Geometric Distribution

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October 15th 2008, 03:23 PM
kid funky fried

Geometric Distribution

The following is a disscussion problem in our stats class. It seems
that several students arrived at several different answers for part b.
So I am curious if I have made an error and if so, where.
(note: this is exactly how the problem reads.)

A certain experiment is to be performed until a successful result is obtained. The trials are independent and the probability of success on a single trial is 0.25. The costof performing the experiment is $25,000; however, if a failure results, it costs $5,000
to "set-up" for the next trial.
a) What is the expected cost of the project? ans. $40,000

b) Suppose that the experimenter has a maximum of $500,000, what is the probability that the experimental work would cost more than this amount?

here is what i did:

$500,000 - $25,000 = $475,000 (to spend on failures)

$475,000 / $5,000 = 95 failures

X: number of trials
So, we wish to find P ( X > 95 ) = (1-.25)^96 which is approx. 0.
October 15th 2008, 04:59 PM
mr fantastic

Quote:

Originally Posted by kid funky fried

The following is a disscussion problem in our stats class. It seems

that several students arrived at several different answers for part b.
So I am curious if I have made an error and if so, where.
(note: this is exactly how the problem reads.)

A certain experiment is to be performed until a successful result is obtained. The trials are independent and the probability of success on a single trial is 0.25. The costof performing the experiment is $25,000; however, if a failure results, it costs $5,000
to "set-up" for the next trial.
a) What is the expected cost of the project? ans. $40,000

b) Suppose that the experimenter has a maximum of $500,000, what is the probability that the experimental work would cost more than this amount?

here is what i did:

$500,000 - $25,000 = $475,000 (to spend on failures)

$475,000 / $5,000 = 95 failures

X: number of trials
So, we wish to find P ( X > 95 ) = (1-.25)^96 which is approx. 0.

The first trial costs $25,000. Then you can afford 95 more trials. So you need success on the 96th trial (1 + 95 = 96).

So you want

$\Pr(X > 96) = 1 - \Pr(X \leq 95) = 1 - \left[ 1 - \left(1 - \frac{1}{4} \right)^{95} \right] = \left( 1 - \frac{1}{4} \right)^{95} = \left(\frac{3}{4} \right)^{95} = 0$ (correct to ten decimal places).
October 15th 2008, 05:53 PM
kid funky fried

Quote:

Originally Posted by mr fantastic

The first trial costs $25,000. Then you can afford 95 more trials. So you need success on the 96th trial (1 + 95 = 96).

So you want

$\Pr(X > 96) = 1 - \Pr(X \leq 95) = 1 - \left[ 1 - \left(1 - \frac{1}{4} \right)^{95} \right] = \left( 1 - \frac{1}{4} \right)^{95} = \left(\frac{3}{4} \right)^{95} = 0$ (correct to ten decimal places).

Mr. Fantastic,

First, I would like to say the I really appreciate all your help.
Your generosity with your time and knowledge means alot to me as well as my fellow students. Thank You.

The way i understand the problem the experiment cost $25000 (as opposed to the first trial) so we are trying to find the probability of having greater than 95 failures(25,000+475000). So , I thought we would use (1-p)^(k+1).
Why do we need success on the 96 trial?
October 15th 2008, 06:02 PM
mr fantastic

Quote:

Originally Posted by kid funky fried

Mr. Fantastic,

First, I would like to say the I really appreciate all your help.
Your generosity with your time and knowledge means alot to me as well as my fellow students. Thank You.

The way i understand the problem the experiment cost $25000 (as opposed to the first trial) so we are trying to find the probability of having greater than 95 failures(25,000+475000). So , I thought we would use (1-p)^(k+1).
Why do we need success on the 96 trial?

My mistake. Replace all 96's with 95 and all 95's with 94.