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[MisterLuffy]

I have a question:

Suppose the expected tensile strength of type-A steel is 105 ksi and the standard deviation of tensile strength is 8 ksi. For type-B steel, suppose the expected tensile strength and standard deviation of tensile strength are 100 ksi and 6 ksi, respectively. Let Xbar = the sample average tensile strength of a random sample of 40 type-A specimens, and let Ybar = the sample average tensile strength of a random sample of 35 type-B specimens.

a. What is the approximate distribution of Xbar? Of Ybar?

b. What is the approximate distribution of Xbar - Ybar? Justify your answer.

c. Calculate (approximately) P(-1 <= (Xbar - Ybar) <= 1).

d. Calculate P( (Xbar- Ybar) >= 10). If you actually observed (Xbar - Ybar) >= 10, would you doubt that u1 - u2= 5?

 

[Harpuia]

Some relatively simply theoretical questions on variance/Statistics. I was wondering if anyone could nudge me in the proper direction, kinda having a bit of an issue trying to understand what the question wants me to solve for.

Let R be a nonnegative integer valued random variable.

A) If Expectation of R = 1, how large can Variance of R be?

B) If R is always positive (nonzero), how large can Expectation [1/R] be?

I know that Var[R] = Ex[R^2] - Ex^2[R], but it turns up to be Ex[R^2] - 1. I'm not sure what to do about that Ex[R^2] though...

And for B, I also wrote out the expectation of R, and got the summation of all outcomes in R, of 1/x * Pr[R = x] where x comes from the range of R.

For B, since we want to know how large Ex[1/R] may be, I decided to observe the limit of the sum of 1/x, which is just 0. I think that's looking in the completely wrong direction though...

 

[MoG EclipsE]

This is an Eco question, hope it counts as a math question :

I know the formula for Annual Equivalent Worth is P(A/P,i%,N) - SV(A/F,i%,N) + (Annual revenues-Annual operating costs)

But how do I use this when Overhaul is a factor at the end of a certain year? Thanks!

 

[Mabef]

Probability question! I have 25 cards, each with numbers on them. Five 1 cards, fifteen 2 cards, and five 3 cards.

1 1 1 1 1   2 2 2 2 2  2 2 2 2 2  2 2 2 2 2   3 3 3 3 3

I randomly draw 5 cards out of here (no repeats). I calculated there to be 53,130 different hands. I used nCr, or n! / (r! * (n-r)!).

I want to determine how likely I am to draw a hand that adds up to amount n.

I figure step one is to find out how many different hands can add up to n. For instance, there are two hands (1,1,1,2,2 or 1,1,1,1,3) that add up to 7. I don't know how to figure this part out, other than by hand.

After I figure out how many hands add up to n, I'll determine those hands' frequency using nCr. For instance, ((5 choose 3) * (15 choose 2)) + ((5 choose 4) * (15 choose 1)) = 1075.

1075/53130= 2.023% chance of drawing a hand that adds up to 7.

 

[cpp_is_king]

Probability question! I have 25 cards, each with numbers on them. Five 1 cards, fifteen 2 cards, and five 3 cards.

1 1 1 1 1   2 2 2 2 2  2 2 2 2 2  2 2 2 2 2   3 3 3 3 3

I randomly draw 5 cards out of here (no repeats). I calculated there to be 53,130 different hands. I used nCr, or n! / (r! * (n-r)!).

I want to determine how likely I am to draw a hand that adds up to amount n.

I figure step one is to find out how many different hands can add up to n. For instance, there are two hands (1,1,1,2,2 or 1,1,1,1,3) that add up to 7. I don't know how to figure this part out, other than by hand.

After I figure out how many hands add up to n, I'll determine those hands' frequency using nCr. For instance, ((5 choose 3) * (15 choose 2)) + ((5 choose 4) * (15 choose 1)) = 1075.

1075/53130= 2.023% chance of drawing a hand that adds up to 7.

