 |  | | Page 3 - OT: I hate it when I'm right.... Discuss OT: I hate it when I'm right..., on Health Forums.
| | 
03-24-2007, 07:20 PM
| | | Re: OT: I hate it when I'm right... spodosaurus <spodosaurus@_yahoo_.com.au> wrote:
> DZ wrote:
>> spodosaurus <spodosaurus@_yahoo_.com.au> wrote:
>>> JMW wrote:
>>>> BTW, when I say it's crap shoot, if Menu Foods produces over a
>>>> billion cans a year, there's a lot higher a probability of them
>>>> turning out some bad cans than the little "premium" companies who
>>>> probably turn out a few hundred thousand.
>>> How do you figure? Given the same probability, they will simply turn
>>> out more bad cans due to their larger production. Why would the
>>> probability change?
>>
>> I read "probability of turning out some bad cans" as meaning that
>> "some" is greater than zero. So, this behaves similarly to the
>> prob. of turning out at least one bad can, which increases with the
>> number of cans produced.
>
> But why does the probability increase? Just off hand, in my very sleepy
> mind, it seems like probability would be constant a 1% probability for
> 1000 cans may equal ten bad 10 cans, and a 1% probability in 1,000,000
> cans may equal 10,000 bad cans. The probability remains constant, but
> the number of instances may increase with increased production.
The prob. of a bad can (1%) is the same. The prob. of "some bad cans"
is the "prob of at least one bad can". In your examples you take it to
be 1 (and it would be close to 1 given the numbers you assume), so
only the number of bad cans changes.
If you produce cans, the prob of a bad can is 0.01.
If you produce 1 can a year, the prob of at least 1 bad
can in a given year is 0.01.
If you produce 100 cans a year, it is 1-(1-0.01)^100 = 64%.
If you produce 1000 cans a year, it becomes almost a certainty,
1-(1-0.01)^1000 = 99.996%.
In reality, the prob. of a bad can (the one that would kill a pet) is
likely much smaller than 1%, so "producing some bad cans" is not a
certainty for any manufacturer.
So, I guess the point was that if the quality is fairly good overall,
a manufacturer that produces lots of goods is more likely to get
flagged for an occasional bad item. | 
03-24-2007, 07:20 PM
| | | Re: OT: I hate it when I'm right... DZ wrote:
> spodosaurus <spodosaurus@_yahoo_.com.au> wrote:
>> DZ wrote:
>>> spodosaurus <spodosaurus@_yahoo_.com.au> wrote:
>>>> JMW wrote:
>>>>> BTW, when I say it's crap shoot, if Menu Foods produces over a
>>>>> billion cans a year, there's a lot higher a probability of them
>>>>> turning out some bad cans than the little "premium" companies who
>>>>> probably turn out a few hundred thousand.
>>>> How do you figure? Given the same probability, they will simply turn
>>>> out more bad cans due to their larger production. Why would the
>>>> probability change?
>>> I read "probability of turning out some bad cans" as meaning that
>>> "some" is greater than zero. So, this behaves similarly to the
>>> prob. of turning out at least one bad can, which increases with the
>>> number of cans produced.
>> But why does the probability increase? Just off hand, in my very sleepy
>> mind, it seems like probability would be constant a 1% probability for
>> 1000 cans may equal ten bad 10 cans, and a 1% probability in 1,000,000
>> cans may equal 10,000 bad cans. The probability remains constant, but
>> the number of instances may increase with increased production.
>
> The prob. of a bad can (1%) is the same. The prob. of "some bad cans"
> is the "prob of at least one bad can". In your examples you take it to
> be 1 (and it would be close to 1 given the numbers you assume), so
> only the number of bad cans changes.
>
> If you produce cans, the prob of a bad can is 0.01.
> If you produce 1 can a year, the prob of at least 1 bad
> can in a given year is 0.01.
> If you produce 100 cans a year, it is 1-(1-0.01)^100 = 64%.
> If you produce 1000 cans a year, it becomes almost a certainty,
> 1-(1-0.01)^1000 = 99.996%.
