Estimate of a 0-1 knapsack problem using probability.

InkyVoyd · 18 days

How does that help though?

infinite8 · 1 month

No, I didn't, I'm not sure how to enter it for bounded intervals. I would just look at the graph and see if it is defined, and if it is, what the answer would approximately be, for a start.

infinite8 · 1 month

http://www.wolframalpha.com/input/?i=ln%28sin+x%29+maximum+minimum

Melchoir · 1 month

Well, it's not that the article was confusing, it's just that I didn't understand most of it because I haven't learned most any of the math it talks about.

imaginarypunctuation · 1 month

Exactly.

imaginarypunctuation · 1 month

Although already mentioned, Jobs himself mentioned that they changed the algorithm for the ipod's shuffle because random seemed to organized. I might want to quote that, but I think I might've paraphrased.

Melchoir · 1 month

I've actually read the same article before. But both times I have failed to understand it.

AutocraticFence · 1 month

Screw this, I'm not making a logical argument and thus making my side look bad. You win.

AutocraticFence · 1 month

The quality of the site would be affected, because it got funding. Even if it couldn't use the funding my point was that it wouldn't get recognition from a major foundation unless that major foundation saw potential. It's a publicity stunt as much as Bill Gates trying to cure malaria is.

Is bandwagon necessarily a fallacy? No. It can be a fallacy, but it isn't always a fallacy. If a mob of reasonable people do something, I would be more likely to agree with them than if no one had opinions on it. Why? Because I'm not a non-conformist. Saying that bandwagons are always correct is the logical fallacy here. Saying that bandwagons are never correct is also a logical fallacy. You can argue that my logic is based on a bandwagon, but can you safely say that yours isn't?

Nice job checking the consistency of something after you say it.

Wasmyfault · 1 month

Well, I can't help you with the intersection part, but I can show you the picture. This is a side view of the picture, note that the legs that cross are actually 12 inches from each other. There's also a third leg, but I don't want to confuse you, so I didn't add it.

Imgur

Beignet · 1 month

I came thinking it was elementary algebra.

:S

AutocraticFence · 1 month

Gates Foundation is not a moot point. I highly doubt Bill and Melinda Gates would go around randomly funding things. If they made a mistake, it would look quite bad.

Cool, it turns out that more than two people are reading this. I wanted to give my opinion because people would realize that the majority of people (three to one instead of two to one) believe than Khan Academy is useful.

According to you - alright, can't argue with your logic according to yourself.

Why the hell would anyone care about what the Khan Academy used to be?

AutocraticFence · 1 month

Wow, I'm not sure if you are trolling. Why did the Gates foundation decide to fund Khan Academy? And, a wall of text, no matter how big or how long, is just not effective for some people. Finally, my personal experience with Khan Academy is that Sal explains things better than most people and textbooks. Not only does he not just write down his thoughts on a subject, he explains the subject in a way that r/learnmath would be proud. For every problem he does, he always goes step by step, and explains each step clearly. Half of the fucking textbooks I've ever read always miss explaining a step that is too easy, and the result is either my classmates are confused as crap, I am, or we both are. Finally, you told us that that was from a while ago. Please do not criticize a site in present tense if you have only been to it in the past. That's like saying Wikipedia sucks balz because it certainly did when it came out.

Unscientific · 1 month

Yea, wouldn't want that unit circle to have an area of 180, would you?

lovely016 · 1 month

Basically, if A and B are disjoint (set A and set B have an intersection of the null set, aka mutually exclusive), then the probability of A or B occurring is the probability of A+B. If A and B do have an intersection, than the probability of A or B happening is the probability of A happening + the probability of B happening - the probability of A and B happening (In most elementary cases the probability of A and B happening is the pA*pB).

If you understand set theory, check out the page on Wikipedia titled "Probability", and check under the heading "Summary of probabilities"

Finally, sometimes we have the problem of finding the probability of r events happening in a total of n trials (without a particular order), given a success rate of p. For example, for your first question, n=3, r=1, p=0.5. Or, let's say you wanted to know the probability of rolling a 3 four times in twenty four rolls of a fair six-sided die. Then n=24, r=4, and p=1/6. When these cases get to complicated to conveniently list them by hand, I usually use the following formula:

P_total = nCr * p^r *(1-p)^n-r

P_total here being the total probability, n being the number of total trials, r being the number of total successes, p being the probability of success. Note that if it is easier to calculate failure than success you can flip the numbers around (success=1-failure).

