Probability problem on umbrellas 
Two absent-minded roommates, mathematicians, forget their umbrellas in some way or another. $A$ always takes his umbrella when he goes out, while $B$ forgets to take his umbrella with probability $1/2$. Each of them forgets his umbrella at a shop with probability $1/4$. After visiting three shops, they return home. Find the probability that they have only one umbrella.

My Attempt
$A_i/\bar{A}_i$ : $A$ remembers or forgets umbrella at shop $i$
$B_i/\bar{B}_i$ : $B$ remembers or forgets umbrella at shop $i$
$B_o/\bar{B}_o$ : $B$ remembers or forgets umbrella at shop home
$A_i,,B_i=3/4;\bar{A}_i,\bar{B}_i=1/4$ and $B_o,\bar{B}_o=1/2$
$$
\text{Req. Prob.}=A_1A_2A_3\bar{B}_o+A_1A_2A_3B_o\bar{B}_1+A_1A_2A_3B_oB_1\bar{B}_2+A_1A_2A_3B_oB_1B_2\bar{B}_3+\bar{A}_1B_oB_1B_2B_3+A_1\bar{A}_2B_oB_1B_2B_3+A_1A_2\bar{A}_3B_oB_1B_2B_3\\
=\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)+\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)\Big(\frac{1}{4}\Big)+\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)\Big(\frac{1}{4}\Big)+\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^2\Big(\frac{1}{4}\Big)+\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^3\\
+\Big(\frac{3}{4}\Big)\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^3+\Big(\frac{3}{4}\Big)^2\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^3\\
=\frac{27}{128}+\frac{27}{128*4}+\frac{81}{128*16}+\frac{243}{128*64}+\frac{27}{128*4}+\frac{81}{128*16}+\frac{243}{128*64}\\
=\frac{3726}{128*64}=\frac{3726}{8192}
$$
But my reference gives the solution $\frac{7278}{8192}$, so what am I missing in my attempt ?
Note: Please check $b$ part of link for a similar attempt to solve the problem to obtain the solution as in y reference.
 A: You want the probability that only one of the two remembers their umbrella at every point.
At the end of the trip, $A$ will have an umbrella if it is remembered at each of the three stores.  This has a probability of $(3/4)^3$
At the end of the trip, $B$ will have an umbrella if it is remembered at home and each of the three stores.  This has a probability of $(1/2)(3/4)^3$
So the probability that exactly one has an umbrella at the end is, indeed:$$\begin{align}&\dfrac {3^3}{4^3}\left(1-\dfrac{3^3}{2\cdot4^3}\right)+\left(1-\dfrac{3^3}{4^3}\right)\dfrac{3^3}{2\cdot4^3}\\[1ex]=&\dfrac{3^4\left(2^5-3^2\right)}{2^{12}}\\[1ex]=&\dfrac{1863}{4096}\\[1ex]=&\dfrac{3726}{8192}\end{align}$$

But my reference gives the solution $7278/8192$, so what am I missing in my attempt ?

The probability that at most one has an umbrella by the end is:$$1-\dfrac{3^3}{2\cdot 4^3}\cdot\dfrac{3^3}{4^3}=\dfrac{7278}{8192}$$
The difference being: the probability that they both forget the umbrella.
A: I get a different answer from either of those.  The probability that a professor who leaves the house with his umbrella returns with it is $$\left(\frac34\right)^3=\frac{27}{64}$$ and the probability that he loses it is $\frac{37}{64}$.  There are two cases.  Both A and B take their umbrellas with them (probability $\frac12$) and exactly one returns with it (probability $2\cdot\frac{27}{64}\cdot\frac{37}{64}$) or only B takes his umbrella (probability $\frac12$,) and he loses it (probability $\frac{37}{64}$.) This gives a total probability of $$
\begin{align}
&\frac12\cdot2\cdot\frac{27}{64}\cdot\frac{37}{64}+\frac12\cdot\frac{37}{64}\\&=\frac12\cdot\frac{37}{64}\left(2\cdot\frac{27}{64}+1\right)\\&=\frac{37}{128}\cdot\frac{121}{64}\\&=\frac{4477}{8192}
\end{align}$$ 
EDIT
The case where B doesn't take his umbrella and $A$ accounts for $\frac{2109}{8192}$ of this probability, so if we exclude it, we get a substantially smaller probability than you did.  It seems to be that are computing the probability that at least one of them leaves his umbrella at a shop, which is the wrong thing, because if both lose their umbrellas, they have zero umbrellas, not one.
A: I think your answer is correct. Each of them can forget the umbrella independently.
So:
Pr{A forgets}= $\frac{1}{4}+\frac{3}{4}.\frac{1}{4}+\frac{3}{4}\frac{3}{4}\frac{1}{4}= \frac{74}{128}$
Pr{B forgets}=$\frac{101}{128}$
Consequently: Pr{one umbrella} = Pr{A forgets}.(1-Pr{B forgets})+(1-Pr{A forgets}).Pr{B forgets}= $\frac{3726}{8192}$
A: A forgets his umbrella along the way with probability $p=1-\left(\frac{3}{4}\right)^3=\frac{37}{64}$. B forgets his umbrella along the way with probability $q=\frac{1}{2}\left(1-\left(\frac{3}{4}\right)^3\right)=\frac{37}{128}$. Therefore, the probability that exactly one umbrella remains when both A and B return home is $$\begin{split}p(1-q)+q(1-p)&=p+q-2pq=\\&=\frac{37}{64}+\frac{37}{128}-\frac{37\cdot37}{4096}=\frac{37\cdot(64+32-37)}{4096}=\frac{37\cdot59}{4096}=\frac{2183}{4096}.\end{split}$$
