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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

767.0. "Probability Problem" by NAC::PICKETT (Plate o' shrimp $1.98) Fri Oct 09 1987 16:46

    Calling all probability fans. I haven't done this type of problem
    in some time now. Surely most of you readers should know how. 
    
    Given:
    
    X is in the range 0 <= X <= 511
    
    P is in the range 0 <= P <= 8191
    
    find the probability that for a random X, P will satisfy:
    
         0 <= P <= (512-X)
    
    I got .03131
    
    dp
T.RTitleUserPersonal
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767.1CLT::GILBERTBuilderFri Oct 09 1987 19:0318
    There are 512*8192 possibilities.  Summing those that have P <= (512-X),

	(P,X) = (0,0),(0,1),(0,2),...,(0,509),(0,510),(0,511)
		(1,0),(1,1),(1,2),...,(1,509),(1,510),(1,511)
		(2,0),(2,1),(2,2),...,(2,509),(2,510)
		(3,0),(3,1),(3,2),...,(3,509)
		...
		(509,0),(509,1),(509,2),(509,3)
		(510,0),(510,1),(510,2)
		(511,0),(511,1)
		(512,0)

    That is, there are 512*(512+1)/2 + 512 possibilities that have P <= (512-X).
    So the probility is:

		(512*(512+1)/2+512) / (512*8192)
		= 515 / 16384
		= 0.0314331
767.2"Job Related?"SQM::HALLYBI sell too soonFri Oct 09 1987 19:101
    This wouldn't have anything to do with fiber optics, would it?
767.3a little higherVINO::JMUNZERFri Oct 09 1987 19:1631
Picture:
		P
		|
		|
	      b |
    		NNNNNNN			0 <= X < a
		NNNNNNN                 0 <= P < b
	      a YNNNNNN
		YYNNNNN                 X + P <= a:  (Y)es or (N)o
		YYYNNNN
		YYYYNNN
		YYYYYNN
		YYYYYYN
		YYYYYYY
	      0	YYYYYYY------- X
		0      a  b
                                                                 
There are (2 + 3 + 4 + ... + a-1 + a + a+1) Y's.
        (a+1) * (a+2)          a * (a+3)
That's  ------------- - 1  or  ---------  Y's
	      2                    2

There are (a * b) total Y's and N's.

                          # Y's         a * (a+3)     a+3
Desired probability = -------------  =  ---------  =  ---
                      # Y's and N's     2 * a * b     2 b

For a = 512, b = 8192, probability = .031433

John
767.4?COMET::ROBERTSPeace .XOR. Freedom ?Fri Oct 09 1987 19:186
    
    The way I read .0 was that X and P are both reals.  .1 appears to
    assume naturals.  Will the author clarify the question, please?
    
    						/Dwayne
    
767.5oopsVINO::JMUNZERFri Oct 09 1987 19:533
    .3 duplicates .1 -- right again, Peter.
    
    John
767.6CLT::GILBERTBuilderMon Oct 12 1987 15:073
If we are using *real* numbers, and take the approach shown in .3, then
the total area is 511*8191, and the area having 0 <= P <= (512-X) is
512*512/2.  Thus, the probability is (512*256)/(511*8191) = 0.031315.
767.7Got the info I needNAC::PICKETTPlate o' shrimp $1.98Mon Oct 12 1987 16:333
    Thanks everyone! The solution I was looking for was in .1
    
    dp