Probability when $A$ and $B$ speak same statement after roll of dice $A$ speaks truth with probability of $1/3$ and $B$ speaks truth with probability of $2/5$. A single die is rolled and both $A$ and $B$ say the same statement simultaneously about the outcome on the top of the die. What is the probability that their statement is true?
I am having difficulty in writing total probability.
In my opinion, 'Either they Speak truth or they speak same lie.'
Probability that they speak truth=$\frac{1}{3} \cdot \frac{2}{5}$
Probability that they speak same lie=$\frac{1}{5} \cdot \frac{2}{3} \cdot \frac{3}{5}$
Multiplied by $1/5$ because among $5$ options for lying they must choose one same number.
But I still feel something is missing. Could someone help me with this?
Edit: The die is fair with possible outcomes $\{1,2,3,4,5,6\}$ and we are concerned with number that appears on the top of the die, i.e. they both say "Outcome is $1$" or "Outcome is $2$" ........ "Outcome is $6$"
 A: If they give the same number, the chances it's correct are 2/15, and incorrect, 2/25, as stated in the question.  The probability of their answer being correct is (2/15)/(2/15 + 2/25): 5/8 or 62.5%. 
Why has no one given an answer in 8 hours? 
Incidentally, Mr. A and  Ms. B are honest persons. I know them both. When they're wrong it's an honest mistake, not a lie, which the question and all the comments assumed. They're both near-sighted. 
A: Edit: I have misread the question and gave a solution in the case where $A$ and $B$ have rolled separate dice each. I shall keep the solution here but note that I have solved a different problem.
Let $A_t, B_t$ be probabilities that $A$ and $B$ speaks the truth respectively. Let $C$ denote the event of $A$ and $B$ saying the same dice value. The question is asking for $P(A_t \cap B_t \mid C)$.
Using Bayes' conditional probabilities, $$P(A_t \cap B_t \mid C)=\frac{P(A_t\cap B_t \cap C)}{P(C)}$$
For the numerator, the only way $A$ and $B$ speaks the truth and speaks the same number is if they indeed got the same number. So the numerator is simply the probability they get the same number and both speak the truth. To calculate this, think that it does not matter what $A$'s number is just that $B$ rolls the same number - all whilst considering the probabilities they speak the truth. So $$P(A_t\cap B_t \cap C)=\frac13 \cdot \frac25 \cdot \frac16=\frac2 {90}$$
To calculate $P(C)$, we can split $C$ into $4$ cases:
$$\begin{cases} C\cap A_t \cap B_t=D_1 \\ C\cap A_t \cap B_t'=D_2 \\ C\cap A_t' \cap B_t=D_3 \\ C\cap A_t' \cap B_t'=D_4\end{cases}$$
And since the four cases partition $C$, we can apply the law of total probability to obtain
$$P(C)=P(D_1)+P(D_2)+P(D_3)+P(D_4)$$
We have already calculated $D_1$.
For $D_2$, $A$ can get any value as long as $B$ doesn't, all the whilst $A$ speaks the truth, $B$ lies and choses to lie the value $A$ gets (we assume that when $B$ lies, they choose a false value with equal probability). So 
$$P(D_2)=\frac13 \cdot \frac56 \cdot \frac35 \cdot \frac15=\frac1{30}$$
$P(D_3)$ is obtained in a similar manner - we reverse the roles of $A$ and $B$:$$P(D_3)=\frac25 \cdot \frac56 \cdot \frac23 \cdot \frac 15=\frac2{45}$$
For $P(D_4)$, we split into further two subcases - $A$ and $B$ get the same value, or they don't.
For the former, say event $E$, $A$ can get any value it wants, but $B$ must match it. Then, $A$ must lie and say any false value, and $B$ must also lie and choose to say the same false value:
$$P(E)=\frac23 \cdot \frac16 \cdot \frac35 \cdot \frac15= \frac1{75}$$
For the second subcase, say event $F$, $A$ can get any value bust must lie, and $B$ must get a different value, must lie, but also choose the same lie. But when they lie, they both cannot choose their own value or the other's value. So when $A$ lies 'first', they have $4$ out of $5$ possible values to lie from. So $$P(F)=\frac23 \cdot \frac45 \cdot \frac56 \cdot \frac35 \cdot \frac15= \frac4{75}$$
So $P(D_4)=\frac1{15}$
Putting this altogether: $$\begin{align*} P(A_t \cap B_t \mid C)&=\frac{P(A_t\cap B_t \cap C)}{P(C)}\\ &=\frac{\frac{2}{90}}{\frac2{90}+\frac1{30}+\frac2{45}+\frac1{15}}
\\ &=\frac2{15}
\end{align*}$$
A: The probabilities calculated in the question are correct under the assumption that with probability $\frac13,$ person $A$ will say that a certain number was rolled and that number will actually be the one rolled, whereas with probability $\frac23$ person $A$ will randomly choose from among the other five numbers with uniform probability and will declare that $A$'s chosen number was the one rolled, and that $B$ will act similarly except that $B$'s chance of naming the number actually rolled is $\frac25.$
This seems a reasonable interpretation of the problem statement.
It is not the only plausible interpretation, but it is one that
makes the question answerable (as opposed to what happens under some other possible interpretations).
My only misgiving is in the reasoning behind the factor $\frac15$ in the product $\frac{1}{5} \cdot \frac{2}{3} \cdot \frac{3}{5}.$
Suppose we also have a third person, $C,$ who speaks truth with probability $\frac12$ and who also declares what number was rolled, and the number declared by $C$ is the same as that declared by $A$ and by $B.$
Is the probability of three simultaneous false statements
$\frac{1}{5} \cdot \frac{2}{3} \cdot \frac{3}{5} \cdot \frac12$ because there are still five possible options for the false statement?
On the contrary, I would say the a priori chance of that all three will make the same false statement is 
$\left(\frac15\right)^2 \cdot \frac23 \cdot \frac35 \cdot \frac12$;
for example, supposing the actual roll is $6,$ there is a probability
$\left(\frac15\right)^3 \cdot \frac23 \cdot \frac35 \cdot \frac12$
that $A,$ $B,$ and $C$ all claim it is $1,$ but we multiply this by $5$ to account for all the possible ways the wrong number could be the same one for all three observers.
The wording in the question is similar enough to this line of reasoning that the question might be based on a correct line of reasoning; my doubts are merely because the reasoning is not spelled out in enough detail to be sure it is correct.
