Re: Probablity questions
- From: "Nick" <tulse04-news1@xxxxxxxxxxx>
- Date: Wed, 15 Nov 2006 14:30:45 -0000
<C6L1V@xxxxxxx> wrote in message
news:1163562714.438722.53770@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Taria wrote:
Neat! I like having good different examples to practice on. It's
rumored that the more practice you get the better you get at it...I'm
still waiting for that. :) I tried making up my own word problems and
solving those but the problem i run into is sometimes I can't solve it.
The three 3's problem was never clarrified to me whether it was an
exact three 3's or at least three 3's. So, let's assume at least three
3's. The answer I have for this problem is the summation for i=1 to n
[(n choose i)(1/10)^i(9/10)^(n-i) = 0.5 I individually did the
equation for each P(1),P(2),P(3) and then added them altogether.
P{number >= 3} = 1-P{number <= 2}, and P{number <= 2} = P(0)+P(1)+P(2).
You see, you need to include P(0) but exclude P(3)!
Thank you very much for your help on the Poisson distribution problem.
I do have a question about your answer that I don't rightly understand.
If P(4 | 3 colds) = NA/P(3 colds) then how does Bayes Theorem fit in?
Bayes thm says:
P(B|A)P(A)
----------------
P(A)
No, that is NOT what it says. It says that P(B|A) = P(A|B)*P(B)/P(A).
In the context of the "colds" problem, this would be
P(type-4|3 colds) =P(3 colds|type 4)*P(type4)/P(3 colds); furthermore,
the denominator is
P(3 colds) = P(3 colds|type-4)*P(type-4)+P(3 colds|type-5)*P(type-5).
This is exactly the same as the answer given in the previous posting
where you imagine a large population of people and then compute
relevant percentages.
sooo if I try to fit what I know into that formula, since P(B|A) = P(A
intersect B) then wouldn't my computation be different from the NA/P(3
colds).
Nope. If you use the _correct_ formula, you will get exactly the same
thing either way.
The numerator is what I"m having difficult understanding. I
understand P(A intersect B) =P(A) + P(B) * P(A|B)
If this is not a typo, then you understand wrongly. The correct result
is P(A intersect B) = P(A|B)*P(B) [also = P(B|A)*P(A) ]. There is no
"P(A)+" term in the correct expression.
for dependent
events. It seems the further I try to expand P(B|A) into something I
can compute, it either evolves into P(A|B) or P(B|A), it keeps on
popping up! Lol. I really thought I understood Bayes formula too, but
I guess not.
Well, to aid in this understanding was the reason for giving the
argument based on calculating percentages in a large population. Some
people find that kind of argument helpful at the beginning (others
don't) , before they feel comfortable with Bayes' Theorem.
I'll review this formula again so I can put the correct
numbers into it to get my answer.
Y'know I thought I was giong to major in math until this course. Lol,
by the looks of it, I don't think so! :)
Please don't let that be the only reason for abandoning math.
Probability is just one branch, and many people get degrees with almost
no probability at all. In my case, I graduated with honours mathematics
and physics, and nowhere in my undergraduate math studies did I take
any probability at all; I did, however, learn the basics of it in
physics courses. Of course, that was back in the Stone Age, so maybe
things are different now. You should investigate further the
requirements at your school before abandoning ship. Besides, the
probability stuff will---guaranteed---get easier as time passes,
especially if you work at it. Not only that, it really is very useful,
and might stand you in good stead later if you decide to do some
graduate work in a related field such as operations research, which
uses mathematics in important ways, but is not actually mathematics
itself.
I have a maths degree and then did a Postgraduate Diploma in Statistics.
Statistics is difficult but there are different levels of every subject.
One can use statistics without understanding every element of it.
The difficult things are often the things worth persevering at. If it was
easy it wouldn't be worth all the effort.
Nick
.
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- Probablity questions
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- Re: Probablity questions
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- Re: Probablity questions
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- Re: Probablity questions
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