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Maths R Fun |
Bardiche
Chief Marshal
Joined: November 16, 2006
Posts: 1247
|  Posted: 2010-11-13 20:59
So hey, here's some critical thinking for you.
Suppose you have two coins. One of these coins, let's call it Coin A, has heads on both sides. The other coin, Coin B, is a regular coin of which one side is heads, and the other is tails.
You randomly grab one of these coins. You flip this coin, and the result is heads.
What is the probability of you having grabbed Coin A?
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Borgie
Chief Marshal
Pitch Black
Joined: August 15, 2005
Posts: 2256
From: close by
| Posted: 2010-11-13 21:06
im sure this is wrong but 50%, since there are 2 coins.
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Darth Offline of the Shadows
Commander
Joined: January 29, 2010
Posts: 12
From: THE BATH TUB
| Posted: 2010-11-13 21:10
id say 70% coz there is 3 heads and 1 tail
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EDELWEISS
Captain
Joined: August 19, 2003
Posts: 22
| Posted: 2010-11-13 21:11
What is the probability of you having grabbed Coin A?
OUTCOMES:
Heads - Coin A
Heads - Coin B
Painless naive guess: 0.5
Taking it as a conditional probability:
P( Choosing Coin A | Randomly chosen coin results heads)
= P((0.5) | (0.75) ----- 0.75 from the 3 possible heads outcomes out of the 4 possible coin face outcomes:
by that, I mean HEADS-A, HEADS-A(2), HEADS-B, TAILS-B
= (0.5*0.75) / 0.75
= 0.5
Seeing as P( Coin A | random heads) = P(Coin A), the odds that you picked coin A is independent of whether or not the coin came up heads. I guess that sits well in my head logically, since the odds of picking coin A from 2 randomly is 0.5, and flipping the coins afterwards won't affect that probability.
Forgive me, it's been a long, loooong time...
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Aradrox
Grand Admiral
Joined: March 12, 2007
Posts: 133
From: Tennessee
| Posted: 2010-11-14 00:34
50% duh for the question what is on the coins has absulotely nothing to do with it. Same question could of been there are 2 pillows Pillow a and pillow b. One has a star on both sides the other has a star on one side. You randomly grab a pillow what is the chance you grabbed pillow A?
50%
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SpaceAdmiral
Grand Admiral
Joined: May 05, 2010
Posts: 1005
| Posted: 2010-11-14 00:47
66.6666666666666666666666666666...%
there are 3 heads
2 of them are on A, 1 on B
chance of getting an A head is 2/3
so probability of getting coin A is 66.66666...%
i don't believe you need more advanced math and
the coin faces do matter, it says you picked a head and one coin has a higher probability of getting heads (100%...)
the above can be seen if they said it was a tails you picked up, then it is a 100% chance of getting B, not 50% because "faces don't matter".
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sonicmeerkat
Joined: August 04, 2010
Posts: 10
| Posted: 2010-11-14 02:43
50% as it's asking what the chance of picking up coin A is.
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SpaceAdmiral
Grand Admiral
Joined: May 05, 2010
Posts: 1005
| Posted: 2010-11-14 02:48
Quote:
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On 2010-11-14 00:34, Grand Admiral Thrawn. wrote:
50% duh for the question what is on the coins has absulotely nothing to do with it. Same question could of been there are 2 pillows Pillow a and pillow b. One has a star on both sides the other has a star on one side. You randomly grab a pillow what is the chance you grabbed pillow A?
50%
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except it says you grabbed a pillow with a star on top.
too many people miss that tiny detail that matters.
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Forger of Destiny
Chief Marshal
We Kick Arse
Joined: October 10, 2009
Posts: 826
| Posted: 2010-11-14 02:58
2 Coins.
4 Outcomes.
3/4 chance for a head to come up.
1/4 chance for head to be on second coin.
2/4=1/2 chance for head to be on first coin.
