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| 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 |
1165.0. "formula needed" by HPSMEG::COLMAN () Mon Dec 18 1989 12:14
I'm NOT a mathematician (as is probably evident in my problem statement),
but I do need a formula that is probably "a piece of cake" for the regulars
of this conference.
If there is 1 chance in a of a particular event occurring and 1 chance in
b of another (unrelated) event occurring, then I BELIEVE the chance of AT
LEAST one of the events occurring is:
a + (b-1)
---------
a * b
Example: 1 chance in 3 of event 1 occurring
1 chance in 4 of event 2 occurring
Therefore, chance of at least one of the events occurring is
3 + (4-1) 6
--------- = -----
3 * 4 12
The formula I seek expresses the chance of at least one event occurring when
there are 4 situations:
1/a, 1/b, 1/c, 1/d
Please help so I don't have to re-invent the wheel (assuming I am even able
to do so).
Thanks,
george
| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1165.1 | look at the other way | UTRUST::DEHARTOG | 925 | Mon Dec 18 1989 12:30 | 5 |
| The way to look at it is that no event occurs at all (that the opposite
of at least one).
So the solution to your problem is:
1 - (1-a)/a * (1-b)/b * (1-c)/c * (1-d)/d
|
| 1165.2 | | HPSMEG::COLMAN | | Mon Dec 18 1989 12:42 | 5 |
| .1: Thanks much!!
regards,
george
|
| 1165.3 | correcting the sign | PULSAR::WALLY | Wally Neilsen-Steinhardt | Mon Dec 18 1989 13:17 | 10 |
| re: <<< Note 1165.1 by UTRUST::DEHARTOG "925" >>>
Your reasoning is good, but your formula
> 1 - (1-a)/a * (1-b)/b * (1-c)/c * (1-d)/d
will give values greater than one when there are an odd number of
factors in the second term. A better formula is
1 - (a-1)/a * (b-1)/b * (c-1)/c * (d-1)/d
|