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Smart? (pg. 10)
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| Flyboy217 |
| quote: | Originally posted by Resnick
ok ok one last point and i promise ill shut up, your right, i havent taken stats
all im arguin is that you cant say something like:
inf/inf = 1
0/inf = 0
0/0= 0
n/inf=0
and stuff like that
one last thing, is stats an engineering course? what faculty is it from and what year course is it? |
It's true, you cannot use infinity as a number in an equation.
It is pretty strange that things with 0% probability of occurring can still occur... lots of things are strange about infinities. Who ever would think that there are "countable" and "uncountable" infinities before learning them?
We had to take stats for Computer Engineering, but I doubt you'll learn about this specifically. I'd suggest learning about measure theory, or some other related branches of probability. |
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| drizzt81 |
| quote: | Originally posted by Resnick
ok ok one last point and i promise ill shut up, your right, i havent taken stats
all im arguin is that you cant say something like:
inf/inf = 1
0/inf = 0
0/0= 0
n/inf=0
and stuff like that
one last thing, is stats an engineering course? what faculty is it from and what year course is it? |
yes you cannot:
inf/ inf is INDETERMINATE, i.e. can be anything and nothing, which relates to the 'size' of infinity. I remember a friend of mine always yelling "some infinities are bigger than others" which is true. in order to find out what inf/ inf is, you need to know the rate at which either functions approach infinity. And if you relate the integers to the irrational numbers, you will see for each integer step you add an INFINITE amount of irrational numbers, therefore the size of the irrational infinity grows much faster than the size of the integer infinity, therefore it must be much bigger. By how much? by infinity of course.
consider:
(3 * (all integers)) / (4 * (all integers)) = 3/4 only because the two infinities are the same.
now someone, please correct me and make me look like a dumb fool
edit spelling |
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| Flyboy217 |
| quote: | Originally posted by drizzt81
OK, sorry for jumping in on this, but i am a curious mind -sometimes and right now everything is better than more 8051 code ;) - so:
you say that by definition 1/(infinity) = 0, and that is the reason why the probability of picking any single number is zero, right?
the only reason we can say that 1/(infinity) is zero is using the limit concept on 1/n as n -> (infinity) ?
Do I understand this correctly? |
That's correct. This is not why it's true for picking a rational out of the reals, however. That has to do with the concept of uncountable infinities being of a higher-order cardinality than the countable ones. |
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| moncster |
lim x->(infinity) 1/x = 0
As X APPROACHES infinity, 1/x APPROACHES 0. What 1/x really equals when x is EQUAL to infinity is irrelevant. |
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| Flyboy217 |
| quote: | Originally posted by drizzt81
yes you cannot:
inf/ inf is INDETERMINATE, i.e. can be anything and nothing, which relates to the 'size' of infinity. I remember a friend of min always yelling "some infinities are bigger than others" which is true. in order to find out what inf/ inf is, you need to know the rate at which either functions approach infinity. And if you relate the integers to the irrational numbers, you will see for each integer step you add an INFINITE amount of irrational numbers, therefore the size of the irrational infinity grows much bigger than the size of the integer infinity, therefore it must be much bigger. By how much? by infinity of course.
consider:
(3 * (all integers)) / (4 * (all integers)) = 3/4 only because the two infinities are the same.
now someone, please correct me and make me look like a dumb fool
edit spelling |
You're wrong, you idiot... it's "friend of mine" not "friend of min"
;-)
Yes, you're right. |
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| Resnick |
k now im confused, let me ask this example again
u have 1 red cup, and infinity blue cups
what is the probability of choosing a red cup?
is the answer dependant on a limit?
and im not trying to argue anything anymore, just wanna see if this has anything to do with the countable/uncountable inifinities (which i havent learned)
everyone feel free to comment |
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| Resnick |
| quote: | Originally posted by Flyboy217
Probability(contains a three) = lim x -> inf (1-.9^x) = 1
Yes, it's a limit, and that limit is *exactly equal* to 1.
Yes, outcomes with 0% probability of occurring can occur when dealing with infinities.
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woah, now this is really confusing (btw sry i just read this)
my whole argument is that its a limit and ppl are saying its not, and now your saying it IS a limit? |
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| Flyboy217 |
| quote: | Originally posted by Resnick
k now im confused, let me ask this example again
u have 1 red cup, and infinity blue cups
what is the probability of choosing a red cup?
is the answer dependant on a limit?
and im not trying to argue anything anymore, just wanna see if this has anything to do with the countable/uncountable inifinities (which i havent learned)
everyone feel free to comment |
The problem is that there is no such thing as "infinity" blue cups. However, we can represent this problem by using the set of integers, where the red cup represents some integer (say, zero), and the blue cups are all the other integers.
