Re: Probability

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From: Hari (excel_hari_at_hotmail.com)
Date: 08/03/04 

Date: Wed, 4 Aug 2004 00:41:55 +0530

Hi MGS,

> Can anyone out there tell me even if I'm on the right track here?

Seems perfect to me.

Regards,
Hari
India

<MGS> wrote in message news:410f727e$0$41790$65c69314@mercury.nildram.net...
> Here's my attempt, but I'm no mathamatician...
>
> The first person picking a number will have a probability of 100/100 (1/1)
> that they will pick a number nobody else has as they are the first. The
> second person has a probability of 99/100 that they will pick another
number
> not chosen previously. The third is then 98/100.. etc.
>
> I guess if you do this through 18 iterations, you have your 18 different
> numbers exactly. The last two people then have an 18/100 chance of picking
> the same number as the previous people (assuming you are allowing both of
> the final people to pick the same number, if not then it would be 18/100
for
> the 19th person and 17/100 for the last person)
>
> You probably then have to multiply all the probabilities together:
>
> 100/100 x 99/100 x 98/100 x 97/100 x 96/100 x 95/100 x 94/100 x 93/100 x
> 92/100 x 91/100 x 90/100 x 89/100 x 88/100 x 87/100 x 86/100 x 85/100 x
> 84/100x 83/100 x 18/100 x 18/100
>
> = 6.36x10^37 / 1x10^40
>
> Take out the exponants to the lowest common denominator is
>
> = 6.36 / 1x10^3, which is 6.36/1000
>
> So the probability would be 157.23 to 1 against that out of 20 people,
> exactly 18 will pick different numbers with 2 picking numbers that have
> previously been chosen.
>
> Can anyone out there tell me even if I'm on the right track here?
>
> MGS
>
> > Hi Everyone, please help me solve this
> >
> > 20 people are invited to pick a random number between 1 and 100. Find
> > the probability that, out of the 20 random numbers selected, that
> > there are exactly 18 different numbers.
> >
> > Thank you so much for solving it
> >
> >
> > ---
> > Message posted from http://www.ExcelForum.com/
>
>

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