Probability that 3 appears on one of the dice at the last throw of the game. 
A and B alternately throw a pair of die until one of them wins. A is
  considered winner if he gets a total of $7$ and B is considered winner
  if he gets total of $5$. If A starts the game, then probability that one
  of the die shows faces $3$ on last throw is:

Attempt: 
Events: E - given event
A - A wins ---> There are $36$ total possibilities out of which $6$ are favourable: $3+4, 2+5, 6+1$ and in different order. 
B - B wins ---> Here only $4$ out of $36$ are favourable: $2+3$, $4+1$
$P(A) = \displaystyle\dfrac 16\sum_{r= 0}^\infty  \left(\dfrac 56 \times \dfrac 8 9 \right)^r \implies P(A) = \dfrac {27}{42}$
$\implies 1- P(A) = P(B) = \dfrac{15}{42}$


*

*For E given A:
only $2/6 = 1/3$ are favourable cases $(3,4)$. 

*For E given B
only $2/4 = 1/2$ are favourable cases $(2,3)$. 


$\Pr(E) = \Pr(E \mid A)\Pr(A) + \Pr(E \mid B)\Pr(B)$
$\implies P(E) = \dfrac{1}{3}\times \dfrac {27}{42}+ \dfrac{1}{2}\times \dfrac{15}{42} = \dfrac{11}{28}$
But answer given is $P(E) = \dfrac{9}{28}$
Please let me know my mistake. 
 A: While you don't say how you calculate $\Pr(A)$, I'm guessing the $r$ in the sum is implicitly conditioning on the number of rounds. Let me show you a nice 'Markovian' way of doing this calculation.
By the Markov property, conditioning on the first (up to) two rolls, we have
\begin{align}
\Pr(A)
&
=
\tfrac16 + \tfrac56 \cdot \Pr(A \mid \text{first roll does not end game})
\\&
=
\tfrac16 + \tfrac56 \cdot \tfrac89 \cdot \Pr(A \mid \text{first two rolls do not end game})
\\&
=
\tfrac16 + \tfrac56 \cdot \tfrac89 \cdot \Pr(A),
\end{align}
and hence $\Pr(A) = \tfrac16 /(1 - \tfrac56 \tfrac 89) = \tfrac9{14}$; hence $\Pr(B) = 1 - \Pr(A) = \tfrac5{14}$.
Using, as you correctly determine, $\Pr(E \mid A) = \tfrac13$ and $\Pr(E \mid B) = \tfrac12$, in exactly the same way as you did, we find that
$$ \Pr(E) = \frac13 \cdot \frac9{14} + \frac12 \cdot \frac5{14} = \frac{11}{28}, $$
which is the answer that you get...

Note that we can calculate $\Pr(E \mid A)$ and $\Pr(E \mid B)$ similarly to $\Pr(A)$.
We have
$$ \Pr(E \mid A)
=
\frac{\Pr(E, A)}{\Pr(A)}.
$$
We need to calculate $\Pr(E, A)$ similarly: noting that the probability of getting both a $3$ and sum $7$ is $\tfrac1{18}$, we have
$$ \Pr(E, A)
=
\tfrac1{18} + \tfrac56 \tfrac89 \cdot \Pr(E, A \mid \text{first two rolls do not end the game}), $$
and hence we obtain $\Pr(E, A) = \tfrac13 P(A)$, as claimed.
