Math/Probability:  In our lab, four machines are on their own network,
isolated from all causal influences except each other. Machine A only
sends packets to Machines B and C (i.e., A only affects B and C).  B
and C do not affect A, but B and C both affect D.

When A is up, the packets B gets from A give it a 90% chance of being
up, otherwise B has only a 1% chance of running (being up). When A is
up, C is up with a 100% probability.  When A is down, C is up with 40%
probability.

D has these probabilities of being up:

    D Up   when  B and C both up:   100%
    D Up   when  only B up:          90%
    D Up   when  only C up:           5%
    D Up   when  neither B or C up:  50%

For twenty dollars, what should the observed frequency for both
combinations of all four machines be?  That is, ABCD,  ABC~D,  AB~C~D,
 A~BCD, etc.?  The probability distribution should cover, of course,
sixteen cases.  You must explain or justify your answer clearly.

Lee Corbin

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Clarification of Question by nobleicer-ga on 10 Jul 2004 01:55 PDT

Oh yeah, I forgot to mention a crucial fact: Machine A has a 30%
probability of being up.

Hello and thank you for your question.

Based on the information that you provide,

A = .3  
~A = .7
B = .9 * .3 + .01 * .7 = .277
~B = .723
C = .3 + .7 * .4 = .58
~C = .42
D =  B*C   + .9 * B*~C + .05 * ~B*C + .5 * ~B*~C
  = .16066 + .104706   +  0.020967  + 0.15183
  = 0.438163
~D  = 0.561837

and so the 16 answers are the indicated products:

ABCD    0.02111858

ABC~D   0.02707942

AB~CD   0.015292765

AB~C~D  0.019609235


A~BCD   0.055121782


A~BC~D  0.070680218


A~B~CD  0.039915773


A~B~C~D 0.051182227


~ABCD   0.049276687


~ABC~D  0.063185313


~AB~CD  0.035683118


~AB~C~D  0.045754882


~A~BCD  0.128617491


~A~BC~D  0.164920509


~A~B~CD  0.093136804


~A~B~C~D 0.119425196


And I'm happy to say that these 16 numbers add up to 1.000000 as they should

No searching on Google for this one; I knew how to do it.

Thanks again for bringing us your question
Richard-ga

Sorry, but your answer seems incorrect to me. (I have
meanwhile figured out the answer myself and have tried
to cancel this question.) The probability of ABCD is
the probability of A times the probability of B given
A times the probability of C given A,B times the probability
of D given A,B,C.  But the probability of C given A must be
the same as the probability of C given A and B, because C
is affected only by A. So ABCD is .27.

You appear to have just assumed independence and
calculated P(ABCD) as the product of P(A), P(B),
P(C), P(D).

Thanks anyway.  If there is a way to tip you for trying, let me know!

Tip for trying?  Regardless of what you rate this, he/she still gets the money!

Thanks, nelson-ga for your remark. At least richard-ga
got the money. I'm only sorry that I sabotaged his rating
with a 1-star evaluation when I didn't know how the system
worked. :-)  I feel a little better.

In the future, you can request an answer clarification so the
researcher can correct the mistake or hone their answer.  If you are
completely dissatisfied you can request a refund.

I agree with Bowler............

Please do not rate until you are totally satisfied with the answer...

Ratings of the number of stars affect the researchers big time........