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?
 A: The probability we pick an odd number first is $\frac{3}{5}$. Then the probability of another odd number being chosen from the remaining numbers is $\frac{2}{4}=\frac{1}{2}$. Put together, the probability is $\frac{3}{5}\times\frac{1}{2}=\frac{3}{10}$.
The probability we pick an even number first is $\frac{2}{5}$. Then the probability of an odd number being chosen from the remaining numbers is $\frac{3}{4}$. Put together, the probability is $\frac{2}{5}\times\frac{3}{4}=\frac{3}{10}$.
The probability that the second number being chosen is odd is the sum of these two probabilities. So $\frac{3}{10}+\frac{3}{10}=\frac{3}{5}$ is the correct answer.

As others have noted, picking randomly could also be thought of as shuffling the numbers like a deck of cards, and then choosing top to bottom. Of course, there's a $\frac{3}{5}$ chance that the second "card" would be an odd number, but it's nice to see that the conditional probability works out as well.
A: There are simpler ways. There is absolutely no preference between the numbers, so each number is equally likely to be the second one drawn. Since 3 of 5 numbers are odd, the probability is 3/5 that an odd number is drawn. 
A: There's another (exhaustive) way of seeing it, and it's counting.
Let us represent the act of picking a number $a$ and then a number $b$ by $(a,b)$, we then have $5\times 4=20$ of this possible "actions". Now, the cases that favors your situation are:
$$(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,5),(4,1),(4,3),(4,5),(5,1),(5,3)\ .$$
Since there are $12$ favor cases, the probablity is $12/20=3/5$. It's not easier, it's just another way of seeing it.
A: 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)
A: 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.