For the second part, you do indeed need to do it by hand. There is no simple expression for integer partitions, which is essentially what you're looking for. I think the easiest, least error prone way is to list each sum from 5 to 15, and then all the triplets that result in that sum, where a triplet (a,b,c) means that you draw a 1's, b 2's, and c 3's.

Sum = 5:
* (5, 0, 0)
Sum = 6:
* (4, 1, 0)
Sum = 7:
* (4, 0, 1)
* (3, 2, 0)
Sum = 8:
* (3, 1, 1)
* (2, 3, 0)
Sum = 9:
* (3, 0, 2)
* (2, 2, 1)
* (1, 4, 0)
Sum = 10:
* (1, 3, 1)
* (0, 5, 0)
* (2, 1, 2)
* (3, 2, 1)

Etc.

From here, it's easy to figure out the number of ways to get each of these results by summing (5; a) * (15;b) * (5;c) over each of the answers.

 

[Kimosabae]

Hey.

Somehow, when evaluating this compound interest problem, solving for t, I end up with 4.13 instead of 4.11. Anyone got any ideas as to what I could be doing wrong?

ln(1.5)/4 * ln(1 + 0.10/4)

No matter what the example, I get the right whole number but my decimals are off. Where am I going wrong in my evaluation?

 

[VT_Brian]

Hey.

Somehow, when evaluating this compound interest problem, solving for t, I end up with 4.13 instead of 4.11. Anyone got any ideas as to what I could be doing wrong?

ln(1.5)/4 * ln(1 + 0.10/4)

No matter what the example, I get the right whole number but my decimals are off. Where am I going wrong in my evaluation?

This is what I got in my calculator and I assume you meant to put parentheses around the (4 * ln(1 + 0.10/4))

 

[The Lamp]

Hey.

Somehow, when evaluating this compound interest problem, solving for t, I end up with 4.13 instead of 4.11. Anyone got any ideas as to what I could be doing wrong?

ln(1.5)/4 * ln(1 + 0.10/4)

No matter what the example, I get the right whole number but my decimals are off. Where am I going wrong in my evaluation?

Sorry everything is kind of out of context for me but...I don't see a t in that expression? When I put it in wolframalpha it gives me 0.0025...

 

[Kimosabae]

This is what I got in my calculator and I assume you meant to put parentheses around the (4 * ln(1 + 0.10/4))

Why the parentheses around the denominator? I've been evaluating inside the denominator's parentheses first, getting 1.025, then hitting 'ln', then multiplying by 4, to get 0.098. The numerator I'd simply hit 1.5, then 'ln' giving me .405. .404/.098 gives me 4.13. Why is that method wrong?


@Lamp: Sorry, t = ln(1.5)/4 * ln(1 + 0.10/4). That's the final part of the evaluation.

 

[Ghizz]

about to head to college and want to touch up everything from basic algebra to calculus 1/2. any recommended resources?

 

[VT_Brian]

Why the parentheses around the denominator? I've been evaluating inside the denominator's parentheses first, getting 1.025, then hitting 'ln', then multiplying by 4, to get 0.098. The numerator I'd simply hit 1.5, then 'ln' giving me .405. .404/.098 gives me 4.13. Why is that method wrong?


@Lamp: Sorry, t = ln(1.5)/4 * ln(1 + 0.10/4). That's the final part of the evaluation.

You're doing the problem right, on your calculator, but you can't write it like that due to the order of operations otherwise it would be ln(1.5)/4 multiplied by ln(1+0.10/4)

Your problem comes from rounding I would expect 0.4054651081/0.0987704504 gives a different answer than .404/0.098

 

[Kimosabae]

You're doing the problem right, on your calculator, but you can't write it like that due to the order of operations otherwise it would be ln(1.5)/4 multiplied by ln(1+0.10/4)

Your problem comes from rounding I would expect 0.4054651081/0.0987704504 gives a different answer than .404/0.098

Ah, you're right. Any ideas on how I can evaluate the denominator on my calculator first without having to write it down in its entirety, clear the answer, evaluate the numerator, write that down in its entirety; then divide both 15 digit decimals to get the most accurate quotient to round to two decimal places? This seems so arbitrary to me. Or am I just rounding certain steps incorrectly or something?

