How can I generate a random vector between 1-9 with 10 numbers and at least one number should be >2.

X=rand(10,1)*(9-1)+1 How should I change this code to make at least one number >2.

Answer 1

[John D'Errico]

Generate 9 (NINE) random numbers with no constraint from 1:9. Then generate a 10th number that is assuredly greater than 2. Randomly permute the set.

R = [ceil(rand(1,9)*9), ceil(rand*7 + 2)];
R = R(randperm(10));

Easy peasy. If you prefer to use randi, it is just as easy, and perhaps easier to read.

R = [randi([1 9],[1 9]), randi([3,9],1)];
R = R(randperm(10));

The general idea is a common one to be found in probability. If you need one number to have some property, then generate it as such.

Edit: to generate real numbers instead of integers...

R = [rand([1 9])*8 + 1, rand(1)*7 + 2];
R = R(randperm(10));

The basic idea remains the same. Note that the rand function generates a value in the strict open interval (0,1), so the result of our 10th number will never be exactly 2. Thus, that value must always be strictly greater than 2.

[Hüseyin]

the numbers does not have to be integer..

[Mustafa]

does instructor mentioned that it must not be integer? I don't remember any necessity about it, right?

[John D'Errico]

Sorry. I presumed (incorrectly) that your numbers be integers given a too fast reading of your question. Even so, it is trivial to change to generate real numbers as I show in the edit.

Answer 2

[Matt J]

One way,

   X=rand(10,1)*(8-1)+1 +ceil(rand(10,1))

Answer 3

[Azzi Abdelmalek]

X=rand(10,1)*8+1;
idx=X>2;
if  nnz(idx)==0
  X(randi(numel(X)))=7*rand+2;
end

[Hüseyin]

I need one row answer sir.