A question in Probability, aces drawn from two halves of a shuffled deck "A deck of cards is shuffled and then divided into two halves of 26 cards each. A card is drawn from one of the halves, it turns out to be an ace. The ace is then placed in the second half-deck. The half is then shuffled and a card is drawn from it. Compute the probability that this drawn card is an ace." 

Source : A First Course in Probability, Sheldon Ross, Chapter 3, Exercise 37 

(My intention was not to be lazy and let the community do all the work for me, but I think that is what has been misunderstood seeing the 2 dislikes this question received. I just thought I should not clutter the question with more text than necessary)

What I've tried : Probability that the second half already contained 0,1,2 or 3 aces before the ace from the first half was added to it. Then, when the new ace was added to it from the first half, we could calculate the probabilities of drawing an ace considering each of these cases, and add them to get the answer. But I am not sure how do I calculate the P(second half contained 0/1/2/3 aces)..
Here are two solutions:
Let $A$ be the event an ace is drawn from the second half
Let $I$ be the event that the interchanged card is selected

$P(A) = P(A|I)\cdot P(I) + P(A|I^c)\cdot P(I^c)$
$P(A) = 1\cdot\dfrac{1}{27} + \dfrac{3}{51}\cdot\dfrac{26}{27}$
$P(A) = \dfrac{43}{459} = 0.094$

Note that $P(A|I^c) = \dfrac{3}{51}$ because there are 3 aces to choose from and all 51 other cards are equally likely to be in the second half

Second Solution:
Let $A$ be the event an ace is drawn from the second half
Let $N_i$ be the event that $i$ aces are in the second half with $0 \leq i \leq 3$

$P(A) = \sum_{i=0}^3 P(A|N_i)\cdot P(N_i)$
$P(A) = P(A|N_0)\cdot P(N_0) + P(A|N_1)\cdot P(N_1) + P(A|N_2)\cdot P(N_2) + P(A|N_3)\cdot P(N_3)$ 
$P(A) = \dfrac{1}{27}\cdot P(N_0) + \dfrac{2}{27}\cdot P(N_1) + \dfrac{3}{27}\cdot P(N_2) + \dfrac{4}{27}\cdot P(N_3)$

Note that there are only 3 aces to choose from and 51 cards total, since one ace is in the first half

$P(N_0) = \dfrac{\dbinom{3}{0}\dbinom{48}{26}}{\dbinom{51}{26}}$
$P(N_1) = \dfrac{\dbinom{3}{1}\dbinom{48}{25}}{\dbinom{51}{26}}$
$P(N_2) = \dfrac{\dbinom{3}{2}\dbinom{48}{24}}{\dbinom{51}{26}}$
$P(N_3) = \dfrac{\dbinom{3}{3}\dbinom{48}{23}}{\dbinom{51}{26}}$  

$P(A) = \dfrac{43}{459} = 0.094$
Let $A$ be the event that an ace is drawn from the first half of the deck.

Let $X$ be the count of aces in the second half-deck before the additional ace is added from the first deck.

$$\mathsf P(X=x\mid A) ~=~ \dfrac{\dbinom{4}{x}\dbinom{48}{26-x}\mathbf 1_{x\in\{0,1,2,3\}}}{\dbinom{52}{26}-\dbinom{48}{26}}$$

Let $B$ be the event of drawing an ace after adding one and shuffling. &nbsp; There are now $X+1$ aces among $27$ cards.

$$\mathsf P(B\mid X=x, A)~=~\dfrac{x+1}{27}$$

By the Law of Total Probability.

$$\begin{align}\mathsf P(B\mid A)~=~&amp; \sum_{x=0}^3 \mathsf P(B\mid X=x, A)\mathsf P(X=x\mid A)
\\[1ex] ~=~&amp; \dfrac{1\cdot \dbinom{3}{0}\dbinom{48}{25}+2\cdot \dbinom{3}{1}\dbinom{48}{24}+3\cdot\dbinom{3}{2}\dbinom{48}{24}+4\cdot \dbinom{3}{3}\dbinom{48}{23}}{27~\left(\dbinom{52}{26}-\dbinom{48}{26}\right)}\end{align}$$

Remark: Since we are given that event $A$ has happened, the probability that we seek is indeed conditional on this event: $\mathsf P(B\mid A)$