Probability that the person speak true (has he drawn two aces when he says so ?) I tried much in this problem But I didn't get my answer correct.
The question states--
A person draws two cards successively without replacement from a pack of 52 playing cards.He tells that both cards are aces,then what is  the probability that both cards are actually aces if there are 60% chances that he speaks truth-----
My attempt -----
$P(T)=0.6$ let$ E1 $be the event that the person says that the first card drawn is ace and $E2 $be the second card drawn is ace said by the person.
Then $$P(E1)=\frac{1}{13}\cdot 0.6+\frac{48}{52}\cdot 0.4$$
Similarly  $$P(E2 \mid E1)=(\frac{7}{15}\cdot 0.6+\frac{95}{51}\cdot 0.4)\cdot (\frac{0.6}{13}+\frac{4.8}{13})$$
So,$$P(E1 \cap E2)=P(E2 \mid E1)\cdot P(E1)$$
But I am not getting the answer.please help me in this regard.
Thank you.
 A: Let's solve a general problem, then check a couple of reasonable outcomes, and finally come back to the specific.
Let $E$ be an event happening with probability $p(E)  $, $S$ be a statement that the event happened made with probability $p(S)$ by a person whose statements are true with probability $p(T)$ 
We want to know what is the probability of the event given that the person states that it did. This is a conditional probability $p(E|S)$ and using the standard formula, $p(E|S) = p(E \& S)/p(S)$
$p(E \& S)$ is easily determined: the probability that the event took place and the person stated it did = the probability that the event took place and the person made a true statement $= p(E).p(T)$
$p(S)$ can't be determined directly. The person can make lots of statements (in this case, he could call two kings, jack high, etc.) and we don't know the relative frequencies of them. But he either makes the statement $S$ or a different one. We can consider all the different statements together as "not S" and $p(S) = 1 - p($not $ S)$. 
We can determine $p($not $ S)$ as the probability of not $E$ and he tells the truth + probability of $E$ and he lies: $p($not $ S) = (1 -p(E)).p(T) + p(E).(1 - p(T) $ 
So, $p(S) = 1 - (1 -p(E)).p(T) + p(E).(1 - p(T)$ and $p(E|S) = p(E \& S)/p(S) = p(E).p(T) / (1 - (1 -p(E)).p(T) + p(E).(1 - p(T))$
We would like to see that if the person always tells the truth, then $p(E|S) = p(E \& S)/p(S) = 1$, so put $p(T) = 1$ and then $p(E|S) = p(E \& S)/p(S) = p(E).1 / (1 - (1 -p(E).1) + p(E).(1 - 1) = p(E)/p(E) = 1 $ (assuming a nonzero probability $p(E)$ for the event). So that works like it should.
We would also like to see that if the event can never happen (i.e. $p(E) = 0$ ) then $p(E|S) = p(E \& S)/p(S) = 0$. Note that in this case the person can only make the statement $S$ if he lies, so $P(T) < 1$. Then $p(E|S) = p(E \& S)/p(S) = 0.p(T) / (1 - (1 -0).p(T) + 0.(1 - p(T)) = 0/(1 -p(T)) = 0$  with $P(T) < 1$. So that works too.
We can also check what happens if the person always lies (i.e. $p(T) = 0$). then we would expect $p(E|S) = p(E \& S)/p(S) = 0$.
$p(E|S) = p(E \& S)/p(S) = p(E).0 / (1 - (1 -p(E)).0 + p(E).(1 - 0) = 0/(1+p(E) = 0 $. That also works.

The aces problem:
$p(E)$ is now that probability of drawing two aces from the deck, and $p(T) = 0.6 = 3/5$. Calculate $p(E)$:
There are $^4C_2$ combinations of two aces out of 4, and $^{52}C_2$ two card draws. So, $p(E)  = ^4C_2/ ^{52}C_2$ $= (4!/2!.2!) / (52!/50!.2!) = (6)/ (52.51/2) = 6/26.51 = 1/13.17 = 1/221$
Plug this into the previous result:
$p(E|S) = (1/221).(3/5) / (1 - (1 -1/221).(3/5) + (1/221).(1 - 3/5))$ $= (3/5)/ (221 - (221 - 1).(3/5) +(1 - 3/5))$ $ = (3/5)/ (221 - 132 + 2/5) = 3/447 = 1/149.$ 

How $p(E|S) $ varies with $p(T)$
As a point of interest I plotted the probability that the person has two aces when he says so against the probability that he tells the truth. For an unlikely event ($p(E) = 1/221$) he has to be very truthful before you believe him: around 99.5% truthful to even have a 50% chance of him holding two aces.

A: So probably he was lying
H1-he has 2 aces
H2-he hasn't 2 aces
p(H1)=(4  2)/(52  2)=1/(13*17)
p(H2)=1-p(H1)=1-1/(13*17)
A-he is saying he has 2 aces
p(A)=0.6*p(H1)+0.4*p(H2)
Your probability is p(H1|A)=p(H1 & A)/p(A)=$\frac{0.6 p(H1)}{p(A)}=1/147$
I'm not really sure about the final result...
A: In your formula you should have the $E_1$ is the probability that $P$ is telling the truth, and $E_2$ is the probability that $P$ draws two aces. Then try again..
And think of the possibility that $P$ lies when he has two aces.