 

[VT_Brian]

Ah, you're right. Any ideas on how I can evaluate the denominator on my calculator first without having to write it down in its entirety, clear the answer, evaluate the numerator, write that down in its entirety; then divide both 15 digit decimals to get the most accurate quotient to round to two decimal places? This seems so arbitrary to me.

You honestly don't need that many digits, haha. What kind of calculator do you have? Even this calculator should let you evaluate the entire expression at one time. You could even just use the calculator built in Google like I linked earlier if all else fails. On a really basic calculator you could store the denominator in memory, then evaluate the numerator, then divide by memory recall

 

[Kimosabae]

You honestly don't need that many digits, haha. What kind of calculator do you have? Even this calculator should let you evaluate the entire expression at one time. You could even just use the calculator built in Google like I linked earlier if all else fails. On a really basic calculator you could store the denominator in memory, then evaluate the numerator, then divide by memory recall

I have an iPhone 4S and am using the standard calc app. Didn't think I needed that many digits but it's the first time since I've been dealing with logs that I've run into this problem: typically, rounding to the 1000th's place has been sufficient. That was the only problem that seems to work only if I use digits passed that. I'm so annoyed right now. Wasted more than an hour trying to figure that out.

 

[Mabef]

For the second part, you do indeed need to do it by hand. There is no simple expression for integer partitions, which is essentially what you're looking for. I think the easiest, least error prone way is to list each sum from 5 to 15, and then all the triplets that result in that sum, where a triplet (a,b,c) means that you draw a 1's, b 2's, and c 3's.

[snip]

From here, it's easy to figure out the number of ways to get each of these results by summing (5; a) * (15;b) * (5;c) over each of the answers.

Thanks bud, integer partitions is exactly what I was looking for. I hate when I know there's a term for the math I'm trying to do but I can't remember what it is.

I'll follow your advice and count out triplets (although I simplified my numbers a bit for GAF, it'll probably take a couple hours to do the real deal).

about to head to college and want to touch up everything from basic algebra to calculus 1/2. any recommended resources?

Do you mean besides illustrious Khan Academy?

 

[Orbis Tabula]

about to head to college and want to touch up everything from basic algebra to calculus 1/2. any recommended resources?

As said above, Khan Academy is a godsend. Absolutely unparalleled in teaching math if you ask me. I actually taught myself calculus using it instead of learning it in high school and I teach calc now, so you could say it works. It's only gotten better since I used to use it. There are tons of videos and interactive examples. You can start an account and work through a whole learning plan with examples and practice sets, watch playlists covering an entire topic, or just skip around and look at whichever videos you'd like for a quixk review. I can't recommend Khan academy enough.



As for my question, I'm having trouble doing Matrix Exponentials. I've got an answer to a problem that I feel pretty confident about, but when I double check it on Wolfram|Alpha, I get different answers depending on how I do it.

I'm given the matrix
[[1,0,0],
[1,2,0], = A
[1,2,3]]
And I need to calculate e^(At).
My procedure was to use diagonalization and get P^-1*A*P and then I could change e^(At) by P*e^(Bt)*P^-1. And e^(Bt) is diagonal, so easy to compute. For my answer, I get something complicated, but it matches what I get from Wolfram|Alpha when I input "matrixexp(t {{1,0,0},{1,2,0},{1,2,3}})" however when I try to do P*e^(Bt)*P^-1 using P matrix I've found, I get a different answer. Could anyone help me figure out which one is correct?

The answer that I think is right is this.

