pedants such as yourself.

MathsManiac wrote: »

I very much doubt it. Can you give a reference (year, paper, question number)?

Seems a bizarre premise to work from.
(But then again, since there were no balls in the question, I guess the probability of picking a ball is 0, and so all balls are equally likely. )


...apart from the ones being made.[/QUOTE]

sigh, no comment

rjt

Mellor wrote: »

If it's LC stuff, then the general rules of LC questions apply.

also, the numbers only come into it if they are visible. Normally, the words at random is used to get around pedants such as yourself.

I probably shouldn't take the bait, but I think this is worth saying. Assuming the OP is dealing with a real world situation (eg. the probability of picking specific boxes in Deal or No Deal, say), then human preference really is a factor. Maybe the OP did just want the absolute, theoretical "all boxes are equal" answer - and that was given by several posters - but in the off chance that this wasn't the case, MathsManiac simply pointed out that this might not be wholly applicable.

I don't really see what the problem is here. I don't think MathsManiac's comment was in any way snide, he was just pointing out something that might be useful to the OP.

MathsManiac

Like rjt, I shouldn't rise to the bait, but I will...

Mellor wrote: »

If it's LC stuff, then the general rules of LC questions apply.

Maybe so, but I have been an LC examiner in the past, and I know that an assumption of equal likelihood is not usually made unless it is clearly justifiable. In LC questions, this assumption is usually explicitly stated when it applies.

Mellor wrote: »

... Normally, the words at random is used to get around pedants such as yourself.

The words "at random" are not inserted so as to satisfy pedants. They are inserted to provide important information. Since this phrase was not used in this case, I believe it was reasonable to point out that such an assumption underlied the solutions that had been offered.

Despite your apparent exasperation, I hope to continue offering helpful comments when I can.