I feel every answer which is given here is wrong (in a way). I know even the official response is D. (10/16). However, Correct answer should be e. None of these. Actually this probability CAN NOT BE CALCULATED AT ALL. First let me explain my answer and then I shall explain where everyone (yeah even the official answer key) is making mistake.
First let me introduce some notations:
Notations:
Amar telling truth will be denoted as event A. β΄ Amar telling lie is denoted by Ac.
Akbar telling truth will be denoted as event B. β΄ Akbar telling lie is denoted by Bc.
India actually winning will be denoted as event W. β΄ India actually losing will be denoted as event Wc.
Now, Lets talk about the sample space. There are 8 members inside the sample space.
1. (India Won, Amar told truth, Akbar told truth) denoted as (W, A, B)
2. (India Won, Amar told truth, Akbar told lie) denoted as (W, A, Bc)
3. (India Won, Amar told lie, Akbar told truth) denoted as (W, Ac, B)
4. (India Won, Amar told lie, Akbar told lie) denoted as (W, Ac, Bc)
5. (India lost, Amar told truth, Akbar told truth) denoted as (Wc, A, B)
6. (India lost, Amar told truth, Akbar told lie) denoted as (Wc, A, Bc)
7. (India lost, Amar told lie, Akbar told truth) denoted as (Wc, Ac, B)
8. (India lost, Amar told lie, Akbar told lie) denoted as (Wc, Ac, Bc)
Let me explain the members of the sample space.
Consider the first member (India Won, Amar told truth, Akbar told truth). What I mean to say is that India actually won the match and then Amar told Akbar that India won the match(truth) and finally Akbar told Anthony that India won the match(truth).
Similarly consider seventh member: (India lost, Amar told lie, Akbar told truth). What this means is that India actually lost the match and then Amar told Akbar that India won the match(lie) and finally Akbar told Anthony that India won the match(truth). Note that Akbar's truth is not on the basis of absolute truth. Akbar's truth is on the basis of Anthony's information only. So here Akbar telling truth means whatever he heard from Anthony, he just says exactly the same thing. And Akbar telling lie is that whatever he heard from Anthony, he tells completely opposite things.
A basic concept: If it is said that given an event A has occurred, what is the probability of B occurring, then it is given by the expression: P (B | A)
Now onto the question:
The question is asking to calculate that given that Akbar told Anthony that India Won, what is the probability of India's actually winning ?
i.e. the problem is asking P(W | Akbar told Anthony that India Won) = ?. Note that the problem does not ask us to find P(W).
Now let us see from our sample space, what are the possible ways in which Akbar can tell that India has won:
1. (W, A, B) . i.e. India can actually win, Amar tells truth (that India won), Akbar tells truth(that India won) .
2. (W, Ac, Bc) . i.e. India can actually win, Amar tells lie (that India lost), Akbar tells lie (that India won) .
3. (L, A, Bc) . i.e. India can actually lose, Amar tells truth (that India lost), Akbar tells lie (that India won) .
4. (L, Ac, B) . i.e. India can actually lose, Amar tells lie (that India won), Akbar tells truth (that India won) .
These 4 are the only possible ways in which Akbar can tell that India has won.
Now the favorable cases are the ones where India has actually won. Thus the favorable cases are:
1. (W, A, B)
2. (W, Ac, Bc)
Thus the problem boils down to :
$P(W | \text{ Akbar told Antony that India Won}) = \LARGE\frac{P(W, A, B)+P(W, A^{c}, B^{c})}{P(W, A, B)+P(W, A^{c}, B^{c})+P(W^{c}, A^{c}, B)+P(W^{c},A, B^{c}))}$
Now we need to calculate the individual probabilities.
P(W, A, B) = P(W) * P(A | W) * P(B | W β© A). Now probability of Amar/Akbar telling truth/lie is independent of anything else. Thus P(A|W) = P(A) and P(B | W β© A) = P(B) β΄ P(W, A, B) = P(W) * P(A) * P(B)
Likewise P(W, Ac, Bc)=P(W) * P(Ac) * P(Bc) . Similarly for the other probabilities.
We know P(A)=P(B)=3/4 and P(Ac)=P(Bc)=1/4. However we dont know P(W) or P(Wc).
So let us assume P(W) = p β΄ P(Wc) = 1-p and hope that while evaluating the above expression all p terms get cancelled.
$\LARGE \frac{P(W, A, B)+P(W, A^{c}, B^{c})}{P(W, A, B)+P(W, A^{c}, B^{c})+P(L, A^{c}, B)+P(L,A, B^{c}))} \\ \LARGE = \frac{p\frac{3}{4}\frac{3}{4} + p\frac{1}{4}\frac{1}{4}}{p\frac{3}{4}\frac{3}{4} + p\frac{1}{4}\frac{1}{4}+(1-p)\frac{1}{4}\frac{3}{4}+(1-p)\frac{3}{4}\frac{1}{4}}\\ \LARGE =\frac{\frac{10}{16}p}{\frac{10}{16}p-\frac{6}{16}p+\frac{6}{16}} =\frac{10p}{4p+6}$
We see that it is dependent on the actual probability of India Winning(p) which is not given anywhere in the question. Thus the answer cannot be found at all.
Interestingly enough, if we assume p=0.5 (Chances of India winning and losing are same), we get that
$\LARGE\frac{10*0.5}{4*0.5+6}=\frac{5}{8}=\frac{10}{16} !!!$ But there is no reason why p should be 0.5. Thus Option (d) 10/16 is NOT the answer.
Now let me explain where other answers have made mistakes:
1. by Anurag Pandey:
He mistakenly assumed the following equation:
$\LARGE P\left(\frac{W}{X}\right) = \frac{P\left(\frac{X}{W}\right)}{P\left(\frac{X}{W}\right) + P\left(\frac{X}{W\neg}\right)}$
The correct Bayes Theorem equation should have been:
$\LARGE P\left(\frac{W}{X}\right) = \frac{P\left(\frac{X}{W}\right)*P(W)}{P\left(\frac{X}{W}\right)*P(W) + P\left(\frac{X}{W\neg}\right)*P(W\neg)}$