[e^t                                0                    0]
[e^(2t)-e^t                         e^(2t)               0]
[(3/2)e^(3t)-2e^(2t)+(1/2)e^t       2e^(3t)-2e^(2t)      0]

 

[Sajjaja]

Okay this is from my Intro to Econometrics class:
__________
a) Suppose you have two random variables, X and Y , with N=2, and so two observations on each, written as x1,x2, y1, y2. (the numbers are subscripts)

Can Summation (i=1 to N) (Xi + Yi) = Summation (i=1 to N) (XiYi) in this case? If so under what conditions?

b) From part a) suppose N can equal any position number. Can the two sides equal? If so, under what general conditions?
___________

For part a), I only see them equating when all variables equal 0, but I feel it's probably a sufficient condition and not a necessary one. Any ideas?

 

[Perfect Cha0s]

I'm back with another stats question. I made a normal distribution sample with a pop. mean of 25, pop. standard deviation of 10, and sample size of n=50.

From that sample I computed a sample mean and sample std dev, and was then asked to created a normal curve with Mu = 25, and Sigma = s/sqrt(N). In this case, since s is approximately the original sigma (which was 10), the new sigma ends up being very small, yielding a very narrow, clustered normal curve.

On this new, narrow normal curve, I'm asked to compute a 95% confidence interval.

I did so using 25 - (1.9599*(sigma/sqrt(50))) < xbar < 25 + (1.9599*(sigma/sqrt(50))), where sigma = s/sqrt(N) but this yields a very, very narrow range of values: 24.6 to 25.4

This 95% interval contains the population mean, but does not contain the sample mean, so I'm confused as to what information exactly this new curve and confidence interval are supposed to convey.

I was also curious as to whether or not the confidence interval formula I used is correct or not; If sigma = s/sqrt(N), and inside the formula I compute sigma/sqrt(N), I'm actually computing mu +/- Z*(s/N) in the formula, which sent up red flags.

Thanks in advance.

 

[Jokab]

I'm back with another stats question. I made a normal distribution sample with a pop. mean of 25, pop. standard deviation of 10, and sample size of n=50.

From that sample I computed a sample mean and sample std dev, and was then asked to created a normal curve with Mu = 25, and Sigma = s/sqrt(N). In this case, since s is approximately the original sigma (which was 10), the new sigma ends up being very small, yielding a very narrow, clustered normal curve.

On this new, narrow normal curve, I'm asked to compute a 95% confidence interval.

I did so using 25 - (1.9599*(sigma/sqrt(50))) < xbar < 25 + (1.9599*(sigma/sqrt(50))), where sigma = s/sqrt(N) but this yields a very, very narrow range of values: 24.6 to 25.4

This 95% interval contains the population mean, but does not contain the sample mean, so I'm confused as to what information exactly this new curve and confidence interval are supposed to convey.

I was also curious as to whether or not the confidence interval formula I used is correct or not; If sigma = s/sqrt(N), and inside the formula I compute sigma/sqrt(N), I'm actually computing mu +/- Z*(s/N) in the formula, which sent up red flags.

Thanks in advance.

What is the sample mean that you calculated? Also, if I'm getting this right, you have the population mean and stddev, so what are you calculating? Or is the task simply to construct a 95% confidence interval?

In any case, you've almost got the confidence interval formula down, except that the last factor is simply s/sqrt, where n is your sample size (50) and s is the population stddev. The population stddev approximation (standard error of the mean) for the sample is calculated as a part of the confidence interval formula, I think you're doing it twice. (I think at least, I'm actually taking statistics intro right now too).

So the calculation would be (where xbar is the sample mean you found):
xbar +/- 1.96*(10/sqrt(50))

Does that seem more correct? Correct me if I'm wrong anyone, as said I'm not experienced in this at all.

EDIT: I see now you used 25 for the calculation, but surely that is not the sample mean you found? That would be an insane sample.

