Your probability distribution is binomial
p = 0.51 q = 0.49
In five trials, the distribution is ( p + q ) ^ 5
= p^5 + 5 p^4q + 10 p^3q^2 + 10 p^2q^3 + 5 pq^4 + q^5
So the probability for one power and four not is 5 pq^4 and
for at least one is 1 - q^5
Arto Huttunen
"Anon." <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> "@Home" wrote:
> >
> > I had the following to solve:
> >
> > 51% of all domestic cars being shipped have power windows. If a lot
contains
> > five such cars:
> >
> > a. what is probability that only one has power windows?
> > b. what is probability that at least one has power windows?
> >
> > I solved each of these problems in two ways, one using std probability
> > theory and one by using a binomial distribution. I seemingly had no
problem
> > w/part b., but in part a. the probability theory did not seem to produce
the
> > correct answer. I have listed these below. What is wrong w/the
probability
> > equation listed below? Also is my answer to part b. correct?
> >
> > a. Randomly Draw Five Samples (Cars)
> >
> > Independent Events Only 1 w/Power Windows
> >
> > P{Only 1 Power} = P (Power) x P (NotPower) x P (NotPower) x P
> > (NotPower) x P (NotPower)
> > 0.51 0.49 0.49 0.49 0.49 =
> >
> What you've got here is the probability that the first car has Power,
> but the rest do not. You also need the probability that the second,
> third, fourth or fifth is the one with the Power.
>
> Bob
>
> --
> Bob O'Hara
> Metapopulation Research Group
> Division of Population Biology
> Department of Ecology and Systematics
> PO Box 17 (Arkadiankatu 7)
> FIN-00014 University of Helsinki
> Finland
>
> NOTE: NEW TELEPHONE NUMBER
> tel: +358 9 191 28779 fax: +358 9 191 28701
> email: [EMAIL PROTECTED]
> To induce catatonia, visit:
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>
> It is being said of a certain poet, that though he tortures the English
> language, he has still never yet succeeded in forcing it to reveal his
> meaning
> - Beachcomber
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