Meaning of drawing any number of balls from an urn containing $n$ balls There is a question as follows:
From an urn containing $n$ balls any number of balls are drawn. Show that the probability of drawing an even number of balls is $\frac{2^{n-1}-1}{2^n-1}$
Firstly what does it mean to draw any number of balls from the urn and how does the sample space look like for this question?

since we need to find the probability of choosing even number of balls it means the probability is
$$ \frac{\binom{n}{2}+\binom{n}{4} + \ .... \ }{\binom{n}{1} + \binom{n}{2} + \binom{n}{3} + \ .... \ \binom{n}{n}} 
 $$
which equals $$ \frac{2^{n-1} \ - \ \binom{n}{0}}{2^n - \binom{n}{0}} $$
which immediatly yeilds your answer.

To find the sum $\binom{n}{2}+\binom{n}{4} + \ .... $ and $\binom{n}{1} + \binom{n}{2} + \binom{n}{3} + \ .... \ \binom{n}{n}$


to find $\displaystyle \binom{n}{1} + \binom{n}{2} + \binom{n}{3} + \ .... \ \binom{n}{n}$ put

$x=1$ in $(1+x)^n$


and to find the second one put $x = -1$and add with the first equation
