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Mathematicians - what's the propability?
(16 posts, started )
Mathematicians - what's the propability?
I've got 2 2 euro coins, 4 1 euro coins and 3 50 cent coins in my pocket (total 2+4+3=9 coins). I take out two coins at the same time. What's the probability for me to able to pay groceries worth of 3 euros? So I need to take out either 2 2euro coins OR one 2 euro and one 1 euro coin.

I know the answer is 1/4 but why? The only way I managet to get the result was: (2 combinations from a set of 3) times (2 combinations from a set of 3) divided with (2 combinations from a set of 9)

Why?
You don't need Maths, you need "Common Sense"...
If you have those coins in your pocket, you _should_ be able to pull out the exact amount EVERY time...because of the difference in sizes you can FEEL what coins are which! (unless you haven't got ANY feeling whatsoever in your fingers, that is!)
First of all you need to figure out what type of mathematician you are because you seem to be stuck between the two; pure or theoretical.

I'm assuming you're approaching this from a theoretical point of view, because you're asking why, when a pure mathematician just understands numbers and reocurring events etc. so on and so forth from negative infinity to infinity and everything between the two.

You're asking why which is a theoretical point of view, but you must understand that the world doesn't go around by fixed numbers, natural events just occur, mathematical equations can explain events to an extent, but not fully.

So really, your question is a good one and I don't know the answer, I just felt like saying something random about the difference between a pure mathematician and a theoretical.
Is this a mathematical or philosophical question?

Mathematically thinking:
Like you said the only way to get the correct amount is either 2e+2e coins or 2e+1e coins.

2e+2e: 2/9 chances to pick a 2e coin first and after that 1/8 chance to pick the last 2e coin = 2/9*1/8 = 2/72 (=1/36)
2e+1e: 2/9 chances to pick a 2e coin and 4/8 chance to pick a 1e coin. But because it is also legal to pick first 1e coin and then 2e coin the chance is double = 2/9*4/8 * 2 = 16/72 (=8/36=2/9)

The combined chance is sum of those: 2/72+16/72=18/72 = 1/4
Quote from Aquilifer :Is this a mathematical or philosophical question?

Mathematically thinking:
Like you said the only way to get the correct amount is either 2e+2e coins or 2e+1e coins.

2e+2e: 2/9 chances to pick a 2e coin first and after that 1/8 chance to pick the last 2e coin = 2/9*1/8 = 2/72 (=1/36)
2e+1e: 2/9 chances to pick a 2e coin and 4/8 chance to pick a 1e coin. But because it is also legal to pick first 1e coin and then 2e coin the chance is double = 2/9*4/8 * 2 = 16/72 (=8/36=2/9)

The combined chance is sum of those: 2/72+16/72=18/72 = 1/4

Hehe, on your avatar pic you definitely look like a mathematic genius
#6 - Gil07
Well that isn't exactly mathematics that need a genius
The probability is near 1 because Euro coins are designed to be identifiable by touch. Mainly for the blind but it's useful for everyone else.
Quote from duke_toaster :The probability is near 1 because Euro coins are designed to be identifiable by touch. Mainly for the blind but it's useful for everyone else.

Wot I said
Use credit card :mr-t:

(what is money)
Quote from Aquilifer :Is this a mathematical or philosophical question?

Mathematically thinking:
Like you said the only way to get the correct amount is either 2e+2e coins or 2e+1e coins.

2e+2e: 2/9 chances to pick a 2e coin first and after that 1/8 chance to pick the last 2e coin = 2/9*1/8 = 2/72 (=1/36)
2e+1e: 2/9 chances to pick a 2e coin and 4/8 chance to pick a 1e coin. But because it is also legal to pick first 1e coin and then 2e coin the chance is double = 2/9*4/8 * 2 = 16/72 (=8/36=2/9)

The combined chance is sum of those: 2/72+16/72=18/72 = 1/4

Yes, it was a mathematical question. Now you've calculated it by presuming that the coins are taken out one at the time, but I clearly said the coins will be out simultaniously. I wanted to know how the comninatoric way works.

Thou it's actually also possible to work it out philosophically: the first coin is or isn't a 2e coin, the probability is 1/2. The other one is either a 1e or 2e coin or it isnt, probability 1/2.... and 1/2*1/2=1/4
Quote from hyntty :Yes, it was a mathematical question. Now you've calculated it by presuming that the coins are taken out one at the time, but I clearly said the coins will be out simultaniously.

It's the same.

(Except if the first coin, after having been drawn, will be put back in the bag. In that case the probability is 2/9 * 2/9 + 2/9 * 4/9 * 2 = 20/72 = 5/18. But that is ludicrous because you were gonna use the coins for a payment. Aquilifer's calculation is correct.)
it should work out as (take with a whole bunch of salt combinatorics was almost my weak link in math)

hope the formula shows up correctly:


as you can either take out 2 of the 2 2-euro coins (yeah i know ...) _or_ you can take 1 of the 2-euros _and_ 1 of the 1-euros
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(speedway) DELETED by speedway
Urgh! And I thought that I could do without this crap after I left school.
Now you have made me want to check your calculations and notation.

BUT I will stay firm! I will fight the temptation of the Mathematics. I will prevail!

Oh bugger! Where's my math book...
Quote from Shotglass :it should work out as (take with a whole bunch of salt combinatorics was almost my weak link in math)

hope the formula shows up correctly:


as you can either take out 2 of the 2 2-euro coins (yeah i know ...) _or_ you can take 1 of the 2-euros _and_ 1 of the 1-euros

Yes, that's it! Thanks!

Quote from speedway :absolutely correct

could save some time though, with:

(2).(5)
(1) (1)
--------
(9)
(2)

take 1 2euro-coin and any of the 4 1€ or 1 2€ coins

havent checked the result, should defenitely work as well

2*5/9ncr2 = 5/18. No, it definately doesn't work
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(speedway) DELETED by speedway
It is very true you can write it down in a much nicer way (shown by Shotglass). I tryed to write it in a different, more logical way how people tend to think it normally with common sense. I thought that was behind the question to make it understandable.

The fact if you pick the coins one at the time or by taking two coins and checking which ones you got, doesn't make any difference (like mentioned already). It cerrtainly doesn't make sense putting the first coin pack to pocket because you cannot use it twice for payment (because the cashier already took the first coin). You can think like you take 2 coins but check them only after taking both.

Mathematicians - what's the propability?
(16 posts, started )
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