2004 Georgia Tech Math Competition questions

[r3dxP]

Could somebody explain this question to me?
A 1/50 scale model of a pond is present. If the volume of the scale model is 4 cm^3, then the volume of the actual pond is..?
the answer is 1/2 m^3. I would like to know HOW.

Also..
A drawer has 6red socks and 6 white socks. If you reach in the drawer and randomly take out two socks, what is the chance(probability) that the two socks will match in color?
answer is 3/8.

 

[cepheid]

A volume of four cubic centimetres can be represented by a cube of edge length (4)^1/3 cm. 1/50 scale means 1 cm on the scale model corresponds to 50 cm in real life. So the edge length of the cube representing the volume of the real pond is: 50(4)^1/3 cm

The volume is then that length cubed:

125 000 (4) = 500 000 cm³

Divide by 1 million to get the volume in m³

Edit: the probabilty question is driving me nuts, but I really need to sleep. Can you please double check the answer? If it is 3/8 then...bummer. see you tomorrow.

 

[Curious3141]

Even easier for the first one, if two objects are similar (in scale) with a factor k,

The ratio of lengths is k.

The ratio of areas is k^2.

The ratio of volumes is k^3.

So you can immediately get the volume of the pond as {(50)^3}*(4) in cm^3, which you can then convert to m^3.

For the second one, I don't see how 3/8 can be correct.

The first sock choice doesn't matter. But after it's made, there are 5 socks of the same color and 6 of the other color in the drawer. The prob of drawing a matching sock is therefore 5/11, which should be the answer.

 

[cepheid]

Even easier for the first one, if two objects are similar (in scale) with a factor k,

The ratio of lengths is k.

The ratio of areas is k^2.

The ratio of volumes is k^3.

So you can immediately get the volume of the pond as {(50)^3}*(4) in cm^3, which you can then convert to m^3.

yes of course...neat!

For the second one, I don't see how 3/8 can be correct.

The first sock choice doesn't matter. But after it's made, there are 5 socks of the same color and 6 of the other color in the drawer. The prob of drawing a matching sock is therefore 5/11, which should be the answer.

YES! That's what I hoped, because I got that answer too. I did it by explicity counting the number of desired outcomes, i.e.:

2*(6 choose 2) / (12 choose 2) = 30/66 = 5/11

 

[moose]

There are 132 possible combinations, 60=pairs
5/11
Are you that it says 3/8 in your answer guide in the very back?
Make sure you have it corresponding to the right test year. Erm, other than that I don't see how it could be 3/8

 

[cepheid]

There are 132 possible combinations, 60=pairs
5/11

As I stated above, I got 66 possible combinations, and 30 of them pairs, using the binomial coefficients indicated. So we got the same answer in the end, but both your values seem to be double mine...

edit...looks like you were permuting...

12*11 = 132 pairings, if the order matters. But since 1st sock red, 2nd white is no different from 1st sock white, 2nd red...

 

[GCT]

You need to consider all possible outcomes. In that case white1, white2 will count as an outcome as well as white2, white1 but not the same.

 

[cepheid]

I disagree...besides, you're drawing out both socks at the same time. There's no way to even distinguish the order. But even if you were drawing them in succession, it still wouldn't make sense to consider the order, because you can get another pair of socks from every distinct pair just by switching the order in which they were drawn, and all that does is needlessly double both your sample space outcomes, and the desired outcomes. Those binomial coefficients describe the outcomes under consideration. There are (12 choose 2) = 66 ways to choose two of those socks from 12. That is the sample space. There are (6 choose 2) = 15 ways to choose two reds, and (6 choose 2) = 15 ways to choose two whites. That is the set of desired outcomes.

If you still don't believe me...then label the white socks w1 through w6. label the reds r1 through r6.

The pair w2, w4 is distinct from the pair w3, w5, but not from the pair w4, w2.

 

[GCT]

Take a look at the answer to this common problem which I encounteredd in the past.

An individual has two siblings and at least one of them is a girl, what is the probability that the individual has two sisters?

The answer is 1/3

hint: consider all possible outcomes

 

[GCT]

That is if there were a variation to this problem in which the question was similar to the above, than your method would have resulted in an incorrect answer.

The essence of probability problems is considering ALL possible outcomes; including all the dimensions of probability.

 

[cepheid]

Two siblings, labelled one and two:


One sibling is certain to be a girl. Let that be sibling #2. We have two possible outcomes:

1 B
2 G

1 G
2 G
__________________________________________________________
Another outcome is that sibling #1 is the girl for sure (after all, we only specified at least one was a girl, we didn't say which). So we got to add this case in:

1 G
2 B

Notice that 1G, 2G, is one out of three cases

Your example is not relevant here. Here, siblings 1 and 2 are different people, so you have to distinguish between them. In the case of the sock problem, I DID consider all possible outcomes. There are (12 choose 2) = 66 of them. stating that there are 132 outcomes possible is WRONG. Because unlike with the siblings, there is NO way to distinguish Sock A first, sock B 2nd from B 1st, A 2nd. So, stating that there are 132 possible outcomes is just as wrong as the stating the following: Say you drew the following sequence of cards in a game of poker:

(jack of hearts) (jack of spades) (some card A) (some card B) (some card C)

What if you had drawn this:

(jack of spades) (jack of hearts) (some card A) (some card B) (some card C)


You would have me say these are two distinct poker hands? Sorry, but no. You have to consider the particular problem in question. The binomial coefficients I used are correct here. On the other hand:

[1] IF we were drawing the socks one after the other, and
[2] IF there were some importance to the order (instead of just whether they matched or not, which is all we care about)

then I would write the number of ways of choosing two reds as (6 choose 1)*(5 choose 1) instead of (6 choose 2) = 15. But notice from the definitions that:

(6 choose 1)*(5 choose 1)

= $$\frac{6!}{(5!)(1!)} \frac{5!}{(4!)(1!)}$$

= $$\frac{6!}{4!}$$

= (6 permute 2) = 6*5 = 30

So in that case we would permute.

