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This question was asked in IITB (RA) admissions 2016.

I have two blue dice, with which I play a game. If I throw a double six (i.e. if I get two six on both the dices) then I win the game. I  separately throw a red dice. If I get a one, then I tell truth about whether I win/loose in the previous game, otherwise I lie. I just rolled the three die. I turn around to you and said, "I won!". What is the probability that I actually won the game?
asked in Probability by Active (2.3k points)   | 121 views

1 Answer

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Best answer

On blue dice throw chances of winning 1/36

chances of lost 35/36

Next chances of getting 1 in red dice is 1/6 i.e. he is telling truth

other 5/6 times he lies

Now there are 4 possibilities

  • he wins (probability 1/36) and telling truth that he wins (probability 1/6).........i
  • he wins (probability 1/36) and telling false that he lost (probability 5/6)
  • he lost (probability 35/36) and telling truth that he lost  (probability 5/6)
  • he lost (probability 35/36) and telling false that he wins (probability 1/6)...........ii

Now we calculated on he tells he wins i.e. i and ii cases


Now he wins and red dice throws , he tells truth that he wins  1/6 ⨉1/36

"     he lost and red dice thrown , he tells lie that he wins 5/6 ⨉ 35/36

Probability of actually winning chance is

= (1/(36⨉6)) / ((1/(6⨉36))+((5⨉35) / (6⨉36))) =1/176

answered by Veteran (49.7k points)  
selected by
isn't question asking probality of actual win not false win ? hence 1/216 is answer?
No , to get actual wining Bayes theorem should be applied
@Srestha I want to know, how you are partitioning the sample space, to apply Bayes Theorem?
@utk now clear?

$P(A)=\frac{1}{6}$

$P(\frac{A}{E1})=\frac{P(A\cap E1)}{P(E1)}=\frac{(\frac{1}{6})(\frac{1}{36})}{(\frac{1}{36})}=\frac{1}{6}$

Similarly, $P(\frac{A}{E2})=\frac{1}{6}$

Then, we have, $P(\frac{E1}{A})=\frac{P(\frac{A}{E1})P(E1)}{P(\frac{A}{E1})P(E1)+P(\frac{A}{E2})P(E2)}=\frac{1}{36}$

 

I am getting this solution. What's wrong? @Srestha

I think u give more priority to blue dice rather than wining probability
Probability for actual win is asked not false win so how is answer 1/176
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