Re: Probabilities and complex numbers.
- From: quasi <quasi@xxxxxxxx>
- Date: Fri, 15 Jul 2005 11:06:11 -0500
"novelo" <novelo@xxxxxxxxx> wrote in news:1120806462.919092.187420
@z14g2000cwz.googlegroups.com:
> I've never seen probability at school and if someone can help me,
> you'll make my day for today! The problem is the following.
>
> A box contains 2 red balls and 3 white balls. A trail is to take one
> ball at random out of the box. Repeat this trial 4 times, without
> putting balls back to the box that have previously been taken out.
>
> 1) The probability for which two red balls are included in the set of 4
> balls taken out is = ?
> 2) Subject to the condition that the ball taken out in the first trial
> is a white ball, the "conditional probability" for which 2 red balls
> are included in the ser of 4 balls taken out is?
>
>
> If someone was able to explain me what is a conditinal probability,
> that will be very helpfull.
>
> ...
A very concrete way to understand it is with a tree diagram showing all
possible ordered drawings of balls. At each stage, there are (potentially)
2 possible outcomes for the ball to be drawn, either R or W, so the tree
has at most 2 branches at each node. The branches are (in general) not
equally likely, but each branch is easily assigned a probability based on
the ratio of the number of balls remaining of that color to the total
number of balls remaining. So for example, at the root node (first ball not
yet drawn), there are 2 branches, labelled R,W, with probability 2/5
assigned to the R branch and 3/5 to the W branch. Each of the 2 new nodes
also has 2 branches but the probabilities for these branches depends on the
new ratios at the node being branched from. After the tree is fully drawn,
the outcomes of interest are the complete paths which satisfy the specified
condition (exactly 2 red balls). For each path, the probability of that
path occurring is the product of the probabilities of the branches on that
path. The probability for the event in question is the sum of the path
probabilities for the paths that satisfy that event. So the path
probabilities are computed as products, and then you sum the resulting
products over all paths that match your event (2 red balls).
This approach is only practical when the number of nodes is reasonably
small. For this problem, since the number stages (balls to be drawn) is 4
and since there are most 2 branches at each stage, the total number of
nodes is at most 1+ 2 + 4 + 8 + 16 which is 31. Actually it will be less
since for some nodes there will be only 1 branch (if only 1 color is left
at that node).
Just to give an example of 1 complete path, take the sequence RWRW. For
that path the branch probabilites are 2/5, 3/4, 2/3, 1 so the probability
for that path is the product 2/5 * 3/4 * 2/3 * 1 = 1/5. If you work out the
probabilities for all the other paths for which the number of red balls is
2 and add the results, you get your answer.
The concept of conditional probability can also be understood with the tree
method. If there is an additonal condition which must be satisfied, then
some of the branches will be eliminated and others will be forced. If a
branch is eliminated (based on the specified condition) it gets a new
probability of 0 (and the branch need not be drawn). On the other hand, if
a branch is forced, then it gets a probability of 1, so in this case, the
ratio is not relevant since the specified condition overrides the ratio and
forces a probability of 1 for that branch.
So for example, for the problem at hand, the path RWRW is eliminated but
the path WRWR has conditional probability 1 * 2/4 * 1/3 * 1/2 = 1/12
(notice the 1 in the first factor) whereas the absolute probability (if no
condition was specified) would have been 3/5 * 2/4 * 1/3 * 1/2 = 1/20.
I think the tree approach, while tedious and not very sophisticated, is key
to understanding the basic concepts, so in the early stages of learning
probability, it's important to get experience with this method, provided
the tree is not too big.
Ok, having said that, there is a sneaky way to avoid doing a 4 stage tree,
by first solving a different problem, Think about the 5th ball. If we
forget the first 4 balls (don't even look at them), what is the probability
that the 5th ball is: (1) red? (2) white? This is easy since effectively
it's the same as just drawing 1 ball. Next think about how the results for
the 5th ball affect the event in question for the first 4 balls. But I
recommend doing the 4 stage tree anyway before trying this shortcut, if
only to verify.
Perhaps these trees will help you see the forest.
-- quasi
.
- References:
- Probabilities and complex numbers.
- From: novelo
- Probabilities and complex numbers.
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