 

[GCT]

ok, let's not complicate things.

Because unlike with the siblings, there is NO way to distinguish Sock A first, sock B 2nd from B 1st, A 2nd.

this is just nonsense; the key error being distinguish; you're appealing to perception here, not reality.

NOTE:There are TWO independent events associated with this problem. You've heard of this term right (independent events)? I suggest that you find it in a probability text and read up on it.

If you disagree with the above proposition, than it seems that you're implicating a very strange form of the problem, frankly one in which I've never encountered and don't believe exists. So...First, describe to me exactly how the two socks are touched exactly at the same "point" in time; remember the person who chooses cannot look.

 

[cepheid]

ok, let's not complicate things.

Should I translate this as "I didn't read all of your last post, cepheid, I just found what I disagreed with and quoted it." :grumpy:

this is just nonsense; the key error being distinguish; you're appealing to perception here, not reality.

NOTE:There are TWO independent events associated with this problem. You've heard of this term right (independent events)? I suggest that you find it in a probability text and read up on it.

If you disagree with the above proposition, than it seems that you're implicating a very strange form of the problem, frankly one in which I've never encountered and don't believe exists. So...First, describe to me exactly how the two socks are touched exactly at the same "point" in time; remember the person who chooses cannot look.

I am well aware of what independent events are, and I am taking a third year course in probability theory, so there is no need to be condescending, or supersilious. We are just having a discussion here, it's nothing personal, so I am sorry if I got overzealous in my last post and capitalized a few words. But seriously...did you even read the argument I put forward? You have asserted that the condition [1] that I wrote must always be satisfied, i.e. the socks have to be drawn, or at least "touched" at two different times. Fine. But condition [2] is still not satisfied as far as I am concerned. We still don't give a damn in which order the socks were selected, only whether they match or not. The number of events in the sample space is the number of distinct pairs of socks that one could make with that set of twelve socks. What if some crazy guy came by, tore the drawer out of the chest of drawers and said, "Show me how many different pairs you can make with these socks! Take a photo each each pair after matching it up." It would be tough to do, and I'd have to label the socks as I described before, but I could do it, and I'd end up with 66 photographs. I'm not going to match up each pair twice...one with one of the socks on the left, and the other one with the other sock on the left. I still don't see what you are talking about.

The independent events you are referring to are the drawings of each sock? Can you enlighten me how that factors in?

Summary:

P(matching pair ) = P(2 reds) + P(2 whites)

= $$\frac{{6 \choose 2} + {6 \choose 2}}{{12 \choose 2}}$$

 

[GCT]

Your answer is correct. My answer and method is certainly not incorrect; it is the purest method of solving probability problems. Even if your method is correct, my method is still correct. Thus your method is a mere simplication, at most. The question is of the basis of your simplification and whether it is justified. It is wrong, simply because it if of the wrong procedure; the problem is simple enough already.

 

[GCT]

I'm just trying to get to the point, hope I didn't sound rude. I read your posts, however, at times you get a bit fuzzy with the details; also I felt that this is a basic probability problem, there's no need to go into thought experiments.

 

[Curious3141]

I'm just trying to get to the point, hope I didn't sound rude. I read your posts, however, at times you get a bit fuzzy with the details; also I felt that this is a basic probability problem, there's no need to go into thought experiments.

Why don't you just use my method - it's simple, direct and completely avoids any of these complications.

Naturally, for a more complex problem such a simple method would not work - but it's fine for this case.

 

[cepheid]

I'm just trying to get to the point, hope I didn't sound rude. I read your posts, however, at times you get a bit fuzzy with the details; also I felt that this is a basic probability problem, there's no need to go into thought experiments.

Okay, well the only reason I introduced the "thought experiment" was to further illustrate the point that only 66 distinct pairs of socks can be formed from the set of socks given. As to why you claim that the sample space of the experiment should be any larger than that...I still do not see. Since you have not addressed any of my points directly, I'm tired of arguing the issue. We will just have to agree to disagree.

 

[mathwonk]

there are 12 socks. the number of pairs is 12 taken 2 at a time, or 66.

the number of pairs of red socks is 6 taken 2 at a time, or 15. the number of chgartreuse scoks taken 2 at a a time is also 15, so there are 30 pairs the same color, ourt of 66, which is 30/66, or 5/11.

hence cepheid is right. however anyone who disagrees has my sympathy, but i decline to argue about it. these matters are notoriously hard to persuade people of.


by the way,
this is a high school level problem. what contest at ga tech had this?

 

[OneSquared]

Did we ever solve the original problem with the 3/8? I don't see how a problem like this could ever have an even number in the denominator...if you use 3 of each, you get 2/5, 4 of each, you get 3/7, if you use 5 of each, you get 4/9, etc., so I like 5/11 as the answer as well.

 

[cepheid]

Thanks for the confirmation mathwonk. I needed to know, one way or another.

Onesquared: the original poster never got back to us regarding the 3/8...so I guess we'll never know where that came from.

 

[abertram28]

Its interesting that someone mentioned this being high school work. Surprise, IT IS a high school competion. It is hosted and run by GATech, but its this question 4 from version A of the Varsity paper, or question 8 from the version B, and indeed, the answer sheet for both gives 5/11 as the answer.

http://www.math.gatech.edu/~pme/hsmc/past/