 

[Perfect Cha0s]

What is the sample mean that you calculated? Also, if I'm getting this right, you have the population mean and stddev, so what are you calculating? Or is the task simply to construct a 95% confidence interval?

In any case, you've almost got the confidence interval formula down, except that the last factor is simply s/sqrt, where n is your sample size (50) and s is the population stddev. The population stddev approximation (standard error of the mean) for the sample is calculated as a part of the confidence interval formula, I think you're doing it twice. (I think at least, I'm actually taking statistics intro right now too).

So the calculation would be (where xbar is the sample mean you found):
xbar +/- 1.96*(10/sqrt(50))

Does that seem more correct? Correct me if I'm wrong anyone, as said I'm not experienced in this at all.

EDIT: I see now you used 25 for the calculation, but surely that is not the sample mean you found? That would be an insane sample.

I don't have the file on me at the moment, but the sample mean was something like 26.8, and sample std dev was 10.3.

 

[Jokab]

I don't have the file on me at the moment, but the sample mean was something like 26.8, and sample std dev was 10.3.

You already have the stddev of the population; there is no need to calculate it for the sample - it'll only be a worse estimator than what you already have. Anyway, plugging 26.8 into the formula:

26.8 +/- 1.96 * (10/sqrt(50)) = [24.02, 29.57].

What this says in practice is that you're 95% sure, from the sampling data, that the true mean of the population is within that interval, which we know it is.

 

[Perfect Cha0s]

You already have the stddev of the population; there is no need to calculate it for the sample - it'll only be a worse estimator than what you already have. Anyway, plugging 26.8 into the formula:

26.8 +/- 1.96 * (10/sqrt(50)) = [24.02, 29.57].

What this says in practice is that you're 95% sure, from the sampling data, that the true mean of the population is within that interval, which we know it is.

Hm, thanks. That stills leaves me with the question of what to do with the whole sigma = s/sqrt thing. Would I further plug that into the sigma/sqrt?

 

[Orbis Tabula]

Questions about systems of ODEs. I've given this problem.

dx/dt=At where A = {{0,2,0},{-2,0,0},{2,0,6}} and I need to solve the system. Can someone just help me get started or tell me if I'm going the right way. I calculated e^(At) by diagonalizing A. Is that correct and where do I go from there?

 

[Jokab]

Hm, thanks. That stills leaves me with the question of what to do with the whole sigma = s/sqrt thing. Would I further plug that into the sigma/sqrt?

Consider a population with mean mu and stddev s. You can create a normal distribution of this population with N(mu, s^2).

Now consider a sample of this population, with mean xbar and stddev sigma. You now have two options for creating a normal distribution for this sample.

If you don't know the stddev of the population, you can approximate it with the sample stddev, using sigma/sqrt.

If you DO know the stddev of the population (which we do in this case, its value is 10), you can use that instead of the sample stddev, so you get s/sqrt. This gives a distribution of N(xbar, s/sqrt).

(Once again, I might be talking out of my ass here, but this thinking has worked for me thus far (exams next thursday, we'll see how correct I am))

 

[Kimosabae]

nvm. I figured it out. My Area equation was wrong.

 

[TUSR]

Questions about systems of ODEs. I've given this problem.

dx/dt=At where A = {{0,2,0},{-2,0,0},{2,0,6}} and I need to solve the system. Can someone just help me get started or tell me if I'm going the right way. I calculated e^(At) by diagonalizing A. Is that correct and where do I go from there?

Find eigen values for your 3x3.

I got lambda = 6, 2i, -2i

Find your eigen vector for each of those. The distinct real root will be easy, and 2i and -2i are okay to find.

x(t) = ...

 

[Memento_Mori15]

Okay, I need help with this problem. Reviewing for my final and I don't get how to solve this:

This was on last year's final. I know how to get the exact number of nodes and that's 132 nodes. However, I don't know how to get the probability of A and B. Can anyone help?

