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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