Matrix probability with just one 1

Lets say I have the following matrix:

[0 0.5 0 0 0.5]

Meaning there is a probability of a 1 on the second or fifth position.

There can only be one 1 on this matrix, so there is only two posible outcomes:

[0 1 0 0 0] or [0 0 0 0 1]

I already solved this problem, but in a very long way, by dividing 1 by the probability (0.5 in this case) and then generate a random number between 1 and the result of the previous division (1 or 2 in this example), and then iterate through the indexes and enter the ones with probability, count them and then set one to 1 and the rest to 0.

I was thinking of a more elegant way to solve this, for example creating a new Matrix [1 1 1 1 1] and then use binornd between this matrix and the probability matrix, but this gives me four possible outcomes instead of two: [0 0 0 0 0], [ 0 1 0 0 0], [0 0 0 0 1] or [0 1 0 0 1].

Is there a way to use binornd or any other function to do what I want in a more elegant way?

Thanks!

[Rik]

Are all the probabilities the same in your real application?

[Marcos Ortego]

Is either two 1/2, or three 1/3 or four 1/4.

Answer 1

[Rik]

This code only works for positions with equal probability.

dist=[0 0.5 0 0 0.5];

valid_pos=find(dist);%select non-zero
out=zeros(size(dist));
out(valid_pos(randi(end)))=1;

[Marcos Ortego]

Thanks!