Also, let me answer your final problem.

"What is the probability of rolling an odd number or a prime number less than 6 on a single roll of a fair twelve-sided die?"

So in this case event A would be rolling a odd number, and event B would be rolling a prime number less than 6. Now are these events mutually exclusive (if you can not possibly do both, then they are mutually exclusive)? Obviously they aren't, because there odd numbers less than 6 that are prime (3 and 5). Thus you find the probability of A (independently) + the probability of B (independently) - the probability of A and B. pA=6/12=1/2, pB=3/12=1/4. The probability of A and B happening is 2/12 (3 and 5; metaphorically, we are essentially adding the area of two circles of a Venn Diagram and subtracting the area of where they intersect). Thus we add pA and pB and subtract p(A and B).

lovely016 · 1 month

For number two, we do something the same thing. List out the possible combinations. (1,5)

(2,4)

(3,3)

That seems like all, right? However, it isn't. This is an easy error to make. It turns out we forgot the other half.

(4,2)

(5,1)

Now how many different possibilities are there? Imagine a tree with 6 branches on the first part, and 6 branches on the second part for each of the former. In other words, 6*6=36, so the total number of possibilities for rolling a die is 36.

Now we divide the two.

EDIT: I stalked you a little, and realize that I don't have to treat you like a 8 year old. Sorry.

lovely016 · 1 month

At least 2 heads means 2 or 3 heads. The probability of OR is usually the two events added up. Thus we find the probability of getting 2 heads, and add it up with the probability of getting 3 heads.

There's a formula for this called binomial probability, but I can't assume that you are learning it right now. Thus I will show you how to do it the old-fashioned way. List out all the probability. Let's say that heads is represented by a "1", and tails is represented by a "0" (because it's hard for me to type all those). Now let's list all of the different situations. I like to do it in the format ( , , ). For example, (1,1,0) means first heads, then heads again (1, 1), then tails (0). Now let's list them all. Remember that we are rolling three times, so we need three numbers each, for each roll.

(1,1,0);(1,0,1);(0,1,1) - Note how all of these are different, yet have 2 heads.

Now that was hard enough (I can explain it further), but let's calculate the probability of getting 3 heads. We can rewrite it into (1,1,1).

Now we add all of the different possibilities of what the question is asking and divide by the total number of possibilities. 3+1=4 for the number the question is looking for. Now to find the number of total possibilities, we can note that for each toss we can get either heads or tails. If you draw a tree, you will find that 222=8; the total number of possible different tosses is 8.

Finally, we divide what it is asking for by what the total is. 4/8=1/2. Therefore the probability of getting at least two heads on three tosses is 1/2.

Anastomosis · 1 month

Sigh. http://www.wolframalpha.com/input/?i=is+cos%28%28271%2F360%29*pi%29%29+transcendental

testname33 · 1 month

My answer is when you are near a bunch of a liberal arts majors.

InkyVoyd · 1 month

Thanks for the picture, but can you label the axis? I know this is r/cheatatmath homework, but this isn't actually a homework question, and it's been a year from when I took algebra two (which talked a little about normal distribution, but not nearly as much as statistics).

Anastomosis · 1 month

If there is a closed form expression of cos(1), why is cos(135.5) transcendental? Does it have to do with the definition of a root of a polynomial (i.e. the length it has to be)

Anastomosis · 1 month

Well, I was just about to write a bunch of arrogant stuff, but then I Googled it, and realized you were right, so here's a cookie for you. 135.5 degrees, apparently, is transcendental though. http://www.wolframalpha.com/input/?i=is+cos%28135.5%29+transcendental

InkyVoyd · 1 month

60+40=100? I would still get close to 25% and 25 points, because the total is 100 right?

So, since they aren't normally distributed, is there an easy way to calculate distribution?

EDIT: Wait, I'm sure they are both normally distributed. Although the density of point distribution per problem is not even, the probability is the same right?

Anastomosis · 1 month

and for proof; http://www.wolframalpha.com/input/?i=is+cos%28135.5%29+transcendental

Also, no problem. Your teacher cannot possibly expect you to find a way to exactly express a transcendental number. That is an outrage.

InkyVoyd

TROPHY CASE

you'll need to login or register to do that

create a new account

login