We're doing a simple exercise and we have the following situation:
if ( random(1-100) == 50 OR random( 1-100 ) == 50 ) THEN [..]
The random function may return any number between 1-100 1 and 100 included with a theoretic complete randomization.
I'd like to know what is the % of that IF statement to return TRUE.
another question is:
Considering the OR statement aqs short-circuit does it affect in any way that %.
Let X = random(1-100) and Y = random( 1-100 ). That is uniformly distributed between the integers $1, 2, 3 \dots 100$
Then $P(X=50 \cup Y=50) = P(X=50)+P(Y=50)-P(X=50 \cap Y=50)$.
If the variables are independent then the probability should be: $0.01+0.01-0.01^2 = 0.0199 $. So, unless I misunderstand your statement, it is true almost 2% of the time.
Also assuming your function can only take integer values (which occur with equal probability), otherwise (if the function can take any real number in the interval) the probability should be 0.
R with n <- 1e5; a <- sample(1:100, size = n, replace = TRUE); b <- sample(1:100, size = n, replace = TRUE); sum(b == 50 | a == 50)/n
$\endgroup$
You can model this as a binomial probability.
$X \sim Binom(n=2, p = 1/100)$
$P(X\geq 1) = 0.0199$
