Probability that coordinate of a dot within a square less than random parameter Z

Question

From square with vertices (0;0), (0;1), (1;1), (1;0) random dot was taken. It has coordinates (a;b). a and b are inside interval [0;1]. For random parameter z that is between [0;1] find probability that P(min(a,b) < 2z).

When I solve this task, I think like this:

P(min(a;b)<2z) = P(a<2z) * P(b<2z). And both of these two probabilities are equally possible. So I draw a square

And find the probability that a<2z and it is 3/4. Symmetrically for b<2z it is 3/4. So the total P(min(a,b) < 2z) = 3/4 * 3/4 = 9/16

But some of my friends distinguish two intervals z=[0;0.5) => Pr(min(a,b) < 2z) = 4z^2 and z=[0.5;1] => Pr(min(a,b) < 2z) = 1

Which solution is right? Thanks.

Answer 1

$P(min(a,b)<2z)\ne P(a<2z).P(b<2z)$