Clarification of Answer by
elmarto-ga
on
04 Feb 2004 11:32 PST
Hi again jgortner!
The reason why P[D intersect S'] = 0.05 is incorrect is given in my
answer. As I said, the problem states that "...5% OF legally marketed
chips are defective". Saying that P[D n S']=0.05 is different from
that statement. P[D n S']=0.05 implies that a randomly selected chip
from the WHOLE universe of chips (legal and illegal ones) has a 5%
chance of being both legal and defective. This is not the same to say
that if we draw a random chip from the subset of *legal* chips, there
is a 5% chance that it is defective.
Let's see why. Imagine there are 200 chips. 100 of them are legal and
100 are not. Of the legal chips, we have 5 defective and 95
functioning. Of the illegal chips, all are defective. So in this case
it's true that P[D|S']=0.05, because from the subset of legal chips
(100) there are 5 defective ones, so the probability of getting a
defective chip is 5/100. However P[D n S']=5/200=0.025 (from the
universe of 200 chips, there are 5 that are legal and defective).
The main mistake in your answer is confusing P[D|S'] with P[D n S'],
which unfortunately throws further calculations in the wrong
direction. Even taking that in account, there are some more mistakes
in your answer:
- You state that P[D n S] = P[D]*P[S]. This is not necessarily true
(and in this case it's definitely not true). The assertion that P[A n
B]=P[A]*P[B] is only true when A and B are "independent" events.
Consider the following distribution of chips:
Defective Non Defective
Stolen 100 0
Non stolen 0 100
There are 200 chips in total. 100 are stolen and defective, 100 are
non-stolen and non-defective. The probability of a random chip (from
the whole universe of chips) being stolen is 100/200=0.5 (100 stolen
chips/200 total chips). Thus P[S]=0.5. The probability of a random
chip being defective is also 0.5 for the same reason. Thus P[D]=0.5.
Now, what's P[S n D]? Looking at the distribution, it's clear that P[S
n D]=0.5 (100 defective-and-stolen chips/200 total chips). However
P[S]*P[D]=0.25. Therefore this step is wrong.
Furthermore, in this step you used the fact that P[D]=0.5. (in the
pdf, you state (Percentage of defective)*(Percentage of
stolen)=0.5*0.01). However, in the next lines, you come to the wrong
conclusion that P[D]=0.055. Although you correctly stated that
P[D]=P[D n S]+P[D n S'], you had wrong values for both P[DnS'] and
P[DnS]. In fact, this equation was not even necessary, since you
already knew that P[D]=0.5. In conclusion, these mistakes caused the
final answer to be wrong.
Regarding the solution using Bayes Theorem, this is exactly what i did
in my answer, although I didn't use the same form of the Bayes Theorem
you state in the first lines of the answer. But do notice that my way
to "attack" this problem was exactly the same as yours; namely, to
find the numerator and denominator in the Bayes Theorem equation. You
stated correctly that the numerator in this equation becomes P[S n D]
(unfortunately you found a wrong solution for P[S n D]). As you can
see in my answer, the steps I did were aimed at finding P[S n D]. Now,
what about the denominator? Again, you stated correctly that the
denominator is:
P[D|S]*P[S] + P[D|S']*P[S']
you also found above that P[D|S]*P[S] = P[D n S]. Using the same
logic, we find that P[D|S']*P[S']=P[D n S']. Therefore we have that
the denominator is:
P[D n S] + P[D n S']
However, as I stated in my answer, defective chips are either stolen
or non-stolen. So the proportion of defective-and-stolen chips plus
the proportion of defective-and-not-stolen chips, is the proportion of
defective chips. Therefore we have that the denominator in the Bayes
Theorem equation becomes simply P[D]. Therefore, the answer simply
comes from the following equation:
P[D n S] / P[D]
Regarding how to tell that a problem requires the usage of Bayes
Theorem, you'll find that almost all problems related to conditional
probability need some form of the Bayes Theorem in order to be solved.
More specifically, you'll recognize that you need to use the Bayes
Theorem in problems that supply (directly or indirectly) P[A|B] and
then ask you to find P[B|A] (or need this value in order to do some
further calculation). Variations could include using the complement of
A and B in the supplied and/or required information. For example, in
this problem, the statement supplies P[D|S'] and then asks you to find
P[S|D]. Whenever you come across a problem that gives you information
on the probability of event A given event B and then asks you to find
something similar to the probability of event B given event A, you'll
positively have to use this theorem in order to solve it.
The mathematical derivation of the Bayes Theorem requires some basic
knowledge of set theory, and can be found, step by step, in the
following page:
http://mathworld.wolfram.com/BayesTheorem.html
I hope this has clarified the answer. Please do ask again if you feel
something's still unclear.
Best wishes!
elmarto
,
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