<s>[INST] 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?
 [/INST] 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.
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