## Prob question

Expand Messages
• Hey guys I m new to this. I m not that big into probability but I have to do some basic stuff as part of my computer degree. It s giving me alot of trouble.
Message 1 of 6 , Aug 7, 2002
Hey guys I'm new to this. I'm not that big into probability but I
have to do some basic stuff as part of my computer degree. It's
giving me alot of trouble. Can any of you tell me the answer to these:

1.Distinguish between statistics of a sample and parameters of a
distribution.

2.An industrial process manufactures components with a 5% defective
rate. Each day products are inspected sequentially until the first
defective is obtained. Let X be the number of inspections carried out
until the first defective is found.

(a) If 5 products have been inspected without finding a defective,
what is the probability that the next 5 inspections will be free of
defectives?

(b) Derrive the expected number of inspections until the first
defective.

I know these questions might be easy to some of you guys but i'm
finding them difficult.
• ... Statistics of the sample is an estimate of the parameter of distribution. ... I think that the production of defectives is independent, so knowing about
Message 2 of 6 , Aug 7, 2002
> 1.Distinguish between statistics of a sample and
> parameters of a
> distribution.

Statistics of the sample is an estimate of the
parameter of distribution.

>
> 2.An industrial process manufactures components with
> a 5% defective
> rate. Each day products are inspected sequentially
> until the first
> defective is obtained. Let X be the number of
> inspections carried out
> until the first defective is found.
>
>
> (a) If 5 products have been inspected without
> finding a defective,
> what is the probability that the next 5 inspections
> will be free of
> defectives?
>

I think that the production of defectives is
independent, so knowing about the quality of some
products doesn't say anything about the following. So
the probability is (0.95)^5.

> (b) Derrive the expected number of inspections until
> the first
> defective.

(b) Derrive the expected number of inspections until
the first
defective.

Number of inspections / probability
1 / 0.05
2 / 0.95 * 0.05
3 / (0.95)^2 * 0.05
4 / (0.95)^3 * 0.05
...
i / (0.95)^(i-1) * 0.05

Expected value =
1 * 0.05 +
2 * 0.95 * 0.05 +
3 * (0.95)^2 * 0.05 +
...
i * (0.95)^(i-1) * 0.05 +
...

= sum(i:1,infinity)[i*0.05*(0.95)^(i-1)]
= 0.05 * sum(i:1,infinity)[i*(0.95)^(i-1)]

Let's compute the sum
sum(i:1,infinity)[i*x^(i-1)] =
= sum(i:1,infinity)[d(x^i)/dx] =
= d[sum(i:1,infinity)(x^i)]/dx =
= d[x/1-x]/dx =
= 1/(1-x)^2

So for x = 0.95 we have
sum(i:1,infinity)[i*(0.95)^(i-1)] = 1/(0.05)^2 = 400
and so
expected value =
= 0.05 * sum(i:1,infinity)[i*(0.95)^(i-1)] =
= 0.05 * 400 =
= 20

__________________________________________________
Do You Yahoo!?
Yahoo! Health - Feel better, live better
http://health.yahoo.com
• Couldn t we have also simply said 1/.05 = 20, and gotten the answer this way? I loooked at the problem as such: 1/.05 = x/1, and then solved for x. I did
Message 3 of 6 , Aug 7, 2002
Couldn't we have also simply said 1/.05 = 20, and

I loooked at the problem
as such: 1/.05 = x/1, and then solved for x. I did
appreciate your approach and maybe it holds for more
complicated situations where a shortcut is not
possible.

Regards,
M.W.