The way I did this before in past hw assignments was with brute force. So I don't know how to approach it

 

[leroidys]

Okay, I need help with this problem. Reviewing for my final and I don't get how to solve this:

This was on last year's final. I know how to get the exact number of nodes and that's 132 nodes. However, I don't now how to get the probability of A and B. Can anyone help?

The way I did this before in past hw assignments was with brute force. So I don't know how to approach it

Can you post your brute force solution? If it's what I think it is, you basically have the answer.

EDIT: So for event A, think about a rooted binary tree, and what different subtrees it can have. The possibilities are:

[B]Key[/B]: (root (left subtree) (right subtree))

[B]a[/B] (1(0)(5))
[B]b[/B] (1(5)(0))
[B]c[/B] (1(1)(4))
[B]d[/B] (1(2)(3))
[B]e[/B] (1(3)(2))
[B][B]f[/B][/B] (1(4)(1))
[B]g[/B] (1(5)(0))

That's all the possibilities, right? And we're only concerned with one of these, namely e. How many ways can e occur? Think about what we just did in the previous step and it should get you close to an answer.

 

[Memento_Mori15]

Can you post your brute force solution? If it's what I think it is, you basically have the answer.

EDIT: So for event A, think about a rooted binary tree, and what different subtrees it can have. The possibilities are:

[B]Key[/B]: (root (left subtree) (right subtree))

[B]a[/B] (1(0)(5))
[B]b[/B] (1(5)(0))
[B]c[/B] (1(1)(4))
[B]d[/B] (1(2)(3))
[B]e[/B] (1(3)(2))
[B][B]f[/B][/B] (1(4)(1))
[B]g[/B] (1(5)(0))

That's all the possibilities, right? And we're only concerned with one of these, namely e. How many ways can e occur? Think about what we just did in the previous step and it should get you close to an answer.

The way I looked at it was there are 5 ways to arrange a 3 node binary tree. If there are 6 nodes, there will be 3 nodes on the left, three nodes on the right. They each can be arranged 5 ways. 5*5=25. Not sure if the logic is correct but the answer is right.

 

[leroidys]

The way I looked at it was there are 5 ways to arrange a 3 node binary tree. If there are 6 nodes, there will be 3 nodes on the left, three nodes on the right. They each can be arranged 5 ways. 5*5=25. Not sure if the logic is correct but the answer is right.

That's not quite right. You're not counting the root as a node, so root + 3 on the left + 3 on the right would be a 7 node binary tree, not a 6 node.

Including the root, if the left subtree has 3, the only arrangements in the right subtree consist of 2-node subtrees. 3 + 2 + 1 = 6. Your combinatorial logic is right though.

For event B, you're solving the same thing but with a more general constraint (yet more specific than the amount of all 6 node binary trees). Specifically, each successive subtree has to have subtrees of 0 and n-1 to achieve that maximum height.

 

[Memento_Mori15]

That's not quite right. You're not counting the root as a node, so root + 3 on the left + 3 on the right would be a 7 node binary tree, not a 6 node.

Including the root, if the left subtree has 3, the only arrangements in the right subtree consist of 2-node subtrees. 3 + 2 + 1 = 6. Your combinatorial logic is right though.

For event B, you're solving the same thing but with a more general constraint (yet more specific than the amount of all 6 node binary trees). Specifically, each successive subtree has to have subtrees of 0 and n-1 to achieve that maximum height.

You would be right, so the answer sheet is wrong. Was wondering why the 7 node binary tree would have the same answer.

For B, the way I thought about it is by drawing it out:

                  o
               o
             o
          o
        o
      o
                   o
               o
             o
          o
        o
          o
                  o
               o
             o
          o
           o
          o
                  o
               o
             o
          o
           o
              o

And as we can see, at height 5, they can alternate. So there are 2 ways to do this for each node so 2^5. Will it always be 2^h?

