<s>[INST] 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?
 [/INST] 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
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