"John Knight" <jwknight at polbox.com> wrote in message
news:1iE89.56385$eb.4167626 at news2.west.cox.net...
> "Bob LeChevalier" <lojbab at lojban.org> wrote in message
> news:j1r5mu07emjtsagjd4c7sigpje9c5id87l at 4ax.com...> > "John Knight" <jwknight at polbox.com> wrote:
> > >"mat" <mats_trash at hotmail.com> wrote in message
> > >news:43525ce3.0208200346.17c4045f at posting.google.com...> > >> mats_trash at hotmail.com (mat) wrote in message
> > >news:<43525ce3.0208031252.370db1e5 at posting.google.com>...
> > >> > No it won't you fool. Answer we this, and don't conveniently avoid
> it
> > >> > like the other difficult parts of my last post:
> > >> >
> > >> > If a coin is flipped twice am I certain to get one head and one
tail
> > >> > (as would be the case according to your logic.
> > >> >
> > >> > If I flip a coin three times, what is the probability of getting at
> > >> > least one head?
> > >> >
> > >> > Just answer those questions and we know where we stand.
Furthermore,
> > >> > if you don't answer you will confirm your inability to comprehend
> > >> > basic mathematics.
> > >>
> > >> Surely not that hard?
> > >
> > >Not only was your question answered a long time ago,
> >
> > Nope.
> >
> > >The probability of getting one head on the first flip is 0.5. The
> > >probability of getting two heads in a row are 0.5 x 0.5, or 0.25. Of
> > >getting three in a row is 0.125, etc.
> >
> > That is not the right question. .125 is the chance of getting THREE
> > heads in a row on three coin tosses. The poster asked you the
> > probability of getting AT LEAST ONE head in three coin tosses. You
> > don't know the answer, do you?
> >
> > The correct answer is .875. Now given the answer, can you tell us WHY
> > that is the correct answer? I've done half the problem for you, and
> > I'll bet you STILL can't solve it.
>> This is TOO funny. Using binomial probability to determine the
probability
> of thousands of students getting ONE correct answer out of 4 multiple
> choices is a far different probability problem than using binomial
> probability to determine "AT LEAST ONE head in three coin tosses".
You can't do either. In fact not only can't you do binomical probablity
you can't
even do common sense. The reason that the probabiltiy of getting at least
one
head is .875 is becasue the probability of getting three tails is .125 and
if you don't
get three tails you get at least one head, therefore the probability is 1 -
.125 =.875.
Thank you Bob for allowinng me to demonstrate my brilliance for my next
trick
I will demonstrate that I passed 6th grade not just attended it.
> >
> > >But that's an entirely different question than the probability of
> thousands
> > >of students (randomly guessing on a four part multiple choice question)
> > >getting the correct answer.
> >
> > You guessed wrong on his question. The probability that you would do
> > so was nearly 1.000, but that has little to do with random selection.
> >
>> Nobody "guessed wrong on *his* question".
The answer is .875 and you said .125. Sounds wrong to me.
> The question is irrelevant to the
> original POINT.
The point is that you don't understand math.
>> And the following suggests that *his* tactic worked--you're both spinning
> around in left field, once again.
>> > >No matter how you slice it, if the guesses are truly random, the larger
> the
> > >sample size, the closer you'll be to 25%.
> >
> > But they aren't random any more than your wrong answer to his problem
> > was random. You misunderstood the question. This makes you no better
> > and probably worse than most of those female students you like to
> > insult.
> >
>> Nobody misunderstood *his* question. Nobody denies you can use binomial
> probability to calculate the answer to his question, and nobody denies
that
> you (amazingly enough) actually calculated it correctly.
>> But *he* is supposed to be answering a *completely* different question.
> This has utterly nothing to do with that question. This distraction from
> the original question accomplishes nothing. The question is, ASSUMING
> random guesses by thousands of students, what is the distribution of the
> responses going to be on a four part multiple choice question. Since
nobody
> seems to *want to* [read: is able to] answer that question, here is the
> ANSWER:
>> 25% plus or minus 2% will select A)
>> 25% plus or minus 2% will select B)
>> 25% plus or minus 2% will select C)
>> 25% plus or minus 2% will select D)
>> No matter how much you confuse yourself with all these other irrelevant
> points, there's no other possible distribution of the responses if they
were
> completely random.
Wrong. The standard deviation for the number of people answering A given
random responses, 4 choices and 1000 respondents is sqr(.75*.25*1000) =
13.7. There is at least a 7% chance of more than 27% of respondents
choosing A.
So 25% +/- 2% is not the only possible distribution.
>> Right?
>> John Knight
>>
