Alice and Bob sometimes lie; a die is thrown and they claim different results; what is the probability that Bob is being honest? 
Alice speaks the truth with probability $3/4$ and Bob speaks the truth with probability $2/3$. A die is thrown and both Alice and Bob observe the number. Afterwards, Alice asserts to Carl (who does not know the number) that the number is $3$ while Bob says (to Carl) the number is $1$. Find the probability that the number is actually $1$.


UPDATE: To clear ambiguity, note that if person decides to lie, he/she will choose a false answer randomly from all the possible false answer ($\{1, 2, \cdots, 6\}$ - $\{\text{The number that actually showed up}\}$). Also, a die is thrown, and then both Alice and Bob will see the number. Then, they will lie/say truth accordingly.

My attempt:
Case $1$: Number $1$ showed up.
Chance of all this happening = $\frac{1}{6} \cdot \frac{2}{3} \cdot \big(\frac{1}{4} \cdot \frac{1}{5}\big) = \frac{1}{180}$
Case $2$: Number $3$ showed up
Chance of all this happening = $\frac{1}{6} \cdot \frac{3}{4} \cdot \big(\frac{1}{3} \cdot \frac{1}{5}\big) = \frac{1}{120}$
Case $3$: Other number showed up
Chance of all this happening = $\frac{4}{6} \cdot (\frac{1}{4} \cdot \frac{1}{5}) \cdot (\frac{1}{3} \cdot \frac{1}{5}) = \frac{1}{450}$
So, total = $\boxed{\frac{\frac{1}{180}}{\frac{29}{1800}} = \frac{10}{29}}$
Is my attempt correct? If not, how to do this problem?

Nice problem!  I think OP has the right answer using a correct method and I have nothing to add in terms of the calculation, but perhaps if we set up some careful notation and also be very precise about our independence assumptions, it will become more clear.  For $k=1,2,3,4,5,6$, define the events

*

*$D_k$ that the die shows $k$;

*$A_k$ that Alice says it is $k$;

*$B_k$ that Bob says it is $k$.

Then the problem asks for $P(D_1\mid A_3\cap B_1)$.  We'll assume the probabilities
$$P(A_l\mid D_k)=\begin{cases}\frac34&\hbox{if $l=k$}\cr \frac14\cdot\frac15&\hbox{if $l\ne k$,}\end{cases}\qquad
  P(B_m\mid D_k)=\begin{cases}\frac23&\hbox{if $m=k$}\cr \frac13\cdot\frac15&\hbox{if $m\ne k$}\end{cases}$$
and also that Alice's and Bob's responses to any roll are independent in the sense that
$$P(A_l\cap B_m\mid D_k)=P(A_l\mid D_k)P(B_m\mid D_k)$$
for all $k,l,m$.
Then we have
$$\eqalign{P(D_1\mid A_3\cap B_1)
  &=\frac{P(D_1\cap A_3\cap B_1)}{P(A_3\cap B_1)}\qquad
    \hbox{(definition of conditional probability)}\cr
  &=\frac{P(D_1\cap A_3\cap B_1)}{\displaystyle \sum_{k=1}^6P(D_k\cap A_3\cap B_1)}\qquad\hbox{(total probability rule)}.\cr}$$
Now using the definition of conditional probability and the above assumptions,
$$\eqalign{P(D_k\cap A_3\cap B_1)
  &=P(A_3\cap B_1\mid D_k)P(D_k)\cr
  &=P(A_3\mid D_k)P(B_1\mid D_k)P(D_k)\cr
  &=\begin{cases}
    \frac14\cdot\frac15\cdot\frac23\cdot\frac16&\hbox{if $k=1$}\cr
    \frac34\cdot\frac13\cdot\frac15\cdot\frac16&\hbox{if $k=3$}\cr
    \frac14\cdot\frac15\cdot\frac13\cdot\frac15\cdot\frac16&\hbox{if $k\ne1,3$,}\end{cases}\cr}$$
and substituting back above gives the answer $\frac{10}{29}$.
