Probability of subsets with different constraints Let $U = \{1, 2, \dots, 1000\}$. We pick a subset from U uniformly at random. That is, every subset has an equal chance of showing up.

*

*What is the probability that the empty set is chosen?

*What is the probability that the subset has at most ten elements?

*What is the probability that the subset contains the numbers 500 or 1000?

*What is the probability that the subset does not contain all the multiples of 10?

*What is the probability that the size of the subset is odd?

My solution:
The total number of subsets of a set with size $n$ is $2^n$

*

*$\frac{1}{2^{1000}}$ (only one possibility)

*$\frac{\dbinom{1000}{10} + \dbinom{1000}{9} + \dots + \dbinom{1000}{1} + \dbinom{1000}{0}}{2^{1000}}$

*$(1 - \frac{2^{1000-1}}{2^{1000}}) + (1 - \frac{2^{1000-1}}{2^{1000}})$ (my understanding, if we want the subset to contain a particular number is the same as $1$ minus probability of the subset does not contain a number and itself is equal to a subset of a set of size $1000-1$)

*$\frac{2^{1000-100}}{2^{1000}} = \frac{2^{900}}{2^{1000}}$ (similar to above. there are 100 multiple of $10$ in $1000$)

*$\frac{\dbinom{1000}{999} + \dbinom{1000}{997} + \dots + \dbinom{1000}{1}}{2^{1000}}$
Am I on the right track?

1 and 2 - OK.


*$(1 - \frac{2^{1000-1}}{2^{1000}}) + (1 - \frac{2^{1000-1}}{2^{1000}})  = (1 - \frac12)  + ( 1 -\frac12) = 1$, but the answer in obviously less than $1$. So no all is good in 4.

In 5 all is OK but there's a closed form.
We know that $(1+x)^n = \sum_{k=0}^{1000} \dbinom{1000}{k}x^k$. Put $x=\pm 1$ and $n=1000$. We got
$$ (1+1)^{1000} = \dbinom{1000}{0} + \dbinom{1000}{1} + \dbinom{1000}{2} + \dbinom{1000}{3} + \dbinom{1000}{4} + \ldots$$
$$ (1 + (-1))^{1000} = \dbinom{1000}{0} - \dbinom{1000}{1} + \dbinom{1000}{2} -\dbinom{1000}{3} + \dbinom{1000}{4} + \ldots$$
Hence $\frac{(1+1)^{1000}  -  (1 + (-1))^{100} }2 = \dbinom{1000}{1} + \dbinom{1000}{3} + \dots + \dbinom{1000}{999}$ and
$$\frac{\dbinom{1000}{999} + \dbinom{1000}{997} + \dots + \dbinom{1000}{1}}{2^{1000}} = \frac12$$
