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?

There is a simpler way.  No digit should have an inherent advantage of being picked second.  So since 3 of the 5 digits are odd, there is a $\frac{3}{5}$ chance of the second number picked being odd.
