Probability of truth in a chain of statements 
Three individuals $A$,$B$ and $C$ tell the truth with probability $1/3$.
(A) $C$ makes a statement and $A$ claims that it is true. What is the probability that the statement is true.
(B) $C$ makes a statement and $A$ tells you that $B$ claims the statement is true. What is the probability that the statement is true.

Let $T_A$ be the event that $A$ tells the truth and similarly $T_B \ \& \ T_C$
My answer in the first case $\frac{1}{3}\cdot \frac{1}{3} = \frac{1}{9}$.

I am confused with part $B$ of the question.
My reasoning: Probability of $C$ telling the truth is $1/3$. Now we don't know what $B$ said so we assume two cases. Therefore we would have the probability as $P(T_A \ T_B \ T_C \cup T_A \ T_B^C \ T_C^C) = \frac{1}{3}\cdot \frac{1}{3} \cdot \frac{1}{3}+\frac{1}{3}\cdot \frac{2}{3} \cdot \frac{2}{3}$.
Is this correct? This problem appears as an exercise to Bayes theorem so I am confused as to how the theorem is applicable.
Thanks in advance for any assistance.
 A: Careful. We need to distinguish between "$X$ tells the truth" and "$X$ claims that $Y$ told the truth". The former always happens with probability $1/3$, while the latter depends on whether or not $Y$ is true or false. Let's revisit part (a) and redefine our events. Define:


*

*$X_C$: $C$ makes a true statement.

*$X_A$: $A$ claims that $C$ makes a true statement.


We're asked to find the probability that $C$ made a true statement, given that $A$ claims that $C$ made a true statement:
\begin{align*}
\Pr[X_C \mid X_A]
&= \frac{\Pr[X_CX_A]}{\Pr[X_A]} \\
&= \frac{\Pr[X_CX_A]}{\Pr[X_CX_A] + \Pr[\overline{X_C}X_A]} \\
&= \frac{\frac{1 \cdot 1}{9}}{\frac{1 \cdot 1}{9} + \frac{2 \cdot 2}{9}} \\
&= \frac{1}{1 + 4} \\
&= \frac{1}{5}
\end{align*}
Can you see how to generalize this for part (b)?

EDIT: Here's part (b). Define:


*

*$X_C$: $C$ makes a true statement.

*$X_B$: $B$ claims that $C$ makes a true statement.

*$X_A$: $A$ claims that $B$ makes a true statement.


We're asked to find the probability that $C$ made a true statement, given that $A$ claims that $B$ made a true statement:
\begin{align*}
\Pr[X_C \mid X_A]
&= \frac{\Pr[X_CX_A]}{\Pr[X_A]} \\
&= \frac{\Pr[X_CX_BX_A] + \Pr[X_C\overline{X_B}X_A]}{\Pr[X_CX_BX_A] + \Pr[X_C\overline{X_B}X_A] + \Pr[\overline{X_C}X_BX_A] + \Pr[\overline{X_C} \, \overline{X_B}X_A]} \\
&= \frac{\frac{1 \cdot 1 \cdot 1}{27} + \frac{1 \cdot 2 \cdot 2}{27}}{\frac{1 \cdot 1 \cdot 1}{27} + \frac{1 \cdot 2 \cdot 2}{27} + \frac{2 \cdot 2 \cdot 1}{27} + \frac{2 \cdot 1 \cdot 2}{27}} \\
&= \frac{1 + 4}{1 + 4 + 4 + 4} \\
&= \frac{5}{13}
\end{align*}
This diagram might help:

A: Hint: Find the conditional probability that the statement is true given the reports, by establishing who must speak true or false for the report to make the statement true or false.
Let $T$ be "the statement is true and the reports were so spoken", and $F$ be "the statement is false and the reports were so spoken".  Find:
$$\mathsf P(T\mid (T\cup F))$$
$1)$ By what was spoken, the statement is true if both $A$ and $C$ spoke true, but it is false if both spoke false.   $T=A\cap C$ and $F = A^\complement\cap C^\complement$.
$$\mathsf P\big(T\mid T\cup F\big) ~{= \dfrac{\mathsf P(A\cap C)}{\mathsf P\big((A\cap C)\cup(A^\complement\cap B^\complement)\big)}}$$
A: We shall use a clear straightforward approach.  There are $2^3=8$ possibilities whether each of $A$, $B$, and $C$ tells the truth. We shall denote each of these possibilities by a triple of letters “T” and “F”, where a letter at the respective place indicates whether the respective person tell the truth. For instance, a triple TFT means that A and C tell the truth, whereas B lies. Assuming that each person says the truth with probability $1/3$ independently of that say (or said) the others, we obtain the following probabilities for triples (for the convenience we use a constant $c=1/27$):


*

*TTT – 1c

*TTF, TFT, FTT – 2c

*TFF, FTF, FFT – 4c

*FFF – 8c


It is easy to check the case when C makes a statement and A tells you that B claims the statement is true corresponds to (and only to) the triples TTT, TFF, FTF, FFT. Then by Bayes formula the probability that the statement is true is 
$$\frac{P(TTT)+P(FFT)}{P(TTT)+P(TFF)+ P(FTF)+P(FFT)}=\frac{1c+4c}{1c+4c+4c+4c}=\frac 5{13}.$$
