Re: question re: problem
@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 EventsOnly 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:
http://www.helsinki.fi/science/metapop/
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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Re: question re: problem
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) x P (NotPower) x P
(NotPower)] +
[P (NotPower) x P (NotPower) x P (NotPower) x P (Power) x P
(NotPower)]+
[P (NotPower) x P (NotPower) x P (NotPower) x P (NotPower) x P
(Power)]
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 EventsOnly 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:
http://www.helsinki.fi/science/metapop/
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
=
Instructions for joining and leaving this list and remarks about
the problem of INAPPROPRIATE MESSAGES are available at
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=
Re: question re: problem
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 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 EventsOnly 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:
http://www.helsinki.fi/science/metapop/
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
=
Instructions for joining and leaving this list and remarks about
the problem of INAPPROPRIATE MESSAGES are available at
http://jse.stat.ncsu.edu/
=
Re: question re: problem
@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 (NotPower) x P (NotPower) x P (Power) x P (NotPower) x P
(NotPower)] +
[P (NotPower) x P (NotPower) x P (NotPower) x P (Power) x P
(NotPower)]+
[P (NotPower) x P (NotPower) x P (NotPower) x P (NotPower) x P
(Power)]
Have a look at Arto's reply, and simple stuff on permutations and
combinations (it's the combinations bit that's relevant).
I assumethat this is homework, so your course notes should help. Or an
elementary textbook on probability and statistics should derive the
binomial distribution for you. But it looks like you've got the basic
idea.
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:
http://www.helsinki.fi/science/metapop/
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
=
Instructions for joining and leaving this list and remarks about
the problem of INAPPROPRIATE MESSAGES are available at
http://jse.stat.ncsu.edu/
=
Re: question re: problem
(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 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 EventsOnly 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 =
Don't forget, you listed only 1 way to get 1 PW (power window) and 4 not.
There are 5 wys you could get this result, if you don't count the order (which
the question doesn't include). C(5,1) = 5!/(4!*1!) = 5.
So: 0.51*0.49*0.49 * 0.49 * 0.49 * 5 = = 0.14700
Also Solve Using BINOMDIST Function in Excel
n 5
? 0.51 Success - PW
x 1
p(x) 0.14700
b. At least 1 w/Power Windows
P {At Least 1} = 1 - P {0}
P {0} = P (NotPower) x P (NotPower) x P (NotPower) x P (NotPower) x
P (NotPower)
0.49 0.49 0.49 0.49 0.49
Prob 0 0.02825
1 - 0.02825
At least 1 0.97175
In this one, all the outcomes are alike, so there is no combination effect.
Also Solve Using BINOMDIST Function in Excel ~ 97%
n 5
? 0.49 Success - No Power
x 0
p(x) 0.02825
1 - 0.028247525
97%
So you got it! Or nearly so.
Cheers,
Jay
--
Jay Warner
Principal Scientist
Warner Consulting, Inc.
North Green Bay Road
Racine, WI 53404-1216
USA
Ph: (262) 634-9100
FAX: (262) 681-1133
email: [EMAIL PROTECTED]
web: http://www.a2q.com
The A2Q Method (tm) -- What do you want to improve today?
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the problem of INAPPROPRIATE MESSAGES are available at
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