| 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 |
When I was in the army ,there was a security guard who used to drink a lot!
Every hour ,during the night,he had to check 10 doors (one different key for
each door).
He was drunk one every 3 nights approximately.
A. When he was drunk he used to try with a key and put it back with the
others if it was the bad one for a door !
B. When he was sober,he was discarding the key temporarily and trying
with another one .
One night we watched him trying to open a door . He tried 6 times without
success and we started laughing at him ,thinking he was drunk !
Was there any/more chance that he was sober at that time ?!
Sante!
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 948.1 | LISP::DERAMO | Daniel V. {AITG,LISP,ZFC}:: D'Eramo | Fri Oct 14 1988 14:29 | 48 | |
Don't be so quick to jump to conclusions next time! :-)
[spoiler follows]
Let A be the event that the guard is sober. We are given
that P(A) = 2/3 ("approximately"). Let B be the event that
the guard's first six attempts at the door fail.
When the guard is sober, the probability of failing in the
first six attempts is
9 8 7 6 5 4 4 2
-- * - * - * - * - * - = -- = -
10 9 8 7 6 5 10 5
So P(B|A) = 2/5. When the guard is drunk, the probability
of failing six attempts in a row is
6
( 9) 531441
(--) = -------
(10) 1000000
So P(B|not A) = 531441/1000000.
We can find the probability of his failing to get the
correct key in the first six attempts as
P(B) = P(B|A)P(A) + P(B|not A)P(not A)
2 2 531441 1
= - * - + ------- * -
5 3 1000000 3
4 177147 1331441
= -- + ------- = -------
15 1000000 3000000
We can now determine the probability that the guard was
sober given that the first six attempts failed; by Bayes
theorem P(AB) = P(A|B)P(B) = P(B|A)P(A) and so
P(B|A)P(A) 2 2 3000000 800000
P(A|B) = ---------- = - * - * ------- = ------- = 0.60085276...
P(B) 5 3 1331441 1331441
So the probability that the guard was sober is 0.60085276....
Dan
| |||||
| 948.2 | My conclusion was:"Look before you leap"! | ULYSSE::ZITTA | ULYSSE in wonderland | Mon Oct 17 1988 06:54 | 14 |
I got to the same conclusion by calculating the probability he was
drunk:
(1/3)*[(9/10)^6]
p(A/C)=-------------------------------- = .39915
(1/3)*(9/10)^6 + (2/3)*(4/10)
Also, when he is drunk he opens the door ,in average,after 10 trials .
When he is sober : after 5.5 trials .
This is an example of how alcohol can significantly reduce a man's
output and how mathematics can help saving someone trouble!
| |||||