Conditional Probability: Bridge Hand given north and south has 8 spades I apologise as I'm pretty certain I could find the actual answer myself on these boards (as in the solution to the problem) but if possible I'd like it if someone can confirm whether my thinking is correct.

In the card game bridge, the entire 52 cards are dealt out equally to
4 players, E/W/N/S if N and S have a total of 8 spades among them,
what is the probability E has 3 of the remaining 5 spades?

This is example 2c on page 60 in "A First Course in probability 8th edition"
and gives the solution as $$\frac{{{5}\choose{3}}{{21}\choose{10}}}{{26}\choose{13}}\approx .339$$
Later in the book, it asks to recalculate this using conditional probability (question 3.3 page 102):

Compute the conditional probability that E has 3 spades given North
and South have a combined total of 8 spades.

My answer is then as follows:
Let E be the event that E has 3 spades. and F the event that N and S have 8 spades.
Then $P(E)=\frac{{{5}\choose{3}}{{21}\choose{10}}}{{26}\choose{13}}$
as it is given that 26 cards have already been dealt to N and S, of which 8 of the 13 spades have also been dealt. This means for E to occur we have to choose 3 of the remaining 5 spades and any combination of 10 cards remaining from the 21 leftovers (that is 26 remaining cards minus the 5 spades). All are divided by the reduced sample space.
(This is one of the parts in which I want to be certain)
As we only care about E we can consider N and S to be one person which will be dealt 26 cards, 8 of which must be spades, giving
$P(F)=\frac{{{13}\choose{8}}{{39}\choose{19}}}{{52}\choose{26}}$
this reasoning seems right to me as the two of them should receive any combination of 26 cards so long as 8 of them are spades. (We don't care if N has all 8 or if S has all 8.)
Then our conditional Probability is
$$P(E|F) = \frac{P(E\cap F)}{P(F)}.$$
My major issue is I'm not sure how I would go about calculating $P(E\cap F)$ in this instance. Logically I argue that the two events P and F are independent of each other, meaning
$$P(E\cap F)=P(E)P(F)$$ and so making our final conditional probability $P(E|F)=P(E)$ which yes does give me the right answer, but unless the question specifically states that these are independent of each other, I would then have to prove that $P(E\cap F)=P(E)P(F)$ holds surely?
Any clarification would be greatly appreciated. thank you.
 A: Calculating $Pr(E\cap F)$, let us define our sample space as all ways of distributing the cards where order within each hand doesn't matter and where north and south's hands are considered collectively as one.  There are $\frac{52!}{26!13!13!}$ such distributions, each of which are equally likely to occur.
Let us count $|E\cap F|$ in this sample space.  To do so, first choose which eight spades north/south got and then which 18 non-spades north/south got.
Then, choose which three spades from those remaining east got and which 10 non-spades from those remaining east got.  All remaining cards will be given to west.
There are then $\binom{13}{8}\binom{39}{18}\binom{5}{3}\binom{21}{10}$ such arrangements implying the probability $Pr(E\cap F)$ is:
$$Pr(E\cap F)=\frac{\binom{13}{8}\binom{39}{18}\binom{5}{3}\binom{21}{10}}{52!/(26!13!13!)}=\frac{15152709}{276092852}\approx 0.05488$$
Continuing on to calculate $Pr(F)$, now we temporarily instead consider the sample space to be just the ways in which we divide the deck in half, i.e. giving 26 cards to north/south collectively where order within the hand is unimportant and giving the remaining cards to east/west.  There are $\binom{52}{26}$ such ways to do so, each of which are equally likely to occur.
We count $|F|$ in this sample space as being $\binom{13}{8}\binom{39}{18}$, thus making $Pr(F)=\frac{\binom{13}{8}\binom{39}{18}}{\binom{52}{26}}=\frac{44681065}{276092852}\approx 0.161833$
Taking the ratio $\frac{Pr(E\cap F)}{Pr(F)}$ will arrive at the same answer as given before after simplifications.
$$Pr(E\mid F)=\frac{Pr(E\cap F)}{Pr(F)}=\frac{39}{115}\approx 0.3391304$$

As an aside, let us look at $Pr(E)$ and inspect whether $E$ and $F$ truly are independent.
We let the sample space be all ways in which East is given a hand ignoring how the rest of the cards are distributed.  There are $\binom{52}{13}$ ways in which this can happen.  Choosing which three spades and which 10 nonspades gives us $\binom{13}{3}\binom{39}{10}$ possible hands and a probability of $Pr(E)=\frac{\binom{13}{3}\binom{39}{10}}{\binom{52}{13}}\approx 0.28633$
(notice that no mention of how spades are distributed among north/south appeared in the calculation of $Pr(E)$.  You seem to have confused calculating $Pr(E)$ with $Pr(E\mid F)$)
We compare $Pr(E\cap F)\approx 0.05488$ to $Pr(E)Pr(F)\approx 0.161833\cdot 0.28633\approx 0.04634$.  Indeed, even if we take the exact values we see that these are different numbers and therefore $E$ and $F$ are not independent as claimed.
A: For calculating $P(E\cap F)$ think of the number of ways that $13$ spades can be divided in this context such that exactly $3$ come in hands of $\bf E$ and exactly $2$ in hands of $\bf W$. 
There are $\binom{13}3\binom{13}2\binom{26}8$ ways, and this on a total of $\binom{52}{13}$ ways. 
This results in: $$P(E\cap F)=\frac{\binom{13}3\binom{13}2\binom{26}8}{\binom{52}{13}}$$
Sortlike you can find $P(F)$ as:$$P(F)=\frac{\binom{26}5\binom{26}8}{\binom{52}{13}}$$
Working this out with application of $P(E\mid F)P(F)=P(E\cap F)$ leads to:$$P(E\mid F)=\frac{13!13!5!21!}{3!10!2!11!26!}=\frac{\binom53\binom{21}{10}}{\binom{26}{13}}$$ 
