Recent content by oriel1

That what I did at first, and I got wrong answer. But I found calculation mistake, now it work and give correct answer. The third part need to be: $$\int_{1.5}^{2}\int_{x-0.5}^{2}f(x,y)dydx$$ Ok, I understand how to set that specific boundaries. But why in the first "1" quadrant area when I...

Sorry, you right! this |x-y| I confused for a moment. I'm trying to calculate this one: $P(|X-Y|<0.5)$. Where do I enter that "plot" command? do I need MATLAB or something? However, I need to find answer. Finding the first quadrant is just part of the process that I think is relevant for...

Hi, I have homework question that I'm trying to solve. But I can't understand the basis. Here is a picture of the question and what I have done: My question is, How do I set the boundaries for the integral? 1. If I want the whole squart. 2. To sum up areas. Assume I want to sum the areas, I...

Thank you very much for that clear explanation. I appreciate that.

There is lucky machine with 3 results (letters): A,B,C P(A)=0.2 (probability to get A) P(B)=0.3 "" "" "" P(C)=0.5 Each round you get just one letter. You played 5 times. What the probability that you got three times "A" and two times "B". i thinked to do that...

Thank you. now i understand it.

Ok Actually R={<1,1>,<2,2>,<3,3>} is identity relation on A for sure. But what prevent from R= {<1,1>,<2,2>} to bo identity on A? It not writed $$\forall$$ x $$\in$$ A.

Let A= {1,2,3}. Let R= {<1,1>,<2,2>}. I(A) (Identity Realtion) on A >(def)> {<x,x>|x $$\in$$ A} So that mean : $$\forall$$ <x,x> x $$\in$$ A (That how I understood it) My question: Is R is identity relation on A ? Thank you !