Recent content by Mandark

I think you'll be unhappy if you start cutting off your friendships, so I wouldn't recommend that. Some of my good friends I only interact with at university, and that's a perfectly fine level of interaction which we're mutually happy with. When your friends invite you to events politely...

Every subset of a discrete topological space has empty boundary. To see this recall the definition of the boundary of a set A as the intersection of the closure of A, and the closure of the complement of A, in a discrete space all subsets are both open and closed, so the closure of A is simply A...

"Here's another self portrait. I'm an only child, so I find ways to amuse myself."

I can induce chills to some extent at will. I've been able to since I can remember and have asked people about this, those who I asked said they could not. For them to be very strong I need an external source though. It starts from the upperback and neck then extends down the back and along the...

Try to prove it by manipulating the general formula I posted, it's not hard.

The formula for general RV X is EX = -\int_{-\infty}^{0} F(x) dx + \int_{0}^{\infty} (1 - F(x)) dx. This formula works in a much more general setting than you might expect. Some distributions don't have densities (singular distributions), for example...

I suspect what is actually meant is \mathbb{C}^3, the space of complex triples, considered as a vectorspace over the reals.

If n is odd then n = 2k + 1 for some integer k. So the sum becomes \sum_{i=0}^{k} (2i + 1). Now find k in terms of n, and hence a closed form.

Having 40 blue balls gives you a 0.846ish chance of having 3 or more blue balls. In general the problem reduces to solving a polynomial equation. Do you know how to set up the equation?

That reasoning is correct for the first group, but try to continue it for the 2nd and 3rd groups (particularly the 3rd), and you'll begin to see where the difficulty lies in that approach (you won't end up with the same product as you wrote). Giving an interpretation to the product you wrote...

Great forum for competition problems: http://www.artofproblemsolving.com/Forum/index.php You can browse the forum or click on "Contests" which has an archive of thousands of problems from different competitions.

In your edit you should be dividing by 4 rather than multiplying because the two boy teams (teams with 3 boys) can be switched and it wouldn't make a difference to the partition, same goes with the two girl teams (3 girls on the team).

You could try some competition maths problems. They tend to require only basic maths but can be challenging nonetheless.

a) That's right. b) Basically correct, I'm assuming you mean find E(X|Y), which is simply E(X|Y=y) evaluated at y = Y.

What makes you say it's indeterminate at 0?