--- Beata Stehlikova <beata_stehlikova@...>
wrote:
> > 1.Distinguish between statistics of a sample and
> > parameters of a
> > distribution.
>
> Statistics of the sample is an estimate of the
> parameter of distribution.
>
> >
> > 2.An industrial process manufactures components
> with
> > a 5% defective
> > rate. Each day products are inspected sequentially
> > until the first
> > defective is obtained. Let X be the number of
> > inspections carried out
> > until the first defective is found.
> >
> >
> > (a) If 5 products have been inspected without
> > finding a defective,
> > what is the probability that the next 5
> inspections
> > will be free of
> > defectives?
> >
>
> I think that the production of defectives is
> independent, so knowing about the quality of some
> products doesn't say anything about the following.
> So
> the probability is (0.95)^5.
>
> > (b) Derrive the expected number of inspections
> until
> > the first
> > defective.
>
>
> (b) Derrive the expected number of inspections until
> the first
> defective.
>
> Number of inspections / probability
> 1 / 0.05
> 2 / 0.95 * 0.05
> 3 / (0.95)^2 * 0.05
> 4 / (0.95)^3 * 0.05
> ...
> i / (0.95)^(i-1) * 0.05
>
> Expected value =
> 1 * 0.05 +
> 2 * 0.95 * 0.05 +
> 3 * (0.95)^2 * 0.05 +
> ...
> i * (0.95)^(i-1) * 0.05 +
> ...
>
> = sum(i:1,infinity)[i*0.05*(0.95)^(i-1)]
> = 0.05 * sum(i:1,infinity)[i*(0.95)^(i-1)]
>
> Let's compute the sum
> sum(i:1,infinity)[i*x^(i-1)] =
> = sum(i:1,infinity)[d(x^i)/dx] =
> = d[sum(i:1,infinity)(x^i)]/dx =
> = d[x/1-x]/dx =
> = 1/(1-x)^2
>
> So for x = 0.95 we have
> sum(i:1,infinity)[i*(0.95)^(i-1)] = 1/(0.05)^2 = 400
> and so
> expected value =
> = 0.05 * sum(i:1,infinity)[i*(0.95)^(i-1)] =
> = 0.05 * 400 =
> = 20
>
>
>
> __________________________________________________
> Do You Yahoo!?
> Yahoo! Health - Feel better, live better
> http://health.yahoo.com
>

__________________________________________________
Do You Yahoo!?
Yahoo! Health - Feel better, live better
http://health.yahoo.com
• Intuitively I would say: Probability 0.05 means that among 100 products there will be on average 5 wrong. So among 20 products there will be on average 1
Message 4 of 6 , Aug 8, 2002
Intuitively I would say:
Probability 0.05 means that among 100 products there
will be on average 5 wrong. So among 20 products there
will be on average 1 wrong. I think you made your
equation in this way.

But I don't know how to say that in this 'among 20 we
expect 1' means 19 will be ok, and 20-th will be
wrong.

Also, how to show that this is what we really want -
expected value (which has its definition).

Any comments on this are appreciated.

