This is definitely a great channel for undergrad maths. And also for having a clear vision of the higher mathematics above it. I am a physics undergrad but I love maths and Cs too. And the book suggestions are great. The problem reviews and course and textbook recommendations are also interesting to me. The content creator is great I was wondering if there is any similar channel for physics too. Anyone any suggestions??
Hi, highschool student from Japan here. I recall seeing the exact same problem on an entrance exam for some university in Japan. I solved it by rephrasing 0≦x,y,z≦1 with x,y,z=0, 1/n, 2/n, …, n/n (n->∞). It takes A LOT of time and calculations, and it is far from elegant compared to your solution, but the answer I got was 1/2. Sorry for the bad English. Love your videos.
Three friends arrange a meeting in front of the Nilton Santos stadium to watch a soccer match. They agreed that each one should arrive at a chosen time between 3:00 pm and 4:00 pm and that none of them would wait more than 30 minutes for the others, within the stipulated time. What is the probability that the three friends meet between 3:00 pm and 4:00 pm?
Currently doing multivariable calculus, having the exam next week. My issue with math is that I enjoy it but I sometimes really don't get the purpose of what I'm doing, however I just accept it and solve the problems even though I really don't know what I am doing. Been having the same issue with earlier match courses in university. I pass with good grades but id like to have a more depth understanding of it. Any suggestions/advice?
I used the following logic Since x y and z are between 0 and 1, once x and y are chosen, the probability of choosing z such that x y and z form a triangle is x+y. Also notice that once x+y is greater than or equal to 1 no matter what value of z is chosen x,y and z will form a triangle. So the probability of forming a triangle can be obtained by integrating [udu] from 0 to 1 where u is x+y. The integral for udu is u^2/2 and applying the limits we get 1/2.
Your claim: "once x and y are chosen, the probability of choosing z such that x y and z form a triangle is x+y" does not sound right. We could obviously have x+y>1, and if x=0.01 and y=0.99 we would need z∈[0.98,1] (which has probability 0.02).
@@vimilsaju9004 You can see my comment, which solve the problem after conditioning on x and y. It's a little more complicated than the way you put it, but just a few lines long.
Three friends arrange a meeting in front of the Nilton Santos stadium to watch a soccer match. They agreed that each one should arrive at a chosen time between 3:00 pm and 4:00 pm and that none of them would wait more than 30 minutes for the others, within the stipulated time. What is the probability that the three friends meet between 3:00 pm and 4:00 pm?
If I can make a humble request.. can you maybe change to using pencil instead of sharpies? The sound the sharpie makes with paper kinda bothers me a bit.. lol
That's a nice problem to exercise understanding of multiple integrals. My solution to a problem: Use triangle's inequality 2max(x,y,z) < x+y+z Let's assume max(x,y,z) = x So yx-y so z can have [x-y;x] Now we integrate on [0;1] x [0;x] x [x-y;x] 1 dz dy dx The result of integral is 1/6 And we multiply it by 3 because we assumed x is max, however y and z also can be max so we have 3 different cases So final result is 3*(1/6) = 1/2
Solution based on simple integrals: Given (x,y) we need z∈[|x-y|,x+y] to form a triangle. The probability that z falls in this range is q(x,y)= min(x+y,1)-|x-y|, since z is uniform on [0,1]. Thus, we need to compute the expected value of q(x,y), which is the difference between two terms. The expected value of the first is E[min(x+y,1)]=E[(x+y)*1(x+y≤1)]+ E[1(x+y>1)], where E[1(x+y>1)]=½ and the first part is ∫_{x∈0,1}∫_{y∈0,1-x}(x+y)dydx=∫_{x∈0,1}x+(1-x)^2dx=2/3. Similar integrations gives E(|x-y|)=2/3 and we can conclude... E[q(x,y)] = 2/3 +1/2 - 2/3 = 1/2.