The probability of picking red is (exactly) zero. It does come from our knowledge of limits. If we had 1 red cup and x blue cups, then the probability would be 1/x. As you know, lim x->inf 1/x = 0. As has been pointed out, it doesn't make sense to say "but what is it when x = infinity" because there is no such number. However, knowing that the cardinality of the real numbers is "countably infinite," we can answer that the probability equals the limit, which is exactly equal to zero.
One problem many people have is when they say that "the limit approaches zero." This is incorrect. The limit doesn't approach zero; the limit IS zero (as the free variable x approaches infinity).
In the cups case, the probability of choosing any item of a finite set from an uncountably infinite one is zero. As noted above, though, this is different from the case of picking a rational from the reals. In this case, the probability of picking an item from a countably infinite-sized one from an uncountably-sized one is zero.
This last point has to do with the "measure" of a set. The measure of a particular interval is equal to the difference of its max and min elements. The probability of picking an integer from [1,2] out of [1,3] is
P(Picking something from [0,1] out of [0,2]) =
Measure [0,1] / Measure [0,2] = 1-0/2-0 = 1/2
The measure of any particular point, x, is x-x = 0. The measure of a countably infinite set (such as the integers or rationals) is actually zero.
P(Picking a rational from the real interval [0,2]) =
Measure (rationals) / Measure [0,2] = 0/(2-0) = 0
To make things weirder, there are sets of uncountably infinite size that also have measure zero (read up on the Cantor set).
Hope this helped. |
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| drizzt81 |
| quote: | Originally posted by Flyboy217
5) You walk into a candy store in a foreign land that has only 2 types of coins. The coins are worth A and B, where A and B are relatively prime (share no common factors other than 1). Assuming you have an infinite number of each coin, what is the price of the most expensive candy that you cannot buy with exact change? That is, find the largest price such that you will require change to purchase it exactly. [20 min.] |
solution approach: you know that you CAN buy any candy of cost C, where C = xA+yB [x,y are integers >= 0]
Let's assume that B < A. This will not loose any generality, since it doesn't matter which one is bigger, but one HAS to be, seeing that they do not share common factors, hence are not the same number.
Therefore anything that costs C, C element of (B, thelesserof(2B, A) cannot be bought.
reconsidering:
B/A = q (a number), we can buy anyting that costs
C=B(y + x/q), we know that q < 1, because A > B
q is the 'step' in our function.
-scrap all that-
ok.. i looked at some examples:
The largest price that you canot pay in correct change is:
the product of the first prime number larger than the big number times the first prime number smaller than the smaller number.
why? Because that number will not be a linear combination of the two numbers that you are using... |
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| Flyboy217 |
| quote: | Originally posted by drizzt81
solution approach: you know that you CAN buy any candy of cost C, where C = xA+yB [x,y are integers >= 0]
Let's assume that B < A. This will not loose any generality, since it doesn't matter which one is bigger, but one HAS to be, seeing that they do not share common factors, hence are not the same number.
Therefore anything that costs C, C element of (B, thelesserof(2B, A) cannot be bought.
reconsidering:
B/A = q (a number), we can buy anyting that costs
C=B(y + x/q), we know that q < 1, because A > B
q is the 'step' in our function.
-scrap all that-
ok.. i looked at some examples:
The largest price that you canot pay in correct change is:
the product of the first prime number larger than the big number times the first prime number smaller than the smaller number.
why? Because that number will not be a linear combination of the two numbers that you are using... |
A = 5, B = 3. Then you get 7*2 = 14. But 14 = 3*3+5*1. Try again :). Think modular arithmetic. |
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| Resnick |
| k cool, thanks for the help flyboy and diginut |
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| DigiNut |
| quote: | Originally posted by Flyboy217
A = 5, B = 3. Then you get 7*2 = 14. But 14 = 3*3+5*1. Try again :). Think modular arithmetic. |
My knowledge of modular arithmetic is extremely limited and I don't even know the correct terminology, so maybe you could help me along here with my assumptions... unless you think it's totally giving it away.
If we have two primes A and B, then our mod base ought to be A+B. That is to say, we *know* that you can make exact change C with coins A and B if you can make exact change for (C % (A+B)). However, the reverse isn't automatically true, so I'm not sure where to go from there... let's say you take 5 and 7, which is mod 12, then you can't make 11, you can't make 23, but you can make 35...
I think that at the very least, the answer must be some function of A+B, and the largest prime less than (A+B-1)? Or maybe I'm on the wrong track... I don't have time to read all about modular arithmetic. :p |
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