 

[leroidys]

You would be right, so the answer sheet is wrong. Was wondering why the 7 node binary tree would have the same answer.

For B, the way I thought about it is by drawing it out:

                  o
               o
             o
          o
        o
      o
                   o
               o
             o
          o
        o
          o
                  o
               o
             o
          o
           o
          o
                  o
               o
             o
          o
           o
              o

And as we can see, at height 5, they can alternate. So there are 2 ways to do this for each node so 2^5. Will it always be 2^h?

I believe that's right for h=n-1

 

[Memento_Mori15]

I believe that's right for h=n-1

Do you consider the root the be part of that height? Because I consider that to be height zero. Then after the second node is inserted, that would be height 1.

Edit: And the answer to the binary tree would be 10. 5 ways to arrange 3. 2 ways to arrange 2. So that logic works then.

 

[leroidys]

Do you consider the root the be part of that height? Because I consider that to be height zero. Then after the second node is inserted, that would be height 1.

Edit: And the answer to the binary tree would be 10. 5 ways to arrange 3. 2 ways to arrange 2. So that logic works then.

Yeah root is considered 0 for height. I usually just count edges to get height so I won't accidentally be off by one.

 

[Memento_Mori15]

Yeah root is considered 0 for height. I usually just count edges to get height so I won't accidentally be off by one.

Just making sure. Thanks for the help.

 

[Memento_Mori15]

Edit: Nvm, solved it. Used the wrong formula Dx

 

[perfectchaos007]

Guys, I made this post in the Lottery thread and at least half the posters think I'm wrong. I don't see how its flawed, but you can see their rebuttals starting with post 254. Post 258 agrees with post 254


This was my post:

Technically it does. Your odds of winning the jackpot go from 1 in ~300 million to 1 in ~150 million if you buy 2 tickets. If you buy 4 tickets, your odds are ~1 in 75 million. If you buy 8 tickets, your odds are 1 in 37.5 million etc.

Think about it on a smaller level and it makes sense. Lets say you're playing a lotto game where you pick 1 number 0-9. Your odds of winning are 1 in 10. If you buy 2 tickets your odds of winning are 1 in 5. If you buy 5 tickets, your odds of winning are 1 in 2. ( all assuming you're picking different numbers of course )

Am I not getting something here?

 

[RevoDS]

Guys, I made this post in the Lottery thread and at least half the posters think I'm wrong. I don't see how its flawed, but you can see their rebuttals starting with post 254. Post 258 agrees with post 254


This was my post:



Am I not getting something here?

I don't think so, I'm with you there. If your ticket is a 1 in 300 million chance, then your second ticket brings your overall chances to 2 in 300 million (or 1 in 150 million, the exact same odds) which is double the chances.

I think people are confused by the infinitesimally small odds. A twice nothing is still nothing kind of thing.

 

[cpp_is_king]

Guys, I made this post in the Lottery thread and at least half the posters think I'm wrong. I don't see how its flawed, but you can see their rebuttals starting with post 254. Post 258 agrees with post 254


This was my post:



Am I not getting something here?

Your'e both right. It does indeed double your odds (assuming you pick different tickets), but #254 is also correct in that the amount that actually increases your odds is extraodinarily miniscule.

Your simplification of the problem is also accurate. If you buy 1 ticket where the odds are 1/10, your odds are 1/10. If you buy 2, your odds are 2/10, which is 1/5.

 

[perfectchaos007]

Your'e both right. It does indeed double your odds (assuming you pick different tickets), but #254 is also correct in that the amount that actually increases your odds is extraodinarily miniscule.

Your simplification of the problem is also accurate. If you buy 1 ticket where the odds are 1/10, your odds are 1/10. If you buy 2, your odds are 2/10, which is 1/5.

Thanks for the clarification guys!

 

[Partial Gamification]

Mega millions has pick five from 75 and one from 15.

