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20 votes

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?

- $\left(\dfrac{9}{16}\right)$
- $\left(\dfrac{6 }{16}\right)$
- $\left(\dfrac{7}{16}\right)$
- $\left(\dfrac{10}{16}\right)$
- None of the above

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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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4

14 votes

Best answer

Option D should be the correct answer.

Consider the following events,

$W$ : India wins,

$W\neg $ : India does not wins (India Lost/ Match Draw/ Match Tie/ Match Suspended etc.)

$X$ : Akbar tells Anthony, "Amar told me that India won"

$X\neg $ : Akbar tells Anthony, "Amar told me that India did not won"

Given $X$, we have to find $W$, that is we have to calculate $P\left(\frac{W}{X}\right)$.

$P\left(\frac{W}{X}\right)$ can be calculated using Bayes's theorem as follow:

$P\left(\dfrac{\text{India Wins}}{\text{Akbar tells Anthony “Amar told me that India won"}}\right)$

$= \dfrac{P\left(\dfrac{\text{Akbar tells Anthony “Amar told me that India won"}}{\text{India Wins}}\right)}{P\left(\frac{\text{Akbar tells Anthony “Amar told me that India won"}}{{\text{India Won}}}\right)\cup P\left(\frac{\text{Akbar tells Anthony “Amar told me that India won"}}{\text{India didn't won}}\right)} $

rewriting same equation using the events defined:

$P\left(\frac{W}{X}\right) = \frac{P\left(\frac{X}{W}\right)}{P\left(\frac{X}{W}\right) + P\left(\frac{X}{W\neg}\right)}\\$

**Calculation of **$P\left(\frac{X}{W}\right) and \ P\left(\frac{X}{W\neg}\right)\\$ :

$P\left(\frac{X}{W}\right) = P\left(\frac{Case \ 1}{W}\right) \cup \ P\left(\frac{Case \ 4}{W}\right)\\$

$P\left(\frac{Case \ 1}{W}\right) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}\\$

$P\left(\frac{Case \ 4}{W}\right) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}\\$

$So \ P\left(\frac{X}{W}\right) = \frac{9}{16} + \frac{1}{16} = \frac{10}{16}\\$

$P\left(\frac{X}{W\neg}\right) = P\left(\frac{Case \ 6}{W\neg}\right) \cup \ P\left(\frac{Case \ 7}{W\neg}\right)\\$

$P\left(\frac{Case \ 6}{W\neg}\right) = \frac{3}{4} \times \frac{1}{4} = \frac{3}{16}\\$

$P\left(\frac{Case \ 7}{W\neg}\right) = \frac{1}{4} \times \frac{3}{4} = \frac{3}{16}\\$

$So \ P\left(\frac{X}{W\neg}\right) = \frac{3}{16} + \frac{3}{16} = \frac{6}{16}\\$

$Hence, \ P\left(\frac{W}{X}\right) = \frac{\frac{10}{16}}{\frac{10}{16} + \frac{6}{16}} = \frac{10}{16}.\\$

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52 votes

2

18 votes

there are two cases in which india win:

1) Akbar tells the truth and Amar tells the truth : 3/4*3/4 = 9/16

2) Akbar tells a lie that India loose and Amar tells lie to anthony that "Akbar told me india win" : 1/4*1/4 = 1/16

So total probability of winning India would be 9/16 + 1/16 = 10/16

Source : http://www.careercup.com/question?id=13438685

I felt this is easy to understand.

16 votes

watch this as well for the better understanding of the solution :

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