The three balls numbered 1 to 3 in an urn are drawn at random one at a
time until the urn is empty. The sequence of the ball numberes is
noted.

a.   Find the sample space.

b.   Find the sets Ak corresponding to the events "ball number k is
selected in the kth draw," for k = 1,2,3.

c.   Find the set (A1 AND A2 AND A3)  and describe the event in words.

d.   Find the set (A1 OR A2 OR A3)  and describe the event in words.

e.  Fine the bar notation of set defined in d and describe the event
in words.

Hello again.

a. Sample Space S = (1,2,3) (1,3,2) (2,1,3) (2,3,1) (3,1,2) (3,2,1)

b. Find the sets Ak corresponding to the events "ball number k is
selected in the kth draw," for k = 1,2,3.
A1  k=1  (1,2,3) (1,3,2)
A2  k=2  (1,2,3) (3,2,1)
A3  k=3  (1,2,3) (2,1,3)

c.  (A1 AND A2 AND A3) = (1,2,3)
     It is the one set where ball#1 was drawn on draw#1, AND ball#2
was drawn on draw#2, AND ball#3 was drawn on draw#3.
     On average this happens 1/6 of the time


d. (A1 OR A2 OR A3) = (1,2,3) (1,3,2) (2,1,3)
     There are three sets where ball#1 was drawn on draw#1 OR  ball#2
was drawn on draw#2, OR ball#3 was drawn on draw#3.
     On average this happens 3/6 = 1/2 the time

e.  P A1 or A2 or A3 (A1 OR A2 OR A3 | (1,2,3) (1,3,2) (2,1,3) )
     Again this is the Probability of getting ball#1 drawn on draw#1
OR  ball#2 drawn on draw#2, OR ball#3 drawn on draw#3,  which is
satisfied in 3 instances of the sample set.

Frankly, it is very unusual to apply "bar notation" in a probability
problem, and I'm not certain this is the usage your instructor has in
mind.  After much searching I was able to find bar notation used in a
highly technical paper on page 3 at
http://trimm.radiology.arizona.edu/~kupinski/BANN.pdf

which is why I believe my answer to 'e' is correct.

If any commentors or other researchers can offer some input on the use
of 'bar notation' in problems of this sort, I'll be very grateful.

Thanks again for letting us help.
Richard-ga

---

Clarification of Answer by richard-ga on 08 Sep 2003 06:08 PDT

Mathtalk's comment (below) includes two important comments on my
answer.

First of all, he's right in pointing out that I managed to get part
"d" wrong.  I missed (3,2,1) which of course is also a 'winner' since
ball number 2 matches draw#2.

So the correct answer to 'd' is

d. (A1 OR A2 OR A3) = (1,2,3) (1,3,2) (2,1,3) (3,2,1)
     There are four sets where ball#1 was drawn on draw#1 OR  ball#2
was drawn on draw#2, OR ball#3 was drawn on draw#3.
     On average this happens 4/6 = 2/3 of the time 

For answer e., Mathtalk's opinion is that "Bar" means "Not", so the
answer to 'e' is not the same as the answer to 'd' but rather "1 minus
'd'"

Since answer 'd' is (1,2,3) (1,3,2) (2,1,3) (3,2,1) which happens 2/3
of the time
that makes answer e  {(2,3,1), (3,1,2)} which happens 1/3 of the time.

Sorry about the confusion, especially about question 'd'.  For
question 'e' as I said originally you should double-check with your
instructor that we are giving the proper meaning to 'bar.'

-R

Hi, Richard-ga:

I have looked at the technical paper to which you provided a link, and
I'm not sure in what sense you mean to say that 'bar notation' is used
there.  There are two possibilities that occur to me.  First, some of
the variables are denoted with arrows over them (which indicate
"vector" quantities).  Second, many of the probabilities and other
expressions involve a vertical stroke.  This is the standard notation
for "conditional" probabilities (and related definitions, such as
conditional expected values).

I think the 'bar notation' spoken of in part e of the customer's post
is a much simpler concept, namely complement of a set.  In elementary
probability theory it is customary to introduce a discrete probability
space as a finite set of "elementary events", to each of which
nonnegative probabilities are assigned in such a manner that they
total up to 1.  "Compound events" are then described as subsets of
this "universal" set, and a familiar law of probability then would
read:

Pr(A bar) = 1 - Pr(A)

meaning that the probability of "not A" is 1 minus the probability of
A.

In order to apply this "complement" interpretation to part e, it is
necessary to back up and correct the answer to part d.  You identified
three of the "elementary events" in the compound event defined by:

(A1 or A2 or A3)

which were (1,2,3), (1,3,2), and (2,1,3).  However you missed (3,2,1).

In words the event of part d is that either ball 1 is drawn first, OR
ball 2 is drawn second, or ball 3 is drawn third.  Since there are
four (equally likely) elementary events that satisfy this, the
probability of the compound event described in part d is 4/6 = 2/3.

The event of part e is then (if my interpretation of bar notation is
correct, a horizontal line being drawn over the top of the event of
part d):
________________
(A1 or A2 or A3)

the complement of this event, i.e. 

(not A1) and (not A2) and (not A3)

Such an outcome, in which no ball is drawn in its "proper" order, is
sometimes called a "derangement".  Here there are but two elementary
events that accomplish this "completely out of order" feat:

(2,3,1) and (3,1,2)

As a final note, it might be best to express answers to these kinds of
problems with set notation.  For example:

d) {(1,2,3), (1,3,2), (2,1,3), (3,2,1)}

e) {(2,3,1), (3,1,2)}

Note that as the chance of the event in part d is 2/3, we have the
complementary probability of part e is 1/3.

regards, mathtalk-ga