Three friends arrange a meeting in front of the Nilton Santos stadium to watch a soccer match. They agreed that each one should arrive at a chosen time between 3:00 pm and 4:00 pm and that none of them would wait more than 30 minutes for the others, within the stipulated time. What is the probability that the three friends meet between 3:00 pm and 4:00 pm?
It was a high school problem for me. Easier way without integral: to get the volume, so the probability, you need to cut down 3 tetrahedrons from the unit cube. One tetrahedron has volume B*h/3=1*1/2*1/3=1/6 volume, since the base is a right triangle with legs 1 and 1, hence the base has B=1*1/2 area and the height of the tetrahedron is h=1. So the probability is 1-3*1/6=1/2.
I saw your logic for the probability problem, and I am certain that it is correct because,I have solved a similar problem ,in the past, which had a group of inequalities that you had to solve for in the 3-d plain. I think another solution to this problem can come from the fact that if you rephrase the 3 inequalities based on 2 variables, and see the 2 - d visualisation of those variables on the x-y plane. You will notice that it will be the 2 -d case for your problem. When you find out the areas of the triangle formed, and input the range [0,1], you will get either 1/2 or 1/3(I think that depends on your parametric inequalities), and you rephrase the parametric inequality into standard form with x,y , and z, you will get 1/6. Love your vids by the way, keep up the hard work:).
Three friends arrange a meeting in front of the Nilton Santos stadium to watch a soccer match. They agreed that each one should arrive at a chosen time between 3:00 pm and 4:00 pm and that none of them would wait more than 30 minutes for the others, within the stipulated time. What is the probability that the three friends meet between 3:00 pm and 4:00 pm?
Ok, try to answer this one. This is one of my favorites. Consider the interval [0,1]. Throw 2 points uniformly and independently at random so the interval is cut into three segments. What is the probability that the three segments form a triangle? (Don’t use order statistics! That’s a boring way to do this.)
the triangle inequality as used in the video is sufficient. A shorter form is |a-b| < c < a+b or 2*max(a,b,c) < a+b+c en.m.wikipedia.org/wiki/Triangle_inequality
@@PhDVlog777 Three friends arrange a meeting in front of the Nilton Santos stadium to watch a soccer match. They agreed that each one should arrive at a chosen time between 3:00 pm and 4:00 pm and that none of them would wait more than 30 minutes for the others, within the stipulated time. What is the probability that the three friends meet between 3:00 pm and 4:00 pm?
as an engineer and non mathematician my first instinct to check your proposed solution is to simulate picking 3 points between 0 and 1 many times and see if the probability of those being a triangle is 1/2, great vid
Not sure that’s correct. The broken stick problem assumes a stick of fixed length, whereas in this case the stick could be any length between 0 and 3. Maybe?
I can relate to the tears on the paper.
This is definitely a great channel for undergrad maths. And also for having a clear vision of the higher mathematics above it.
I am a physics undergrad but I love maths and Cs too. And the book suggestions are great. The problem reviews and course and textbook recommendations are also interesting to me. The content creator is great
I was wondering if there is any similar channel for physics too. Anyone any suggestions??
Andrew dotson
Tibees
For math book recommendations and math advices you have The Math Sorcerer.
Thanks
@@tamalpanda9834 you're welcome 🤗
Your answer to the probability problem sounds perfectly fine to me :)
Hi, highschool student from Japan here.
I recall seeing the exact same problem on an entrance exam for some university in Japan.
I solved it by rephrasing 0≦x,y,z≦1 with x,y,z=0, 1/n, 2/n, …, n/n (n->∞).
It takes A LOT of time and calculations, and it is far from elegant compared to your solution, but the answer I got was 1/2.
Sorry for the bad English. Love your videos.
Three friends arrange a meeting in front of the Nilton Santos stadium to watch a soccer match. They agreed that each one should arrive at a chosen time between 3:00 pm and 4:00 pm and that none of them would wait more than 30 minutes for the others, within the stipulated time. What is the probability that the three friends meet between 3:00 pm and 4:00 pm?
@@raels5535 3/4 is it??