75 options first pick, 74 seond, ..., 71 for the fifth ball, and 15 for the sixth ball.

75*74*73*72*71*15 = 31,066,902,000 ways to pick winning numbers.

The lottery has been reffered to as a tax on people that don't know math.

One-in-300M is about one out of the US population. 1-in-31B is one of fivefour(+)-times-the-population-of-the-planet.

edit: check my math, it could be wrong.

 

[Memento_Mori15]

Mega millions has pick five from 75 and one from 15.

75 options first pick, 74 seond, ..., 71 for the fifth ball, and 15 for the sixth ball.

75*74*73*72*71*15 = 31,066,902,000 ways to pick winning numbers.

The lottery has been reffered to as a tax on people that don't know math.

One-in-300M is about one out of the US population. 1-in-31B is one of fivefour(+)-times-the-population-of-the-planet.

edit: check my math, it could be wrong.

I thought it would be choose 75 5 *choose 15 1=17259390 *15=258890850. I could be wrong since I suck at counting.But I thought in how many ways can I choose 5 numbers out of 75? 1 from 15? And thus that.

 

[Partial Gamification]

I thought it would be choose 75 5 *choose 15 1=17259390 *15=258890850. I could be wrong since I suck at counting.But I thought in how many ways can I choose 5 numbers out of 75? 1 from 15? And thus that.

Yeah 'counting without counting' is a bitch. I think dividing by 5! might be the rigth odds. there is no repetition, and order is not important... I'm too scatter-brained.

 

[Kieli]

Can someone clarify what this means?

Concerning the 2D-coordinate system, I need to find a "circle in a plane centered at (x,y)".

What the hell does the part "in a plane" mean?

 

[Leezard]

Can someone clarify what this means?

Concerning the 2D-coordinate system, I need to find a "circle in a plane centered at (x,y)".

What the hell does the part "in a plane" mean?

http://www.thefreedictionary.com/plane

It's on a flat 2D surface.

 

[Kieli]

http://www.thefreedictionary.com/plane

It's on a flat 2D surface.

Thanks, but I already know what plane means. ;__;

I guess I should rephrase my question. We are only working in 2 dimensions (the question states R^2). Is there any special meaning to stating the that the circle must lie in a plane when the only possible plane of interest is the xy-plane (or whatever we call our two axes)?

Edit: Also, is it possible for a derivative of a vector function to be in three-dimensions while it, itself, is only 2D?

 

[alatif113]

Can someone clarify what this means?

Concerning the 2D-coordinate system, I need to find a "circle in a plane centered at (x,y)".

What the hell does the part "in a plane" mean?

Can you state the specific question word for word as stated by your instructor or textbook?

 

[Ourobolus]

Thanks, but I already know what plane means. ;__;

I guess I should rephrase my question. We are only working in 2 dimensions (the question states R^2). Is there any special meaning to stating the that the circle must lie in a plane when the only possible plane of interest is the xy-plane (or whatever we call our two axes)?

Edit: Also, is it possible for a derivative of a vector function to be in three-dimensions while it, itself, is only 2D?

Doesn't sound like there is a difference as far as the plane is concerned. If it was a 3D figure it should have 3 coordinates for the origin.

Also I don't believe you can derive something 3D from a 2D vector.

 

[RevoDS]

Can someone clarify what this means?

Concerning the 2D-coordinate system, I need to find a "circle in a plane centered at (x,y)".

What the hell does the part "in a plane" mean?

Since we're talking about R^2, I think it's just a poorly written question. By definition, anything in R^2 IS in a plane, you just can't specify which one because otherwise you're adding a third coordinate and therefore it becomes part of R^3.

 

[TUSR]

Since we're talking about R^2, I think it's just a poorly written question. By definition, anything in R^2 IS in a plane, you just can't specify which one because otherwise you're adding a third coordinate and therefore it becomes part of R^3.

To keep it simple, yeah thats exactly what it means.