WTF is happening????? I did sign in. how come it is showing me as a guest???? I know i am a member of this board!!
Help me moderator (or somebody)
This is a discussion on Test your math ability within the A Brief History of Cprogramming.com forums, part of the Community Boards category; WTF is happening????? I did sign in. how come it is showing me as a guest???? I know i am ...
WTF is happening????? I did sign in. how come it is showing me as a guest???? I know i am a member of this board!!
Help me moderator (or somebody)
I am the Alpha and the Omega!!!
You have to use the user cp if you want to sign in (well most people do...)
I'd have to go with that answer aswell.According to your wording, all five people own all five hats, so the probability is 1 in 1.
First, let's see ...
Probability = (Favourable Outcomes) / (Total Number of outcomes)
Favourable Outcomes....
it's when all the five have their own
hats... no other combination is acceptable... so only 1 favourable
outcome
Total Outcomes ....
Five hats can be given to 5 people in 5! ways
5! = 5 * 4 * 3 * 2 * 1 = 120
Probability (P) = 1/120
Have fun
now you ask me why 5! ways ?
observe the pattern carefully... if you can follow
till 3, you'll understand...
1 person ... 1 hat ... 1 way ... ( 1! )
2 persons .. 2 hats ... 2 ways ... ( 2! )
3 persons ... 3 hats ... 6 ways ... ( 3! )
4 persons ... 4 hats ... 24 ways ... ( 4! )
5 persons ... 5 hats ... 120 ways .... ( 5! )
from the above pattern,
n persons ... n hats ... n! ways
here's another way of looking at it (if you're still interested)
5 hats - 1 2 3 4 5
5 girls - A B C D E
probability of A getting the right hat (out of 5 hats)
1/5
probability of B getting the right hat (out of remaining 4 hats)
1/4
probability of C getting the right hat (out of remaining 3 hats)
1/3
probability of D getting the right hat (out of remaining 2 hats)
1/2
probability of E getting the right hat (out of remaining 1 hat)
1/1
Now if A gets the right hat, no one else gets A hat (sounds
dumb, but see...) ... it's the same for B C D & E
so these "events" tell us that it is conditional probability.
so, the probability of everybody getting their own hats is
1/5 * 1/4 * 1/3 * 1/2 * 1/1 = 1/5!
someone here said 1/3125
that answer will be correct if the person "picks" the right hat
and puts it back again ... and the rest do the same ("pick" the
right hat and put it back again)
Math is fun
1/5! = 1/120
Simple stats question.
Try not.
Do or do not.
There is no try.
- Master Yoda
Odds of contracting and being affected by the West Nile virus in Illinois: 1/30000
7 more cases found today, one critical, lucky us.Originally posted by BMJ
Odds of contracting and being affected by the West Nile virus in Illinois: 1/30000
meh? Oh... yea, heh; Lucky
I just saw a bunch of stats, so I just felt like posting something
I am just glad I live downtown..I don't see to many bugs here.
Okay , early i said 1/120 , I am sticking with that answer. Here is a little reasoning. Lets assume you have 5 objects. There are only a certain number of permuations for this set. 2 Permuations are
1,2,3,4,5
1,2,3,5,4
Just for example. To get the total number of permuations you do 5! which equals 120. Only 1 of those is right however because each of the 5 owners only owns 1 hat. So 1/120 is the correct answer. It doesn't matter if you pass out the hats all at once OR one by one, the answer holds.
"...the results are undefined, and we all know what "undefined" means: it means it works during development, it works during testing, and it blows up in your most important customers' faces." --Scott Meyers
Is it a permutations answer or a combinations answer? I don't want to think about, but my guess after reading was 5! or 1 in 120 that each individual would recieve their own hat.
probabilities are usually based on permutations.
combinations come into place when it doesn't matter
which way you give the hat as long as they get hats..
Eg.
The number of combinations of 3 girls sitting together in
3 seats is 1.
(doesn't care about arrangements.. as long as they sit
together, it's a combination.. different ways of arrangements
of the same are considered as repitition)
The number of permutations of 3 girls sitting together
in 3 seats is 6.
(does care about arrangements)
Now why would we want to study combinations then?
It's when we want to cancel out repititions.
For example, the number of ways of "choosing" 3 girls
from 12 girls is 12 C 3 (doesnt matter which way you
choose the girls, as long as you choose 3 of them)
The number of ways of "arranging" 3 specific girls out
of 12 is 12 P 3 (it matters which spice girl you pulled out
1st, 2nd and 3rd )
Last edited by moonwalker; 08-21-2002 at 10:19 AM.
Most of you put what was expected: 1/25.
But the correct answer is: 1/120.
Thanks Again