--- The Webmaster <maintainer_wiz@...> wrote:
> Couldn't we have also simply said 1/.05 = 20, and
> gotten the answer this way?
>
> I loooked at the problem
> as such: 1/.05 = x/1, and then solved for x. I did
> appreciate your approach and maybe it holds for more
> complicated situations where a shortcut is not
> possible.
>
> Regards,
> M.W.
>
>
> --- Beata Stehlikova <beata_stehlikova@...>
> wrote:
> > > 1.Distinguish between statistics of a sample and
> > > parameters of a
> > > distribution.
> >
> > Statistics of the sample is an estimate of the
> > parameter of distribution.
> >
> > >
> > > 2.An industrial process manufactures components
> > with
> > > a 5% defective
> > > rate. Each day products are inspected
> sequentially
> > > until the first
> > > defective is obtained. Let X be the number of
> > > inspections carried out
> > > until the first defective is found.
> > >
> > >
> > > (a) If 5 products have been inspected without
> > > finding a defective,
> > > what is the probability that the next 5
> > inspections
> > > will be free of
> > > defectives?
> > >
> >
> > I think that the production of defectives is
> > independent, so knowing about the quality of some
> > products doesn't say anything about the following.
> > So
> > the probability is (0.95)^5.
> >
> > > (b) Derrive the expected number of inspections
> > until
> > > the first
> > > defective.
> >
> >
> > (b) Derrive the expected number of inspections
> until
> > the first
> > defective.
> >
> > Number of inspections / probability
> > 1 / 0.05
> > 2 / 0.95 * 0.05
> > 3 / (0.95)^2 * 0.05
> > 4 / (0.95)^3 * 0.05
> > ...
> > i / (0.95)^(i-1) * 0.05
> >
> > Expected value =
> > 1 * 0.05 +
> > 2 * 0.95 * 0.05 +
> > 3 * (0.95)^2 * 0.05 +
> > ...
> > i * (0.95)^(i-1) * 0.05 +
> > ...
> >
> > = sum(i:1,infinity)[i*0.05*(0.95)^(i-1)]
> > = 0.05 * sum(i:1,infinity)[i*(0.95)^(i-1)]
> >
> > Let's compute the sum
> > sum(i:1,infinity)[i*x^(i-1)] =
> > = sum(i:1,infinity)[d(x^i)/dx] =
> > = d[sum(i:1,infinity)(x^i)]/dx =
> > = d[x/1-x]/dx =
> > = 1/(1-x)^2
> >
> > So for x = 0.95 we have
> > sum(i:1,infinity)[i*(0.95)^(i-1)] = 1/(0.05)^2 =
> 400
> > and so
> > expected value =
> > = 0.05 * sum(i:1,infinity)[i*(0.95)^(i-1)] =
> > = 0.05 * 400 =
> > = 20
> >
> >
> >
> > __________________________________________________
> > Do You Yahoo!?
> > Yahoo! Health - Feel better, live better
> > http://health.yahoo.com
> >
>
>
> __________________________________________________
> Do You Yahoo!?
> Yahoo! Health - Feel better, live better
> http://health.yahoo.com
>

__________________________________________________
Do You Yahoo!?
HotJobs - Search Thousands of New Jobs
http://www.hotjobs.com
• ... I agree. It s not at all clear without a derivation. Intuitively, if one out of twenty will be defective, we ought to expect the defect to occur somewhere
Message 5 of 6 , Aug 14, 2002
--- In probability@y..., Beata Stehlikova <beata_stehlikova@y...>
wrote:
> Intuitively I would say:
> Probability 0.05 means that among 100 products there
> will be on average 5 wrong. So among 20 products there
> will be on average 1 wrong. I think you made your
> equation in this way.
>
> But I don't know how to say that in this 'among 20 we
> expect 1' means 19 will be ok, and 20-th will be
> wrong.

I agree. It's not at all clear without a derivation. Intuitively, if
one out of twenty will be defective, we ought to expect the defect to
occur somewhere around the tenth observation.

The fact that the "tail" (the very large waiting times with very
small probability) is compensating things in just the right way so as
to make the expected value 20 is not at all obvious.

