TIFR CSE 2012 | Part A | Question: 1

Question

Amar and Akbar both tell the truth with probability $\dfrac{3 } {4}$ and lie with probability $\dfrac{1}{4}$. Amar watches a test match and talks to Akbar about the outcome. Akbar, in turn, tells Anthony, "Amar told me that India won". What probability should Anthony assign to India's win?

• A. $\left(\dfrac{9}{16}\right)$
• B. $\left(\dfrac{6 }{16}\right)$
• C. $\left(\dfrac{7}{16}\right)$
• D. $\left(\dfrac{10}{16}\right)$
• E. None of the above

Comments

Looks like an ambiguous question.  Antony can assign probability based on both actuality and Akbars' information. Later one is answered in this thread. Former one seems to make it 1/2.

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I just want to know that in this ques and in this https://gateoverflow.in/18499/tifr2010-a-19-tifr2014-a-6 ,we would not consider the case of tie unless specified??

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Yes. Mostly in any question about win/loss tie is ignored. If game is cricket whichever team hits more boundaries will be the winner 😉

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I don’t know the probability to the Event asked in the Question. But is know that the Probabillity that you will go mad solving these question model is 1.😶

Answer 1

All the answers given here are not entirely true. The question given is not solvable because "the probability of india winning " is not given in the question. 

Let e1 : India wins ; e2 : Akbar says that "Amar told me that India won"

now in the Numerator(e1 "AND " e2) there will be two cases : 

case (i) :- India actually winning and both amar and akbar telling the truth (i.e. IndiaWin AND AmarTruth AND AkbarTruth)

case (ii) :- India actually winning and both amar and akbar telling lies.       (i.e. IndiaWin AND AmarFalse AND AkbarFalse)

Denominator will be the reduced sample space of event e2 i.e. ("Akbar telling anthony that ''Amar told him that india won'").

Here we need to consider the cases where india winning and also india losing which will give us 4 cases :

case (i) :- India actually winning and both amar and akbar telling the truth (i.e. IndiaWin AND AmarTruth AND AkbarTruth)

case (ii) :- India actually winning and both amar and akbar telling lies.       (i.e. IndiaWin AND AmarFalse AND AkbarFalse)

case (iii) :- India actually losing and amar telling truth and akbar telling lie (i.e. IndiaLose AND AmarTruth AND AkbarFalse)

case (iv) :- India actually losing and amar telling lie and akbar telling truth (i.e. IndiaLose AND AmarFalse AND AkbarTruth)

Now writing all of them in equation, we get one which I have written in the image

But to find the solution, we need P(India winning) is not given, so no sufficient data to solve it.  

Note :- Everyone are considering only the cases where india wins. But we need to consider the cases where the india loses also, since the sample space of event e2 contains both india winning and also india losing..

I am attaching also the tree diagram which may make it easier to understand.

Also we can't just assume that "arjun and amar telling lies are independent of the outcome of the match or not". So, in either way, this question is unsolvable. There's just no sufficient information for anthony to come to a conclusion.

Answer 2

https://gateoverflow.in/?qa=blob&qa_blobid=17508858049422517468

Answer 3

Upon watching the test match, either Amar tells Akbar that India won the match or Amar tells that India lost the match. Next, Akbar tells Anthony that Amar told him India won the match. Anthony concludes that India can win in two mutually exclusive and exhaustive events: Either both Amar and Akbar tells the truth or both of them lies. Therefore, 
P(W) = P(TAm) * P(TAk) + P(FAm) * P(FAk) = (3/4 * 3/4) + (1/4 * 1/4) = 10/16, where P(W) is the probability Anthony assigns to India's win, P(TAm) the probability of Amar telling the truth, and so on.
Note that Amar's credibility, Akbar's credibility, and India's likelihood of winning are independent of each other.