Conditional Probability of a Uniform Random Subset. 
Question:
Let X = $({1,2,3,..,10})$ Let $Y$ be a uniformly random subset of $X$. Define the events:
A = “$Y$ contains at least $4$ elements",
B = “all elements of $Y$ are even".
What is $Pr(A|B)$?

Answer: 0.1875
Attempt:
I know that P(A $\bigcap$ B) / P(B) is what I have to ultimately find.
For $P(B)$ = $\frac{5}{10}$ = $\frac{1}{2}$
For P(A) => Must have exactly 4, 5, 6, 7, 8, 9, 10 elements
So, I used the binomial for this by doing:
P(A) =
$10\choose4$$.$ $\frac{1}{10}$$^4$$.$$(1-\frac{1}{10}$)$^6$ +
$10\choose5$$.$ $\frac{1}{10}$$^5$$.$$(1-\frac{1}{10}$)$^5$ +
$10\choose6$$.$ $\frac{1}{10}$$^6$$.$$(1-\frac{1}{10}$)$^4$ +
$10\choose7$$.$ $\frac{1}{10}$$^7$$.$$(1-\frac{1}{10}$)$^3$ +
$10\choose8$$.$ $\frac{1}{10}$$^8$$.$$(1-\frac{1}{10}$)$^2$ +
$10\choose9$$.$ $\frac{1}{10}$$^9$$.$$(1-\frac{1}{10}$)$^1$ +
$10\choose10$$.$ $\frac{1}{10}$$^{10}$$.$$(1-\frac{1}{10}$)$^0$ +
= $0.012795$
P(A $\bigcap$B) = $\frac{1}{2}*0.012795$
Pr(A|B) =  $\frac{\frac{1}{2}*0.012795}{2}$
Pr(A|B)  = $0.012795$
Where did I go wrong with this approach? I find finding P(A) very time consuming, I feel like there has to be a much more simpler approach.
 A: $5$ elements of $X$ are even. 
To have at least $4$ elements and all elements are even, you have either $4$ elements ($5$ ways, choose one to leave out) or $5$ elements ($1$ way).
$$\frac{1+5}{2^{5}}=\frac{6}{32}$$
Remark: 


*

*We are choosing uniformly over all subsets. $P(B)= \frac{2^5}{2^{10}} \ne \frac{5}{10}$.

*We don't have to find $P(A)$, what is of interest is $P(A \cap B)$.

*We do not have $P(A \cap B)=P(A)P(B)$ in general, we need independence.
A: *

*Why are you using binomial distribution to evaluate the probabilities for event $A$, when $Y$ is a uniformly random subset of $X$?   That means that every element (ie subset of $X$) of the powerset of $X$ (the sample space) has an equal probability of being selected.   Just count the favoured and total spaces.

*$\mathsf P(A\cap B)=\mathsf P(B)~\mathsf P(A)$ is only true for independent events, of which these are not.   Evaluate $\mathsf P(A\cap B)$ using a similar method as you should have evaluated $\mathsf P(A)$.

*$\mathsf P(B) \neq \tfrac 12$.   The event of "all the elements of $Y$ are even" is not exactly half of the sample space, which may be partitioned by "all the elements of $Y$ are even", "all of the elements of $Y$ are odd", and "some of the elements of $Y$ are even, and some are odd."   Rethink.
