Set A = {1,2,3,4,5} Pick randomly one digit and remove it. What is the prob. that we pick an odd digit the 2nd time. The probability that we pick any number for the first time is $\dfrac{1}{5}$
the sample space of sample spaces after the first event is then
{2,3,4,5}
{1,3,4,5}
{1,2,4,5}
{1,2,3,5}
{1,2,3,4}
prob. to pick an odd from the 1st sample space is $\dfrac{1}{2}$
prob. to pick an odd from the 2nd sample  space is $\dfrac{3}{4}$
prob. to pick an odd from the 3rd sample  space is $\dfrac{1}{2}$
prob. to pick an odd from the 4th sample  space is $\dfrac{3}{4}$
prob. to pick an odd from the 5th sample  space is $\dfrac{1}{2}$
The final result is:
$\dfrac{1}{5}$ * $\dfrac{1}{2}$ + $\dfrac{1}{5}$ * $\dfrac{3}{4}$ + $\dfrac{1}{5}$ * $\dfrac{1}{2}$ + $\dfrac{1}{5}$ * $\dfrac{3}{4}$ + $\dfrac{1}{5}$ * $\dfrac{1}{2}$ = $\dfrac{3}{5}$
Is this reasoning correct? Are there any simpler ways to solve this problem?

You are probably familiar with "tree diagrams" to solve these types of problems. For the first pick you have a (2/5) chance to remove an even, and a (3/5) chance to remove an odd.
When you remove an odd first, there are only (2/4) odds left. Meanwhile, if you remove an even first, there are now (3/4) odds left.
Finally (2/5)(3/4)+(3/5)(2/4) = (6/20)+(6/20) = (12/20) = (3/5)
