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,⋯,6} - {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 = $\frac{1}{6}\cdot \frac{2}{3}\cdot {\textstyle (}\frac{1}{4}\cdot \frac{1}{5}{\textstyle )}=\frac{1}{180}$

Case 2: Number 3 showed up

Chance of all this happening = $\frac{1}{6}\cdot \frac{3}{4}\cdot {\textstyle (}\frac{1}{3}\cdot \frac{1}{5}{\textstyle )}=\frac{1}{120}$

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 = $\overline{){\displaystyle \frac{\frac{1}{180}}{\frac{29}{1800}}=\frac{10}{29}}}$

Is my attempt correct? If not, how to do this problem?

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,⋯,6} - {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 = $\frac{1}{6}\cdot \frac{2}{3}\cdot {\textstyle (}\frac{1}{4}\cdot \frac{1}{5}{\textstyle )}=\frac{1}{180}$

Case 2: Number 3 showed up

Chance of all this happening = $\frac{1}{6}\cdot \frac{3}{4}\cdot {\textstyle (}\frac{1}{3}\cdot \frac{1}{5}{\textstyle )}=\frac{1}{120}$

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 = $\overline{){\displaystyle \frac{\frac{1}{180}}{\frac{29}{1800}}=\frac{10}{29}}}$

Is my attempt correct? If not, how to do this problem?