>
Unless the probability for each bad can is independent of all other cans
produced. Thus, the probability of any individual can being bad would
still be 0.01. Only if we assume that the probability is additive would
the percentage of getting at least one bad can rise....right? (still
exhausted, want to sleep, but baby is sleeping on me at the moment)
> In reality, the prob. of a bad can (the one that would kill a pet) is
> likely much smaller than 1%, so "producing some bad cans" is not a
> certainty for any manufacturer.
>
Unless rat poison is introduced...isn't that what turns out really happened?
> So, I guess the point was that if the quality is fairly good overall,
> a manufacturer that produces lots of goods is more likely to get
> flagged for an occasional bad item.
--
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Many people around the world are waiting for a marrow transplant. Please
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03-25-2007, 12:48 AM
| | | Re: OT: I hate it when I'm right... spodosaurus <spodosaurus@_yahoo_.com.au> wrote:
> DZ wrote:
>> spodosaurus <spodosaurus@_yahoo_.com.au> wrote:
>>> DZ wrote:
>>>> spodosaurus <spodosaurus@_yahoo_.com.au> wrote:
>>>>> JMW wrote:
>>>>>> BTW, when I say it's crap shoot, if Menu Foods produces over a
>>>>>> billion cans a year, there's a lot higher a probability of them
>>>>>> turning out some bad cans than the little "premium" companies who
>>>>>> probably turn out a few hundred thousand.
>>>>> How do you figure? Given the same probability, they will simply turn
>>>>> out more bad cans due to their larger production. Why would the
>>>>> probability change?
>>>> I read "probability of turning out some bad cans" as meaning that
>>>> "some" is greater than zero. So, this behaves similarly to the
>>>> prob. of turning out at least one bad can, which increases with the
>>>> number of cans produced.
>>> But why does the probability increase? Just off hand, in my very sleepy
>>> mind, it seems like probability would be constant a 1% probability for
>>> 1000 cans may equal ten bad 10 cans, and a 1% probability in 1,000,000
>>> cans may equal 10,000 bad cans. The probability remains constant, but
>>> the number of instances may increase with increased production.
>>
>> The prob. of a bad can (1%) is the same. The prob. of "some bad cans"
>> is the "prob of at least one bad can". In your examples you take it to
>> be 1 (and it would be close to 1 given the numbers you assume), so
>> only the number of bad cans changes.
>>
>> If you produce cans, the prob of a bad can is 0.01.
>> If you produce 1 can a year, the prob of at least 1 bad
>> can in a given year is 0.01.
>> If you produce 100 cans a year, it is 1-(1-0.01)^100 = 64%.
>> If you produce 1000 cans a year, it becomes almost a certainty,
>> 1-(1-0.01)^1000 = 99.996%.
>
> Unless the probability for each bad can is independent of all other cans
> produced. Thus, the probability of any individual can being bad would
> still be 0.01. Only if we assume that the probability is additive would
> the percentage of getting at least one bad can rise....right? (still
> exhausted, want to sleep, but baby is sleeping on me at the moment)
The above formula does assume independence. If there is dependency,
one can either simply redefine the problem and talk about "some
batches of bad cans", instead of "some cans", or deal with the
dependency. But since the dependency is going to be positive and local
(cans produced close to each other in time are more likely to be both
bad), the effect of this is a reduction of K (the actual number of
cans produced) in 1-(1-p)^K to Ke less than K which can be termed "an
effective number of cans". The 1-(1-p)^Ke is an approximation. But if
you flip a coin long enough, you're going to get a tail - even if
flipping results are somehow dependent.
Whenever there is a seeming additivity in probabilities p1 and p2, a
more likely result is something like p1+p2 - p1*p2.
