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?
 A: 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
A: There are $2^{n-1}$ ways to choose the balls, because there are $2^n$ subsets of $n$ balls, but I believe that drawing $0$ balls is not allowed. That's the denominator.
Then, for the numerator, we must choose an even number of balls. There are $2^{n-1}$ ways to do this, because $2^n$ is the total, and there are an even number of odd and even possibilities, so we divide by $2$, and $\frac{2^n}{2}=2^{n-1}.$ Again, we can't have $0$ balls, so we subtract one.
Therefore, the probability is $$\frac{2^{n-1}-1}{2^n-1}.$$
