Conditional Probability - truth/lie The probability that two poker players speak truth is 3/5 and 2/3.  A card is drawn from the pack and they both agree that it is a queen.  What is the probability that it is really a queen ?
My approach:-
4 cases possible for A and B based on one saying Truths and Lies (TT,TF,FT,FF)
Case 1:- Both speaking truth and it is really a queen = 3/5 * 2/3 * 4/52
Case 2 and 3:- One person saying truth and other lie will not be possible as it would lead to an inconsistency because it is given that both agree it to be queen
Case 4:-Both speaking a lie, and it is not a queen = 2/5 * 1/3 * 48/52 = 8/65
Answer= 12/65
Answer given in book is 1/5
Can this be solved using probability tree diagram ?

If, as lulu suggested, A and B choose their lie with a uniform distribution, we wind up with a vastly different answer. The problem can be broken down as follows:
Define $Q, A_Q, B_Q$ to be the events that the drawn card is a queen, $A$ says the card is a queen, and $B$ says the card is a queen respectively.
Then, you have the following truth table:
$$\begin{array}{c|c|c|c}Q & A_Q & B_Q & \text{Prob} \\ \hline T & T & T & \tfrac{1}{13}\cdot \tfrac{3}{5}\cdot \tfrac{2}{3} = \tfrac{2}{65} \\ T & T & F & \tfrac{1}{13} \cdot \tfrac{3}{5} \cdot \tfrac{1}{3} = \tfrac{1}{65} \\ T & F & T & \tfrac{1}{13} \cdot \tfrac{2}{5}\cdot \tfrac{2}{3} = \tfrac{4}{195} \\ T & F & F & \tfrac{1}{13}\cdot \tfrac{2}{5}\cdot \tfrac{1}{3} = \tfrac{2}{195} \\ F & T & T & \tfrac{12}{13}\cdot \left(\tfrac{2}{5}\cdot \tfrac{1}{12}\right)\cdot \left(\tfrac{1}{3}\cdot \tfrac{1}{12}\right) = \tfrac{1}{1170} \\ F & T & F & \tfrac{12}{13}\left(\tfrac{2}{5}\cdot \tfrac{1}{12}\right)\left(1-\tfrac{1}{3}\cdot \tfrac{1}{12}\right)=\tfrac{7}{234} \\ F & F & T & \tfrac{12}{13}\left(1-\tfrac{2}{5}\cdot \tfrac{1}{12}\right)\left(\tfrac{1}{3}\cdot \tfrac{1}{12}\right)=\tfrac{29}{1170} \\ F & F & F & \tfrac{12}{13}\left(1-\tfrac{2}{5}\cdot \tfrac{1}{12}\right)\left(1-\tfrac{1}{3}\cdot \tfrac{1}{12}\right) = \tfrac{203}{234}\end{array}$$
Note: I am assuming a uniform distribution of lies So, if the card is not a queen, and they say it is a queen, not only is it the probability of them lying, it is also the probability that their lie happened to be a queen. So, $P(Q^C\cap A_Q^C)$ is the probability that it is not a queen and $A$ says a different card. $A$ will tell the truth with probability $\tfrac{3}{5}$. $A$ will say it is a queen with probability $\tfrac{2}{5}\cdot \tfrac{1}{12}$. $A$ will lie and say it is another card (that is not a queen) with probability $\tfrac{2}{5}\cdot \tfrac{11}{12}$. For example, if the actual card is a king, he will say it is not a king with probability $\tfrac{2}{5}$, but the chosen lie (ace through queen) is uniformly distributed over that event. So, each individual card that is not a king has its own $\tfrac{2}{5}\cdot \tfrac{1}{12} = \tfrac{1}{30}$ chance of being told.
Next, we add up the probabilities to ensure we have a total probability space:
$$\dfrac{2}{65}+\dfrac{1}{65}+\dfrac{4}{195}+\dfrac{2}{195}+\dfrac{1}{1170}+\dfrac{7}{234}+\dfrac{29}{1170}+\dfrac{203}{234} = 1$$
And indeed we do. So, unless I made a calculation error, we have:
$$P(Q|A_Q\cap B_Q) = \dfrac{P(Q\cap A_Q\cap B_Q)}{P(A_Q \cap B_Q)} = \dfrac{\left(\dfrac{2}{65}\right)}{\left(\dfrac{2}{65}\right)+\left(\dfrac{1}{1170}\right)} = \dfrac{36}{37}$$
Another note: This also assumes all events are independent (players choose to lie independently of reality and each other). This seems like another valid answer to the original problem since it was not clearly worded.
