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A man is known to speak truth 3 out of 4 times. He throws [#permalink]
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13 Aug 2008, 10:47
A man is known to speak truth 3 out of 4 times. He throws die and reports that it is a 6. The probability that it is actually a 6 is A) 3/4 B) 5/8 C) 2/5 D) 3/5 E) 4/5 == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.
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Re: probabilty [#permalink]
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13 Aug 2008, 19:18
nice trap question. agree with 3/4



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Re: probabilty [#permalink]
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13 Aug 2008, 19:39
Prob that its a 6 and he speaks truth : 3/4 * 1/6 = 3/24
Now what is the prob of reporting a 6 on a dice when it is a lie = 1/4 (speaking a lie) * 5/6 (prob of no. other than 6) * 1/5 (lieing the number as 6 and not telling any othe number except the actual no.) = 5/120 = 1/24
Thus the prob of reporting no. as 6 (total lie + truth) = 3/24 + 1/24 = 4/24 = 1/6
Thus prob of actually having 6 when reported so is = 3/24/1/6 = 3/4
Answer A.



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Re: probabilty [#permalink]
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13 Aug 2008, 20:11
actually, i just thought of it like this: if he is telling the truth 3 out of 4 times, that means hes telling the truth 75% of the time. That means on any give statement, he is telling the truth 75% of the time .... i.e. 0.75, which is still just 3/4



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Re: probabilty [#permalink]
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13 Aug 2008, 21:43
good question what if i change the question to "A man is known to speak truth 3 out of 4 times. He throws a die. What is the probability that he'll report a 6."



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Re: probabilty [#permalink]
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13 Aug 2008, 21:50
durgesh79 wrote: :) good question
what if i change the question to
"A man is known to speak truth 3 out of 4 times. He throws a die. What is the probability that he'll report a 6." Prob that its a 6 and he speaks truth : 3/4 * 1/6 = 3/24 Now what is the prob of reporting a 6 on a dice when it is a lie = 1/4 (speaking a lie) * 5/6 (prob of no. other than 6) * 1/5 (lieing the number as 6 and not telling any othe number except the actual no.) = 5/120 = 1/24 Thus the prob of reporting no. as 6 (total lie + truth) = 3/24 + 1/24 = 4/24 = 1/6



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Re: probabilty [#permalink]
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13 Aug 2008, 21:55
abhijit_sen wrote: durgesh79 wrote: :) good question
what if i change the question to
"A man is known to speak truth 3 out of 4 times. He throws a die. What is the probability that he'll report a 6." Prob that its a 6 and he speaks truth : 3/4 * 1/6 = 3/24 Now what is the prob of reporting a 6 on a dice when it is a lie = 1/4 (speaking a lie) * 5/6 (prob of no. other than 6) * 1/5 (lieing the number as 6 and not telling any othe number except the actual no.) = 5/120 = 1/24 Thus the prob of reporting no. as 6 (total lie + truth) = 3/24 + 1/24 = 4/24 = 1/6 i knew that you'll catch the trap I missed it (highlighted) the first time and ended up with 1/3 .. then i thought the sum of all porbable cases ( reporting 1,2,3,4,5 or 6 ) should be 1 .... whatever is the % chances of speaking truth ... so the probability should be 1/6... and you proved it mathematically ... thanks



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Re: probabilty [#permalink]
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14 Aug 2008, 12:01
Could someone please clarify what the trap was in this question and detail any other such probability traps that may be out there. It would be very helpful.
Thank you, Vin



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Re: probabilty [#permalink]
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14 Aug 2008, 12:10
There's an assumption here that when this guy lies, he tells a plausible lie. When he lies, why wouldn't he say 'I rolled a 41', or 'I didn't roll the die at all'? (I'm halfjoking, of course)
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Re: probabilty [#permalink]
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14 Aug 2008, 12:45
I didn't quite understand the solution. Should the total probability not be 1? What I mean is, that the Probability of Truth, let's say, Pt and the Probability of Lie, say, Pl, should add up to 1. In your solution Pt is 3/24 and Pl is 1/24. They add up to only 4/24 or 1/6. Isn't that right?
I am not sure but I feel the answer should be different. By my logic I am getting the answer as 4/5. I could be wrong, but if I am, please correct me. From the question, it seems there are 3 truths and 1 false in every 4 times. Probability of False is 1*1*1*1/2^4 i.e., 1/16 (This is the same as tossing a coin, since there are only 2 options). Therefore, the probability of truth is 11/16 = 15/16 for a single event. Given Probability for a single event is 3/4. Therefore, the total Probability is 3/4 / 15/16 = 4/5.



