# uneven probability problem

I was encountered with a probability problem, and here I tried to explain it in a easier way.

Now we have a square(w*l), and I am going to randomly put a white point on the surface of this square, and the probabilitythat the point would turn to black is based on its coordinate, let say P = 1/(x+y) (x+y>1), and this probabilityis independent.

The question is now I am going to put 10 points randomly on the surface of this square, what is the P (all points turn to black)?

I first tried to apply double integrations $$P=\frac{1}{wl}\int_{0}^{w} \int_{0}^{l} \frac{1}{x+y}dxdy$$

but after simulation by matlab, It is wrong, but the probabilityis not even on the surface, how to calculate the probability.

This self learning

• $1/(x+y)$ can be greater than $1$. Commented Jul 10, 2016 at 5:46

Let $B(x,y)$ be the probability of a white point turning black, for any point $(x,y)\in[0;w]{\times}[0;l]$.
Let $\mathop{\big\{(X_k, Y_k)\big\}}_{k\in\{1..10\}}$ be a sample of ten points chosen independently from a uniform distribution in the rectangle.
The probability that all points turn black will be:$$P=\left(\frac{1}{wl}\int_0^w\int_0^l B(x,y)\operatorname d y\operatorname d x\right)^{10}$$
Note: As AlphaGo commented, the selection of the function $B$ is the issue.   The function you gave is inadmissible as $1/(x+y)$ may be greater than $1$ for some points in the rectangle, and hence not a valid probability function.   That is at any $(x,y): x+y<1$.   For example at $(\tfrac 13, \tfrac 14)$ you would have: $\tfrac{1}{\tfrac 1 3+\tfrac 1 4}=\frac{12}7$.