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We are given three coins: one has heads in both faces, the [#permalink]
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23 Jul 2004, 13:57
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11. We are given three coins: one has heads in both faces, the second has tails in both faces, and the third has a head in one face and a tail in the other. We choose a coin at random, toss it, and it comes heads. What is the probability that the opposite face is tails?
(a) 1/4
(b) 1/3
(c) 2/3
(d) 1/2
(e) 2/5



CIO
Joined: 09 Mar 2003
Posts: 463

hi boksana,
Great question. is it d? 1/2?
If we already know it's one of the two coins with heads, then there's just a one in two chance that it's the one with heads and tails, and not the one with two heads...



Manager
Joined: 16 Jan 2004
Posts: 64
Location: NJ

1/2
Conditional prob
P(head) = 1/2 (total 6 faces 3 heads)
P(tail)= 1/2 (same)
p(head) and p(tail) = 1/4
P(tail) give p(head) =1/4/1/2 =1/4 *2/1=1/2
Hope I am correct I have the exam next week.By the way great question



Director
Joined: 05 May 2004
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Location: San Jose, CA

Re: PS coins [#permalink]
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23 Jul 2004, 16:31
We dont need to calculate anything.
There is only one way that we can have Head in one side and Tail in the other. Hence Ans is 1/3



CIO
Joined: 09 Mar 2003
Posts: 463

i disagree. You said we know we get a heads first. So now there are only 2 coins to choose from. The fact that we pulled it from a pool of 3 coins is irrelevant. There's no chance at all that it's the coin that's two tails, so we shouldn't think of it as part of the probability any more.
anyone want to discuss this?



Manager
Joined: 16 May 2004
Posts: 64
Location: columbus

good problem boksana. this problem is an application of bayes' theorem.
This theorem shold be applied when the outcome is already determined . for more info refer to the following link
http://mathforum.org/library/drmath/view/56622.html
so in the problem we need to calculate all possible ways a head could show up.
we have HH TH TT
probability of head showing up on the first coin is chance of first coin getting picked * chance of head showing up = 1/3 * 1 = 1/3
simlarly for the 2nd coin the probablility is 1/3 *1/2
for the 3rd since it doesnt have a head we dont really care
now to calculate the prob that the head came from the 2 nd coin applying bayes theorem
p(H) = 1/3(1/2) / 1/3 + 1/3(1/2) = 1/3



CIO
Joined: 09 Mar 2003
Posts: 463

I see your point, but I still think it's open to interpretation. If we are sitting there looking at a heads, and you ask me what's the chance there's a tails under there, I'm going to say, "well, there are two coins that have a heads on it, and only one of them has a tails, too. So it's got to be one of those two coins".
Isn't probability supposed to take place in the absence of knowledge? The more we know, the more refined our choices are. The difference between this and the one that you referenced, smandalika, is that we know something about this coin. It's not the all tails one. The one in your example (on the other website) doesn't give us that. After all is said and done, the coin could still be any of those 6 coins.
So that's my position on this one...



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Joined: 16 May 2004
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ian7777 wrote: I see your point, but I still think it's open to interpretation. If we are sitting there looking at a heads, and you ask me what's the chance there's a tails under there, I'm going to say, "well, there are two coins that have a heads on it, and only one of them has a tails, too. So it's got to be one of those two coins".
Isn't probability supposed to take place in the absence of knowledge? The more we know, the more refined our choices are. The difference between this and the one that you referenced, smandalika, is that we know something about this coin. It's not the all tails one. The one in your example (on the other website) doesn't give us that. After all is said and done, the coin could still be any of those 6 coins.
So that's my position on this one...
wow its not that complicated. I'm agreeing with you to a good extent when u say that only 2 coins are to be considered.
Now that we have eliminated one coin seems like the logical answer would be 1/2 but in this case the answer would be different because between the 2 coins we have more heads than tail so its no longer split right in the middle .It is rigged (I forget the exact term for it ) and hence u get a different result . Assume if the HH coin had HT instead then the ans would have been different.
it would have been 1/3(1/2)/ 1/3(1/2) + 1/3(1/2) = 1/2
But you bring up a good point of prob being calculated in the absence of knowledge but problems using bayes theorem are a little different because the outcome is already determined and its like we are trying to back calculate and see the probability of the occurence of individual event that ultimately lead to the outcome
does this make sense



CIO
Joined: 09 Mar 2003
Posts: 463

Hi, smandalika, yes, it does make sense. Thank you for taking the time to explain it.
Quote: Assume if the HH coin had HT instead then the ans would have been different.
To answer this point, though, if the HH coin had been HT also, then I would assume there was a 100% chance that the other side was a tails, since there are only two coins and both now are HT.
Either way, from the GMAT perspective, this is all of little importance. These kinds of problems aren't on the test. And I can say that with 100% certainty.



Senior Manager
Joined: 01 May 2004
Posts: 335
Location: USA

Ian you cannot make hasty generalizations! You are wrong. Similar problems are in GMAT. My friends had problems in real exam where it was NECESSARILY to know Bayes theorem! So it is FORMAT of GMAT!



CIO
Joined: 09 Mar 2003
Posts: 463

Boksana,
I have taken the test every 36 months for the past 2 years, and I've been teaching it for 5 years. I have taken it times when I've gotten everything right and gotten the hardest questions on the test, and times where I've missed up to 12 at a time to see what kinds of questions people get all around. I've scored everywhere from 40  51 quant.
Nothing like this has ever come up on any exam I've ever seen, and none of my students has ever reported having had a question like this.
There is such a fear of this test all around the internet it's amazing. People tap into the highest levels of combinatorics and probability, using the most random theories around. But the test doesn't need it.
A collegue of mine took the test 5 years ago and got a question involving bournulli's formula. I turned around and taught that crap to 2 years' worth of students. You know how many of them ever used it after I forced it down their throats? Not one. So I stopped. It's a waste of time.
My point: of course you can use higher level math to solve problems, but you don't have to. Would you use trig to solve a 306090 triangle? You could, but it would be a waste of time. That's why I don't use formulas for combinations. All this 8C3 stuff that's all over the internet is too much  you don't need to know it to still get every question right. None of my students do, and they can answer every question I put in front of them.
If your friend saw a problem like this, then in my opinion he has either reported it incorrectly, or it was experimental, or there was another way to do it. Either way, it's not necessary, for the typical GMAT taker, to spend countless days trying to learn all these nuances. There are more important things to spend time on, like the 1012 algebra questions they're going to get. And if you think you can't do well by missing one (not you, Boksana, but anyone reading this), you should know that I missed 12 full questions and still scored a 48. So I wouldn't fret over the hardest probability questions.



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I didn't use such a complicated method though....
Since it came out heads, it can only be either the coin with the head on one side and a tail on the other, or the coin with both heads. To get a face on the other side, you will have to pick the coin with a both head and tail faces. Now, since there're three coins, and you're picking at random, there's only a 1/3 chance you'll pick the coin with 2 different faces.



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Joined: 07 Jul 2004
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Sorry, forgot the log in when i replied. The post by 'guest' was actually me.