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Re: probabilty [#permalink]
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14 Aug 2008, 17:52
durgesh79 wrote: abhijit_sen wrote: durgesh79 wrote: :) good question
what if i change the question to
"A man is known to speak truth 3 out of 4 times. He throws a die. What is the probability that he'll report a 6." Prob that its a 6 and he speaks truth : 3/4 * 1/6 = 3/24 Now what is the prob of reporting a 6 on a dice when it is a lie = 1/4 (speaking a lie) * 5/6 (prob of no. other than 6) * 1/5 (lieing the number as 6 and not telling any othe number except the actual no.) = 5/120 = 1/24 Thus the prob of reporting no. as 6 (total lie + truth) = 3/24 + 1/24 = 4/24 = 1/6 i knew that you'll catch the trap I missed it (highlighted) the first time and ended up with 1/3 .. then i thought the sum of all porbable cases ( reporting 1,2,3,4,5 or 6 ) should be 1 .... whatever is the % chances of speaking truth ... so the probability should be 1/6... and you proved it mathematically ... thanks This question, in its current form is vague. I would be tempted to say that the 1/6*3/4 only. Bcos the solution assumes that his lie is restricted to numbers 16. His lies can actually be limitless...



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Re: probabilty [#permalink]
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14 Aug 2008, 19:14
durgesh79 wrote: abhijit_sen wrote: durgesh79 wrote: :) good question
what if i change the question to
"A man is known to speak truth 3 out of 4 times. He throws a die. What is the probability that he'll report a 6." Prob that its a 6 and he speaks truth : 3/4 * 1/6 = 3/24 Now what is the prob of reporting a 6 on a dice when it is a lie = 1/4 (speaking a lie) * 5/6 (prob of no. other than 6) * 1/5 (lieing the number as 6 and not telling any othe number except the actual no.) = 5/120 = 1/24 Thus the prob of reporting no. as 6 (total lie + truth) = 3/24 + 1/24 = 4/24 = 1/6 i knew that you'll catch the trap I missed it (highlighted) the first time and ended up with 1/3 .. then i thought the sum of all porbable cases ( reporting 1,2,3,4,5 or 6 ) should be 1 .... whatever is the % chances of speaking truth ... so the probability should be 1/6... and you proved it mathematically ... thanks the explanation given by abhijit_sen is the most apt mathematically. I was just wondering if the probability of man saying a truth or lie really matters for Durgesh79's question. For predicting any one number on a die the probability will be 1/6 Just wanted to discuss, I may be "perfectly" wrong



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Re: probabilty [#permalink]
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14 Aug 2008, 20:56
OK, I'll correct the "my version" of question.
A man is known to speak truth 3 out of 4 times. He throws a die. What is the probability that he'll report a 6 provided that he can not only report a possible outcome.
alpha : you are right, the probablity will be 1/6 only, it doesnt depend on the probability of speaking truth...



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Re: probabilty [#permalink]
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15 Aug 2008, 13:52
durgesh79 wrote: OK, I'll correct the "my version" of question.
A man is known to speak truth 3 out of 4 times. He throws a die. What is the probability that he'll report a 6 provided that he can not only report a possible outcome.
alpha : you are right, the probablity will be 1/6 only, it doesnt depend on the probability of speaking truth... Are the outcome of the die and the guy's reporting independent here ? Am confused because his lying or speaking truth is independent of die roll, but his actual report doesnt seem to be independent of the outcome of the die roll.



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Re: probabilty [#permalink]
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15 Aug 2008, 20:57
bhushangiri wrote: durgesh79 wrote: OK, I'll correct the "my version" of question.
A man is known to speak truth 3 out of 4 times. He throws a die. What is the probability that he'll report a 6 provided that he can not only report a possible outcome.
alpha : you are right, the probablity will be 1/6 only, it doesnt depend on the probability of speaking truth... Are the outcome of the die and the guy's reporting independent here ? Am confused because his lying or speaking truth is independent of die roll, but his actual report doesnt seem to be independent of the outcome of the die roll. i dont know if i understand your doubt but look at it this way x is the probability of speaking truth, then 1x is the probablity of not speaking truth. probability of reporting a 6 = when actual is six and he is reporing 6 + when actual is some thing else (1,2,3,4,5) and he is not speaking truth and reporting 6. = 1/6 * x + 5/6 * (1x) * 1/5 = 1/6 > doesnt depend on x this'll only work when he always reports one of the possible outcome .... that is clarified in the question. == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.










