"@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 the
Thanks alot - it worked. How would you compose a short formula depicting:
P {Only 1} = [P (Power) x P (NotPower) x P (NotPower) x P (NotPower)
x P (NotPower)] +
[P (NotPower) x P (Power) x P (NotPower) x P (NotPower) x P
(NotPower)] +
[P (NotPower) x P (NotPower) x P (Power)
Your probability distribution is binomial
p = 0.51q = 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 PRO
"@Home" wrote:
>
> Thanks alot - it worked. How would you compose a short formula depicting:
>
> P {Only 1} = [P (Power) x P (NotPower) x P (NotPower) x P (NotPower)
> x P (NotPower)] +
> [P (NotPower) x P (Power) x P (NotPower) x P (NotPower) x P
> (NotPower)] +
> [P (NotPo
(sending to all - @Home is a non-functioning address) - Jay
"@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 a
