there are 5 different people who ordered 5 different drinks
what is the probability that all 5 people will get THE WRONG DRINK?
(its not combination - order matters so its PERMUTATION; denominator is 5!=120; i don't know what the numerator is )
2007-03-30 20:15:56 · 9 answers · asked by =D 1 in Science & Mathematics ➔ Mathematics
please note- the question seems easy but is not
my teacher says he gets it wrong sometimes;
i asked 2 friends who are math teachers and they got the question wrong
2007-03-30 20:51:08 · update #1
it is NOT COMBINATION. i tried it already and got it wrong.
i asked dr.math and they couldnt even give me the exact answer
PLEASE HELP!
2007-03-30 21:00:03 · update #2
Everyone so far is wrong...
Using permutations:
Since order matters, as the questioner notes, the denominator is 120.
Well there are 4 ways for person A to have the wrong cup - say he has person B's.
There is 1 way he has person A's, which results in 2 ways for C, E, and D.
Or else, there are 3 ways he has person C's.
There is 1 way C has person A's which results in 1 way for D and E.
Or else, there are 2 ways he has person D's.
D must have person E's.
4(1 *2 + 3(1 * 1 + 2 * 1))
= 4(2 + 3 * 3)/120
= 11/30
Helmut has illustrated what I have said.
However, BCEDA does not work... and therefore his answer is wrong.
It should be 11 * 4 = 44
44/120 = 11/30
2007-03-30 20:59:08 · answer #1 · answered by Jeffrey W 3 · 1⤊ 2⤋
3 of 5
2007-03-31 03:41:34 · answer #2 · answered by d c 1 · 0⤊ 2⤋
Let me try :
Asuming the server does not know who ordered the drinks.
First person served : 4/5 chance to be served wrong
Second person : 1/5 chance his drink was first served or 3/4 on the 4/5 left chance that he will be served a wrong one
So 1/5+3/4 x 4/5 = 4/5 chance
third : 2/5 chance his drink is already served or 2/3 on the 3/5 chance that he will be served a wrong one
so 2/5+2/3 x 3/5 = 4/5 chance
same on the 4th and fifth
1 : 4/5
2: 1/5+3/4x4/5= 4/5
3: 2/5+2/3x3/5=4/5
4: 3/5+1/2x2/5=4/5
5: 4/5
The probability that all theses events will takes place at the same time is
4/5 x 4/5 x 4/5 x 4/5 x 4/5 = 0.32768
2007-03-31 04:58:24 · answer #3 · answered by sedfr 3 · 1⤊ 2⤋
I think it's 24/120 or 1/5.....Chances 1st person gets wrong drink is 4/5...2nd person is 3/4....3rd person is 2/3....4th person is 1/2...5th person is 1/1. Multiply the chances together and you get 24/120.
I'm really bad at probability though, so I probably shouldn't have attempted to answer this one. :-P
2007-03-31 03:22:24 · answer #4 · answered by Ms.ADJ 2 · 1⤊ 0⤋
1st person: Wrong drink 4/5
2nd: 3/4
3rd: 2/3
4th: 1/2
5th 1
So its 4/5 x 3/4 x 2/3 x 1/2 = 1/5
2007-03-31 03:49:33 · answer #5 · answered by kirlia7755 3 · 1⤊ 0⤋
Cannot be determined from the information given, as it depends on the skill of the bartender. If the bartender is a hack, your situation can arise if all five people are seated around a table and each gets the drink ordered by the person on his left.
2007-03-31 03:32:40 · answer #6 · answered by Anonymous · 2⤊ 1⤋
Every one get correct drink with combination = 5 C 1 = 5
Prob. = 5/120 = 0.041
So, Probability that no one will get a correct drink = 1 - 0.041
=0.958
2007-03-31 03:30:01 · answer #7 · answered by Peter_Jackson_Fan 4 · 0⤊ 3⤋
i fail to see what the challenge is
most scientific calculators have combination and permutation functions built-in, so you don't even have to remember the formula, and there are plenty of websites to look it up from if you really need to know.
after you've calculated the total possible outcomes (let's call it C), then the requested probability is: (C-1)/C
2007-03-31 03:20:53 · answer #8 · answered by Anonymous · 0⤊ 2⤋
A B C D E
---------------
B A D E C
B A E C D
B C A E D
B C D E A
B C E A D
B C E D A
B D A E C
B D E A C
B D E C A
B E A C D
B E D A C
B E D C A
. . . . . . . . .
p = 48/120 = 0.4
2007-03-31 04:43:59 · answer #9 · answered by Helmut 7 · 0⤊ 2⤋