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• 08-20-2002
Jet_Master
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)
• 08-20-2002
Eibro
You have to use the user cp if you want to sign in (well most people do...)

Quote:

According to your wording, all five people own all five hats, so the probability is 1 in 1.
I'd have to go with that answer aswell.
• 08-20-2002
moonwalker
Ask the master if it's math
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 :)
• 08-20-2002
moonwalker
and....
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
• 08-20-2002
moonwalker
another explanation...
here's another way of looking at it (if you're still interested)

5 hats - 1 2 3 4 5
5 girls :D - 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 :)
• 08-20-2002
Cshot
1/5! = 1/120
Simple stats question.
• 08-20-2002
DISGUISED
Quote:

1/5 * 1/4 * 1/3 * 1/2 * 1/1 = 1/5!
Ditto
• 08-20-2002
BMJ
:D

Odds of contracting and being affected by the West Nile virus in Illinois: 1/30000
• 08-20-2002
DISGUISED
Quote:

Originally posted by BMJ
:D

Odds of contracting and being affected by the West Nile virus in Illinois: 1/30000

7 more cases found today, one critical, lucky us.
• 08-20-2002
BMJ
meh? Oh... yea, heh; Lucky :)

I just saw a bunch of stats, so I just felt like posting something :p
• 08-20-2002
DISGUISED
I am just glad I live downtown..I don't see to many bugs here.
• 08-20-2002
MrWizard
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.
• 08-21-2002
Troll_King
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.
• 08-21-2002
moonwalker
hmm
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 :D 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.

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 ;) )
• 08-21-2002
Unregistered
Most of you put what was expected: 1/25.
But the correct answer is: 1/120.

Thanks Again
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