>
> Also, how to show that this is what we really want -
> expected value (which has its definition).
>
> Any comments on this are appreciated.
>
>
> --- The Webmaster <maintainer_wiz@y...> wrote:
> > Couldn't we have also simply said 1/.05 = 20, and
> > gotten the answer this way?
> >
> > I loooked at the problem
> > as such: 1/.05 = x/1, and then solved for x. I did
> > appreciate your approach and maybe it holds for more
> > complicated situations where a shortcut is not
> > possible.
> >
> > Regards,
> > M.W.
> >
> >
> > --- Beata Stehlikova <beata_stehlikova@y...>
> > wrote:
> > > > 1.Distinguish between statistics of a sample and
> > > > parameters of a
> > > > distribution.
> > >
> > > Statistics of the sample is an estimate of the
> > > parameter of distribution.
> > >
> > > >
> > > > 2.An industrial process manufactures components
> > > with
> > > > a 5% defective
> > > > rate. Each day products are inspected
> > sequentially
> > > > until the first
> > > > defective is obtained. Let X be the number of
> > > > inspections carried out
> > > > until the first defective is found.
> > > >
> > > >
> > > > (a) If 5 products have been inspected without
> > > > finding a defective,
> > > > what is the probability that the next 5
> > > inspections
> > > > will be free of
> > > > defectives?
> > > >
> > >
> > > I think that the production of defectives is
> > > independent, so knowing about the quality of some
> > > products doesn't say anything about the following.
> > > So
> > > the probability is (0.95)^5.
> > >
> > > > (b) Derrive the expected number of inspections
> > > until
> > > > the first
> > > > defective.
> > >
> > >
> > > (b) Derrive the expected number of inspections
> > until
> > > the first
> > > defective.
> > >
> > > Number of inspections / probability
> > > 1 / 0.05
> > > 2 / 0.95 * 0.05
> > > 3 / (0.95)^2 * 0.05
> > > 4 / (0.95)^3 * 0.05
> > > ...
> > > i / (0.95)^(i-1) * 0.05
> > >
> > > Expected value =
> > > 1 * 0.05 +
> > > 2 * 0.95 * 0.05 +
> > > 3 * (0.95)^2 * 0.05 +
> > > ...
> > > i * (0.95)^(i-1) * 0.05 +
> > > ...
> > >
> > > = sum(i:1,infinity)[i*0.05*(0.95)^(i-1)]
> > > = 0.05 * sum(i:1,infinity)[i*(0.95)^(i-1)]
> > >
> > > Let's compute the sum
> > > sum(i:1,infinity)[i*x^(i-1)] =
> > > = sum(i:1,infinity)[d(x^i)/dx] =
> > > = d[sum(i:1,infinity)(x^i)]/dx =
> > > = d[x/1-x]/dx =
> > > = 1/(1-x)^2
> > >
> > > So for x = 0.95 we have
> > > sum(i:1,infinity)[i*(0.95)^(i-1)] = 1/(0.05)^2 =
> > 400
> > > and so
> > > expected value =
> > > = 0.05 * sum(i:1,infinity)[i*(0.95)^(i-1)] =
> > > = 0.05 * 400 =
> > > = 20
> > >
> > >
> > >
> > > __________________________________________________
> > > Do You Yahoo!?
> > > Yahoo! Health - Feel better, live better
> > > http://health.yahoo.com
> > >
> >
> >
> > __________________________________________________
> > Do You Yahoo!?
> > Yahoo! Health - Feel better, live better
> > http://health.yahoo.com
> >
>
>
> __________________________________________________
> Do You Yahoo!?
> HotJobs - Search Thousands of New Jobs
> http://www.hotjobs.com
• I looked at it this way. If Prob(x is defective) = .05, then on average each observation has a 5% defective part associated with it. Of course, we don t look
Message 6 of 6 , Aug 15, 2002
I looked at it this way. If Prob(x is defective) =
.05, then on average each observation has a 5%
defective part associated with it. Of course, we
don't look at it this way and we want to know when a
"whole" observation will be defective. So, when might
I get one whole defective unit? I said take 1 and
divide by .05 to get 20, so that on average, I should
expect to see a wholey defective unit on the twentieth
observation.

BTW, can anyone give me a counterexample where this
approach is not correct, or another problem of the
exact same nature where my back-of-the-envelope didn't
work? Remember, we aren't using conditional
probability here.

Regards,
M.W.