@@oreki8707 not.
@@raels5535 OOO my bad it's 3 freinds
I did the calculation assuming it's 2
@@oreki8707 Yes there are three friends is a generalization of the problem with two.
Currently doing multivariable calculus, having the exam next week. My issue with math is that I enjoy it but I sometimes really don't get the purpose of what I'm doing, however I just accept it and solve the problems even though I really don't know what I am doing. Been having the same issue with earlier match courses in university. I pass with good grades but id like to have a more depth understanding of it. Any suggestions/advice?
Whoa. That's a lot To Do. [ Plus a channel Q & A 💪 :) ] I have faith in you. You've got this!
Cool problem you shared with us today.
☆☆☆☆☆
Love geometry related problems like the one in the video. Really fun to think about.
Three Points form a Triangel if: A is set, B isnt A and C is Not an Element of the linear equation Containing A and B
First of all you need to specify if the three variables are UNIFORMLY distributed
I used the following logic
Since x y and z are between 0 and 1, once x and y are chosen, the probability of choosing z such that x y and z form a triangle is x+y. Also notice that once x+y is greater than or equal to 1 no matter what value of z is chosen x,y and z will form a triangle. So the probability of forming a triangle can be obtained by integrating [udu] from 0 to 1 where u is x+y. The integral for udu is u^2/2 and applying the limits we get 1/2.
Your claim: "once x and y are chosen, the probability of choosing z such that x y and z form a triangle is x+y" does not sound right. We could obviously have x+y>1, and if x=0.01 and y=0.99 we would need z∈[0.98,1] (which has probability 0.02).
@Peter Hansen I think in that case, we can switch the variables so that y has the probability of x+z. I'm not sure if this is rigorous, though.
@@vimilsaju9004 You can see my comment, which solve the problem after conditioning on x and y. It's a little more complicated than the way you put it, but just a few lines long.
Three friends arrange a meeting in front of the Nilton Santos stadium to watch a soccer match. They agreed that each one should arrive at a chosen time between 3:00 pm and 4:00 pm and that none of them would wait more than 30 minutes for the others, within the stipulated time. What is the probability that the three friends meet between 3:00 pm and 4:00 pm?
If I can make a humble request.. can you maybe change to using pencil instead of sharpies? The sound the sharpie makes with paper kinda bothers me a bit.. lol
A bit disappointed that this wasn't just your non-math todo-list and the probability that you'll procrastinate more than 50% of them.
That's a nice problem to exercise understanding of multiple integrals. My solution to a problem:
Use triangle's inequality 2max(x,y,z) < x+y+z
Let's assume max(x,y,z) = x
So yx-y so z can have [x-y;x]
Now we integrate on [0;1] x [0;x] x [x-y;x] 1 dz dy dx
The result of integral is 1/6
And we multiply it by 3 because we assumed x is max, however y and z also can be max so we have 3 different cases
So final result is 3*(1/6) = 1/2
I ran 10000000 simulations 100 times and the average proportion of triangles is 0.4999922340000002, so it seems right.
your dad is a math professor ?
let us know if it’s right
but x=y+z is not a triangle, its a straight line! all three points must be colinear, that doest make a triangle.
its a plane
As a 10th grader, I can't understand more than 10% of the things
What pen is that, that you use to write here??
I believe it is correct... Always loved these geometric probabilities
Solution based on simple integrals: Given (x,y) we need z∈[|x-y|,x+y] to form a triangle. The probability that z falls in this range is q(x,y)= min(x+y,1)-|x-y|, since z is uniform on [0,1]. Thus, we need to compute the expected value of q(x,y), which is the difference between two terms.
The expected value of the first is E[min(x+y,1)]=E[(x+y)*1(x+y≤1)]+ E[1(x+y>1)], where E[1(x+y>1)]=½ and the first part is ∫_{x∈0,1}∫_{y∈0,1-x}(x+y)dydx=∫_{x∈0,1}x+(1-x)^2dx=2/3. Similar integrations gives E(|x-y|)=2/3 and we can conclude...
