basic stats question

2001-02-26 Thread James Ankeny

 Hello, 
   I have a question regarding basic probability and statistics. If I
understand correctly, the definition of independence holds for two events
that are subsets of the same sample space. In other cases, we may need to
construct a new sample space, such as with the flipping of a coin twice.
Here, we construct the new sample space S=S1xS2={HH,HT,TH,TT}, where
Si={H,T} for i=1,2. This way, two events that are independent, such as
A="head on first toss" and B="tails on second toss," are subsets of the same
sample space. 
   Now, the problem that I have is that, while it is not difficult to
construct sample spaces intuitively for textbook problems, it is difficult
to do so using these basic definitions of probability. For ex., consider a
problem where a manufacturer has five seemingly identical computers, though
two are really defective and three are good. An order calls for two of the
computers, and we want the probability of the event A="order is filled with
two good computers."
Intuitively, it is obvious that if D1 and D2 are the bad computers, and
G1-G3 are the good computers, then
S={D1D2,D1G1,D1G2,D1G3,D2G1,D2G2,D2G3,G1G2,G1G3,G2G3}. Thus, P(A)= 0.30.
However, I cannot think of any way of constructing the sample space using
definitions like the cartesian product. Perhaps this is because the second
computer chosen depends on which computer is chosen first. Yet, another
similar problem in my textbook states that the probabilities of a computer
being good and defective (from a particular manufacturer) are 0.90 and 0.10,
respectively. Then, if we want to test five computers, we may construct the
sample space S=S1xS2xS3xS4xS5, where Si={G,D} for i=1,...,5. Hence, if
A="all five computers tested are good," P(A)=(0.90)^5. Why is that we can
use the Cartesian product in this case but not in the other case? Is it that
in the first case we are not performing an experiment, but just sampling?
Perhaps I am thinking about this too much, but it would be nice to be able
to construct these sample spaces for problems using some sort of formulaic
method, as opposed to intuition (perhaps this isn't the right way to view
this subject?). Any help would be greatly appreciated. 





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Re: basic stats question

2001-02-26 Thread Donald Burrill

Perhaps jthis is too superficial -- no time to think more deeply just 
now.  But I suspect the difference between your two scenarios below is 
that with exactly 5 computers to deal with (i.e., population size = 5) 
you are sampling without replacement (which is only sensible, for the 
background scenario!);  whereas with the textbook problem you are 
assuming that the probabilities do not change (and in any case they 
aren't the probabilities that correspond to your N=5 situation!), which 
is equivalent to sampling _with_ replacement (or, what is much the same 
thing, assuming the number of entities available to sample from is 
infinite -- which is probably _not_ sensible for any real-life 
scenario!). 
-- Don.

On Mon, 26 Feb 2001, James Ankeny wrote in part:

> ... consider a problem where a manufacturer has five seemingly 
> identical computers, though two are really defective and three are 
> good.  ... we want the probability of the event A="order is filled with 
> two good computers." ... then
> S={D1D2,D1G1,D1G2,D1G3,D2G1,D2G2,D2G3,G1G2,G1G3,G2G3}. Thus, P(A)= 0.30.

<  snip  >

> ...  Yet, another similar problem in my textbook states that the 
> probabilities of a computer being good and defective (from a particular 
> manufacturer) are 0.90 and 0.10, respectively.  Then, if we want to test 
> five computers, we may construct the sample space S=S1xS2xS3xS4xS5, 
> where Si={G,D} for i=1,...,5. Hence, if A="all five computers tested are 
> good," P(A)=(0.90)^5.  Why is that we can use the Cartesian product in 
> this case but not in the other case? 

 --
 Donald F. Burrill[EMAIL PROTECTED]
 348 Hyde Hall, Plymouth State College,  [EMAIL PROTECTED]
 MSC #29, Plymouth, NH 03264 (603) 535-2597
 Department of Mathematics, Boston University[EMAIL PROTECTED]
 111 Cummington Street, room 261, Boston, MA 02215   (617) 353-5288
 184 Nashua Road, Bedford, NH 03110  (603) 471-7128



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