| 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 |
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.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 767.1 | CLT::GILBERT | Builder | Fri Oct 09 1987 16:03 | 18 | |
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
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| 767.2 | "Job Related?" | SQM::HALLYB | I sell too soon | Fri Oct 09 1987 16:10 | 1 |
This wouldn't have anything to do with fiber optics, would it? | |||||
| 767.3 | a little higher | VINO::JMUNZER | Fri Oct 09 1987 16:16 | 31 | |
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
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| 767.4 | ? | COMET::ROBERTS | Peace .XOR. Freedom ? | Fri Oct 09 1987 16:18 | 6 |
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
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| 767.5 | oops | VINO::JMUNZER | Fri Oct 09 1987 16:53 | 3 | |
.3 duplicates .1 -- right again, Peter.
John
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| 767.6 | CLT::GILBERT | Builder | Mon Oct 12 1987 12:07 | 3 | |
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. | |||||
| 767.7 | Got the info I need | NAC::PICKETT | Plate o' shrimp $1.98 | Mon Oct 12 1987 13:33 | 3 |
Thanks everyone! The solution I was looking for was in .1
dp
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