--- jason1990 <jason1990@...> wrote:
> --- In probability@y..., Beata Stehlikova
> <beata_stehlikova@y...>
> wrote:
> > Intuitively I would say:
> > Probability 0.05 means that among 100 products
> there
> > will be on average 5 wrong. So among 20 products
> there
> > will be on average 1 wrong. I think you made your
> > equation in this way.
> >
> > But I don't know how to say that in this 'among 20
> we
> > expect 1' means 19 will be ok, and 20-th will be
> > wrong.
>
> I agree. It's not at all clear without a derivation.
> Intuitively, if
> one out of twenty will be defective, we ought to
> expect the defect to
> occur somewhere around the tenth observation.
>
> The fact that the "tail" (the very large waiting
> times with very
> small probability) is compensating things in just
> the right way so as
> to make the expected value 20 is not at all obvious.
>
> >
> > Also, how to show that this is what we really want
> -
> > expected value (which has its definition).
> >
> > Any comments on this are appreciated.
> >
> >
> > --- The Webmaster <maintainer_wiz@y...> wrote:
> > > Couldn't we have also simply said 1/.05 = 20,
> and
> > > gotten the answer this way?
> > >
> > > I loooked at the problem
> > > as such: 1/.05 = x/1, and then solved for x. I
> did
> > > appreciate your approach and maybe it holds for
> more
> > > complicated situations where a shortcut is not
> > > possible.
> > >
> > > Regards,
> > > M.W.
> > >
> > >
> > > --- Beata Stehlikova <beata_stehlikova@y...>
> > > wrote:
> > > > > 1.Distinguish between statistics of a sample
> and
> > > > > parameters of a
> > > > > distribution.
> > > >
> > > > Statistics of the sample is an estimate of the
> > > > parameter of distribution.
> > > >
> > > > >
> > > > > 2.An industrial process manufactures
> components
> > > > with
> > > > > a 5% defective
> > > > > rate. Each day products are inspected
> > > sequentially
> > > > > until the first
> > > > > defective is obtained. Let X be the number
> of
> > > > > inspections carried out
> > > > > until the first defective is found.
> > > > >
> > > > >
> > > > > (a) If 5 products have been inspected
> without
> > > > > finding a defective,
> > > > > what is the probability that the next 5
> > > > inspections
> > > > > will be free of
> > > > > defectives?
> > > > >
> > > >
> > > > I think that the production of defectives is
> > > > independent, so knowing about the quality of
> some
> > > > products doesn't say anything about the
> following.
> > > > So
> > > > the probability is (0.95)^5.
> > > >
> > > > > (b) Derrive the expected number of
> inspections
> > > > until
> > > > > the first
> > > > > defective.
> > > >
> > > >
> > > > (b) Derrive the expected number of inspections
> > > until
> > > > the first
> > > > defective.
> > > >
> > > > Number of inspections / probability
> > > > 1 / 0.05
> > > > 2 / 0.95 * 0.05
> > > > 3 / (0.95)^2 * 0.05
> > > > 4 / (0.95)^3 * 0.05
> > > > ...
> > > > i / (0.95)^(i-1) * 0.05
> > > >
> > > > Expected value =
> > > > 1 * 0.05 +
> > > > 2 * 0.95 * 0.05 +
> > > > 3 * (0.95)^2 * 0.05 +
> > > > ...
> > > > i * (0.95)^(i-1) * 0.05 +
> > > > ...
> > > >
> > > > = sum(i:1,infinity)[i*0.05*(0.95)^(i-1)]
> > > > = 0.05 * sum(i:1,infinity)[i*(0.95)^(i-1)]
> > > >
> > > > Let's compute the sum
> > > > sum(i:1,infinity)[i*x^(i-1)] =
> > > > = sum(i:1,infinity)[d(x^i)/dx] =
> > > > = d[sum(i:1,infinity)(x^i)]/dx =
> > > > = d[x/1-x]/dx =
> > > > = 1/(1-x)^2
> > > >
> > > > So for x = 0.95 we have
> > > > sum(i:1,infinity)[i*(0.95)^(i-1)] = 1/(0.05)^2
> =
> > > 400
> > > > and so
> > > > expected value =
> > > > = 0.05 * sum(i:1,infinity)[i*(0.95)^(i-1)] =
> > > > = 0.05 * 400 =
> > > > = 20
> > > >
> > > >
> > > >
> > > >
> __________________________________________________
> > > > Do You Yahoo!?
> > > > Yahoo! Health - Feel better, live better
> > > > http://health.yahoo.com
> > > >
> > >
> > >
> > >
> __________________________________________________
> > > Do You Yahoo!?
> > > Yahoo! Health - Feel better, live better
> > > http://health.yahoo.com
> > >
> >
> >
> > __________________________________________________
> > Do You Yahoo!?
> > HotJobs - Search Thousands of New Jobs
> > http://www.hotjobs.com
>
>

__________________________________________________
Do You Yahoo!?
HotJobs - Search Thousands of New Jobs
http://www.hotjobs.com
Your message has been successfully submitted and would be delivered to recipients shortly.
»
«