That's what happens above too, e.g. 1 - (1 - p)^2 = p+p - p*p. | 
03-25-2007, 12:48 AM
| | | Re: OT: I hate it when I'm right... DZ <10505@138322757.2474316411.11078.19175.30444> wrote:
>spodosaurus <spodosaurus@_yahoo_.com.au> wrote:
>> DZ wrote:
>>> spodosaurus <spodosaurus@_yahoo_.com.au> wrote:
>>>> JMW wrote:
>>>>> BTW, when I say it's crap shoot, if Menu Foods produces over a
>>>>> billion cans a year, there's a lot higher a probability of them
>>>>> turning out some bad cans than the little "premium" companies who
>>>>> probably turn out a few hundred thousand.
>>>> How do you figure? Given the same probability, they will simply turn
>>>> out more bad cans due to their larger production. Why would the
>>>> probability change?
>>>
>>> I read "probability of turning out some bad cans" as meaning that
>>> "some" is greater than zero. So, this behaves similarly to the
>>> prob. of turning out at least one bad can, which increases with the
>>> number of cans produced.
>>
>> But why does the probability increase? Just off hand, in my very sleepy
>> mind, it seems like probability would be constant a 1% probability for
>> 1000 cans may equal ten bad 10 cans, and a 1% probability in 1,000,000
>> cans may equal 10,000 bad cans. The probability remains constant, but
>> the number of instances may increase with increased production.
>
>The prob. of a bad can (1%) is the same. The prob. of "some bad cans"
>is the "prob of at least one bad can". In your examples you take it to
>be 1 (and it would be close to 1 given the numbers you assume), so
>only the number of bad cans changes.
>
>If you produce cans, the prob of a bad can is 0.01.
>If you produce 1 can a year, the prob of at least 1 bad
>can in a given year is 0.01.
>If you produce 100 cans a year, it is 1-(1-0.01)^100 = 64%.
>If you produce 1000 cans a year, it becomes almost a certainty,
>1-(1-0.01)^1000 = 99.996%.
>
>In reality, the prob. of a bad can (the one that would kill a pet) is
>likely much smaller than 1%, so "producing some bad cans" is not a
>certainty for any manufacturer.
>
>So, I guess the point was that if the quality is fairly good overall,
>a manufacturer that produces lots of goods is more likely to get
>flagged for an occasional bad item.
That was my point. Or in the alternative, a company that produces a
lot of goods has a lot of employees. They are more likely to get an
occasional disgruntled one. Who poisons goods. With aminopterin. | 
03-25-2007, 04:14 PM
| | | Re: OT: I hate it when I'm right...
"JMW" <jmwilliams@enforcergraphics.f2s.com> wrote in message
news:mg1b031t9m397lbh5qtm3jn71k6q9adrng@4ax.com...
> DZ <10505@138322757.2474316411.11078.19175.30444> wrote:
>>So, I guess the point was that if the quality is fairly good overall,
>>a manufacturer that produces lots of goods is more likely to get
>>flagged for an occasional bad item.
>
> That was my point. Or in the alternative, a company that produces a
> lot of goods has a lot of employees. They are more likely to get an
> occasional disgruntled one. Who poisons goods. With aminopterin.
Since it is reportedly only available in the US and Canada as an anti-cancer
drug and not as a rodenticide, that seems unlikely. More likely that "Menu
Foods Income Fund" sources the cheapest possible wheat gluten from China,
without an adequate investigation. | 
03-25-2007, 04:14 PM
| | | Re: OT: I hate it when I'm right... John M. Williams wrote:
[...]
> That was my point.
Riiiiiiiiiiiiiiiiight. THAT was your point.
> Or in the alternative, a company that
> produces a lot of goods has a lot of
> employees. They are more likely to
> get an occasional disgruntled one.
> Who poisons goods. With aminopterin.
So, a pet food manufacturer has employees who don't simply key the
boss's car, but instead take on the guise of evil secret agent by
engaging in deliberate poisoning of the company's product?
RIIIIIIIIIIIIIIGGGGHHHHHHHHHT!!!!!!
Stfu, Em.
I'm as much of a conspiracy theorist as the next guy, but I doubt
you're correct on this one. Were you serious? http://en.wikipedia.org/wiki/Menu_Foods#Recall
--
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