Re: calculte probility
- From: matt271829-news@xxxxxxxxxxx
- Date: 14 Jul 2006 18:30:36 -0700
Bopar wrote:
matt271829-news@xxxxxxxxxxx wrote:
Bopar wrote:
i have 2 bag of n and m colored balls, can u tell me what is the
probility to pick upa 2 balls of the same color from bag 1 and bag 2 ?
Not unless you (i) specify how many balls of each colour are in each
bag, and (ii) explain the selection procedure (e.g. one ball is chosen
at random from each bag, or whatever it might be).
if i know there are X(=M+N) colors (N colors in bag 1, M in bag 2), any
better chances for me to pick up balls of the same color ?
What is the way to reason this ?
Is this supposed to be a new question or a clarification of the first
question? Whatever, it's impossible to answer without knowing how the
colours are distributed across the two bags, and how the selection is
to be made. Anyway, if there N (different) colours in bag 1, M
(different) colours in bag 2, and M + N (different) colours in total,
then there can't be any ball in bag 1 that is the same colour as any
ball in bag 2. So if you're choosing one ball at random from each bag
then the probability of getting two of the same colour is zero. I
assume this isn't what you meant.
Basically I think you need to explain the problem more clearly.
By the way, yes, it is my homework, I don't have any idea how to solve
it, thanks anyway
Thanks matt,
-----------------------
(1) The colors are repeated, and also there are colors that one bag
does not have but the other does. Picker picks ball up at random with
replacement-return it back to the bag. That's for the first question
You still haven't properly explained the selection procedure. You
haven't said how many balls are selected, and from which bags. The
obvious guess is that one ball is selected randomly from each bag (two
balls in total), and you want the probability that both balls are the
same colour. But since you now specify "with replacement" it seems that
more than one ball is selected from at least one bag, so I am confused.
Write an explanation of the procedure, then read what you have written,
and ask yourself if someone could carry out the procedure using only
your explanation and without having to guess at anything.
Assuming that the selection procedure is explained, this still leaves
the question of the distribution of colours - in other words, how many
balls of each colour each of the bags contains. The answer will depend
on this distribution of colours. I am not clear if you are expecting a
specific answer (in which case you haven't given enough information for
that to be possible), or if you are asking for a general procedure that
will work for any arbitrary distribution of colours.
-----------------------
(2) The second question is applied to a situation when the number of
colors is known, X or N and M for each bag in this case.
supposing that bag1=color1+color2+....+colorN
bag2=color5+color2+color4+color1+....colorM
color difference =|M-N|.
That's what I meant for the second problem
------------------------
I really don't know how to solve this problem. thanks again
.
- References:
- calculte probility
- From: Bopar
- Re: calculte probility
- From: matt271829-news
- Re: calculte probility
- From: Bopar
- calculte probility
- Prev by Date: Re: Easier solution to wiles´ one to the fermat congruence
- Next by Date: Re: Attempts to Refute Cantor's Uncountability Proof?
- Previous by thread: Re: calculte probility
- Next by thread: Re: calculte probility
- Index(es):
Relevant Pages
|
