Suppose that cards are picked at random from a full deck of 52 cards
What is the probability that exactly two jacks, two queens, and two kings have been picked from the deck when the second ace turns up?
NO. of ways to select 2 jack:$ 4\choose 2$
NO. of ways to select 2 queen: $4\choose 2$
NO. of ways to select 2 king : $4\choose 2$
NO. of ways to select 2 ace : ${4\choose 1} * {3\choose 1}$
$\therefore \text{ required probability is} \frac{{4\choose 2}*{4\choose 2}*{4\choose 1}*{3\choose 1}}{52\choose 8}$
Is this answer correct??
Your answer is not correct.
We only have to look at jacks, queens, kings and aces. The other cards may be neglected.
Also no distinction is relevant between e.g. the jack of aces and the jack of spades.
So actually we are dealing with equiprobable strings of length $16$ with the letters letters J,K,Q,A each $4$ times.
There are: $$\frac{16!}{4!4!4!4!}=63063000 $$ strings in total.
It remains to find the number of strings with A on spot $8$, and next to that exactly twice a J, twice a Q, twice a K and once an A on the spots $1,2,3,4,5,6,7$.
That number is: $$\frac{7!}{2!2!2!1!}\times\frac{1!}{1!}\times\frac{8!}{2!2!2!2!}=1587600 $$
So we end up with probability:$$\frac{1587600}{63063000}=\frac{15876}{630630}\approx0,0252 $$
Actually it is the probability that among the first $8$ cards there are exactly $2$ of each sort divided by $4$ (because the ace, jack, queen and king have equal probability to be the card on spot $8$).
So you could write the probability also as:$$\frac14\times\frac{\left(\frac{8!}{2!2!2!2!}\right)^2}{\frac{16!}{4!4!4!4!}}=\frac14\times\frac{\binom42^4}{\binom{16}8}$$
In its looks that comes closer to the (wrong) answer that you gave in your question.