E[q(x,y)] = 2/3 +1/2 - 2/3 = 1/2.
Three friends arrange a meeting in front of the Nilton Santos stadium to watch a soccer match. They agreed that each one should arrive at a chosen time between 3:00 pm and 4:00 pm and that none of them would wait more than 30 minutes for the others, within the stipulated time. What is the probability that the three friends meet between 3:00 pm and 4:00 pm?
What kind of fluids are on the paper?
Tears my man
It is water.
tears
@@PhDVlog777 tears are a subset of water, yes?
fluids 😭😭😭
It was a high school problem for me. Easier way without integral: to get the volume, so the probability, you need to cut down 3 tetrahedrons from the unit cube. One tetrahedron has volume B*h/3=1*1/2*1/3=1/6 volume, since the base is a right triangle with legs 1 and 1,
hence the base has B=1*1/2 area and the height of the tetrahedron is h=1.
So the probability is 1-3*1/6=1/2.
I saw your logic for the probability problem, and I am certain that it is correct because,I have solved a similar problem ,in the past, which had a group of inequalities that you had to solve for in the 3-d plain.
I think another solution to this problem can come from the fact that if you rephrase the 3 inequalities based on 2 variables, and see the 2 - d visualisation of those variables on the x-y plane. You will notice that it will be the 2 -d case for your problem. When you find out the areas of the triangle formed, and input the range [0,1], you will get either 1/2 or 1/3(I think that depends on your parametric inequalities), and you rephrase the parametric inequality into standard form with x,y , and z, you will get 1/6.
Love your vids by the way, keep up the hard work:).
Three friends arrange a meeting in front of the Nilton Santos stadium to watch a soccer match. They agreed that each one should arrive at a chosen time between 3:00 pm and 4:00 pm and that none of them would wait more than 30 minutes for the others, within the stipulated time. What is the probability that the three friends meet between 3:00 pm and 4:00 pm?
@@raels5535 Essentially the same problem, the numbers are slightly different though this time.
Ok, try to answer this one. This is one of my favorites.
Consider the interval [0,1]. Throw 2 points uniformly and independently at random so the interval is cut into three segments. What is the probability that the three segments form a triangle? (Don’t use order statistics! That’s a boring way to do this.)
Undergrad Math student who loves probability. That proof was absoultely beautiful. Was the inspiation I needed. Thank you
Can you prove: iff x,y,x doesn’t make a triangle then max > sum of other 2. Then use order statistics to compute the probability?
thanks for the content. I have an interest in math but I have to learn it on my own atm.
Would you ever make a day in the life kind of a video? that would be awesome
the triangle inequality is necessary but not sufficient. you have to consider the following 3 equations:
(1) |x-z|
the triangle inequality as used in the video is sufficient. A shorter form is |a-b| < c < a+b or 2*max(a,b,c) < a+b+c
en.m.wikipedia.org/wiki/Triangle_inequality
I think you probably did it right.
why dont use an ipad
I like to write with pen and paper.
@@PhDVlog777 legendary
@@PhDVlog777 Three friends arrange a meeting in front of the Nilton Santos stadium to watch a soccer match. They agreed that each one should arrive at a chosen time between 3:00 pm and 4:00 pm and that none of them would wait more than 30 minutes for the others, within the stipulated time. What is the probability that the three friends meet between 3:00 pm and 4:00 pm?
I'm first !!
as an engineer and non mathematician my first instinct to check your proposed solution is to simulate picking 3 points between 0 and 1 many times and see if the probability of those being a triangle is 1/2, great vid
simulating does converge to 1/2
This problem is known as the broken stick problem and the probability is, in fact, 1/4. I'd love to see more on probability!
Not sure that’s correct. The broken stick problem assumes a stick of fixed length, whereas in this case the stick could be any length between 0 and 3. Maybe?
Nope
@@Pommes736 The comment we all needed. David Wilson already pointed out my fallacy.