Probability of coin A = 50%
A. Thats because you had as much chance of flipping a tails outcome as for any unique head outcome. The question asks : Whats the chance of the outcome received to be from coin A. This can be re-interpreted to create : Whats the chance for coin A to give an outcome.
66.7% would hold if you couldn't roll a tails.
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Mint Ice Cream[+R]
Grand Admiral
Joined: December 05, 2009
Posts: 43
| Posted: 2010-11-14 04:25
u said maths r fun right 
my turn:
is this true??
"Theorem: 4 = 5
Proof:
-20 = -20
16 - 36 = 25 - 45
4^2 - 9*4 = 5^2 - 9*5
4^2 - 9*4 + 81/4 = 5^2 - 9*5 + 81/4
(4 - 9/2)^2 = (5 - 9/2)^2
4 - 9/2 = 5 - 9/2
4 = 5"
[ This Message was edited by: Mint Ice Cream[+R] on 2010-11-14 04:37 ]
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*Obsidian Shadow*
Grand Admiral
Joined: January 03, 2010
Posts: 316
| Posted: 2010-11-14 05:13
1=2 :>
a=b
a^2=ab
a^2+a^2 = a^2 + ab
2a^2 =a^2 +ab
2a^2 - 2ab = a^2 - 2ab
2a^2 - 2ab = a^2 - ab
this can be written as
2(a^2-ab) = 1(a^2-ab)
and cancelling the (a^2-ab) from both sides gives 1=2
In essence, this proof boils down to saying "1 times 0 equals 2 times 0, therefore 1 equals 2". The fallacy is that, just because two numbers give you the same answer (zero) after you multiply them each by zero, doesn't necessarily mean that the two numbers are the same, because anything when multiplied by zero gives zero.
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Forger of Destiny
Chief Marshal
We Kick Arse
Joined: October 10, 2009
Posts: 826
| Posted: 2010-11-14 05:15
Quote:
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is this true??
Theorem: 4 = 5
Proof:
-20 = -20
16 - 36 = 25 - 45
4^2 - 9*4 = 5^2 - 9*5
4^2 - 9*4 + 81/4 = 5^2 - 9*5 + 81/4
(4 - 9/2)^2 = (5 - 9/2)^2
4 - 9/2 = 5 - 9/2
4 = 5"
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Based on the (-1)^2 = 1^2 principle? Not Funny.
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Forging legends and lives outside till naught remains inside.
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Bardiche
Chief Marshal
Joined: November 16, 2006
Posts: 1247
| Posted: 2010-11-14 05:31
Unfortunately, for the crew insisting that it's 50%, that's wrong. That's not how probabilities work. Sure, at first glance that may seem to be the truth, but as someone else explained.
There are four possibilities.
Either you:
Grabbed Coin A and flipped Heads. Grabbed Coin A and flipped the other Heads. Grabbed Coin B and flipped Heads. Grabbed Coin B and flipped Tails.
So A-H1, A-H2, B-H, B-T.
You flip heads. You immediately eliminate B-T. Leaves A-H1, A-H2, B-H. Two out of three remaining possibilities belong to A-, therefore A- has a 2/3rds probability of being the coin you picked.
For more detailed information on this problem, please consult the following page and be marveled at the mysteries of mathematics:
http://mathproblems.info/prob16s.htm
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Gejaheline
Fleet Admiral
Galactic Navy
Joined: March 19, 2005
Posts: 1127
From: UGTO MUNIN HQ, Mars
| Posted: 2010-11-14 06:47
I think this would be clearer if you said something along the lines of "you flip both coins, and they both come up heads. You then select one at random. What are the odds of choosing the one with two heads?"
This randomises the system before you select the coin, as opposed to after, which makes the fact that both coins show heads relevant in the selection process.
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Bardiche
Chief Marshal
Joined: November 16, 2006
Posts: 1247
| Posted: 2010-11-14 07:11
Unfortunately, Gejaheline, this is how the problem was presented. You first pick, then observe. Textbook math problem.
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