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A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually six is 1. 1/8 2. 2/8 3. 3/8 4. 1/2

I think that the answer is indeed 3/8 (of course in case every time he lies he says that he got a 6, if it's not the case then the answer is simply 3/4), but the above solution is not right.

P=Favorable outcomes/Total # of possible outcomes.

Favorable outcome is that it's actually six and he's telling the truth \(=\frac{3}{4}*\frac{1}{6}=\frac{1}{8}\)

Total # of possible outcomes is: either it's six and he's telling the truth OR it's not six and he's telling the lie \(=\frac{3}{4}*\frac{1}{6}+\frac{1}{4}*\frac{5}{6}=\frac{1}{3}\)

Bunuel...I think the answer is still incorrect. I had a lengthy fight with a friend on this (admittedly he won ) and though the mathematics on this question is right logically the answer still feels should be 3/4 (though it isn't an option here). Applying probability.... p= (Number of time when it is actually 6)/(Number of times when it is and isn't a 6). And according to his truth factor of 3/4 he should have said 6 exactly 3 times out of 4. Still can't solve it mathematically using your equation though

I think that the answer is indeed 3/8 (of course in case every time he lies he says that he got a 6, if it's not the case then the answer is simply 3/4), but the above solution is not right.

P=Favorable outcomes/Total # of possible outcomes.

Favorable outcome is that it's actually six and he's telling the truth \(=\frac{3}{4}*\frac{1}{6}=\frac{1}{8}\)

Total # of possible outcomes is: either it's six and he's telling the truth OR it's not six and he's telling the lie \(=\frac{3}{4}*\frac{1}{6}+\frac{1}{4}*\frac{5}{6}=\frac{1}{3}\)

\(P=\frac{\frac{1}{8}}{\frac{1}{3}}=\frac{3}{8}\)

No, the answer should be 3/4. When he rolls a '1', and lies, he won't always say he rolled a '6'. He'll say he rolled a '6' one fifth of the time; the other four fifths of the time he'll claim he rolled a 2, 3, 4 or 5. So in your calculation of favorable outcomes, you need to multiply the (1/4)(5/6) by (1/5), and the answer becomes (3/24)/(4/24) = 3/4.

That shouldn't be surprising; no matter what he rolls, he's as likely to claim he rolled a '6' as he is to claim he rolled any other number. He's telling the truth 3/4 of the time, so the answer is 3/4.

All of that said, the question is very badly worded; I'm assuming above he tells a 'plausible lie' when he does lie - that is, he chooses one of the other numbers on the die. Of course, if he rolls the die and he's lying, there's no reason why he couldn't say 'I rolled a 41', or 'I rolled a banana', or 'What are you talking about, I didn't roll the die at all!'

The solution above is right - 3/8 is the answer - if every time he lies, he claims he rolled a 6. I don't see any reason to assume that, however.
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I think that the answer is indeed 3/8 (of course in case every time he lies he says that he got a 6, if it's not the case then the answer is simply 3/4), but the above solution is not right.

P=Favorable outcomes/Total # of possible outcomes.

Favorable outcome is that it's actually six and he's telling the truth \(=\frac{3}{4}*\frac{1}{6}=\frac{1}{8}\)

Total # of possible outcomes is: either it's six and he's telling the truth OR it's not six and he's telling the lie \(=\frac{3}{4}*\frac{1}{6}+\frac{1}{4}*\frac{5}{6}=\frac{1}{3}\)

\(P=\frac{\frac{1}{8}}{\frac{1}{3}}=\frac{3}{8}\)

No, the answer should be 3/4. When he rolls a '1', and lies, he won't always say he rolled a '6'. He'll say he rolled a '6' one fifth of the time; the other four fifths of the time he'll claim he rolled a 2, 3, 4 or 5. So in your calculation of favorable outcomes, you need to multiply the (1/4)(5/6) by (1/5), and the answer becomes (3/24)/(4/24) = 3/4.

That shouldn't be surprising; no matter what he rolls, he's as likely to claim he rolled a '6' as he is to claim he rolled any other number. He's telling the truth 3/4 of the time, so the answer is 3/4.

All of that said, the question is very badly worded; I'm assuming above he tells a 'plausible lie' when he does lie - that is, he chooses one of the other numbers on the die. Of course, if he rolls the die and he's lying, there's no reason why he couldn't say 'I rolled a 41', or 'I rolled a banana', or 'What are you talking about, I didn't roll the die at all!'

The solution above is right - 3/8 is the answer - if every time he lies, he claims he rolled a 6. I don't see any reason to assume that, however.

So Ian, even in your 'all of that said...' part the underlying assumption that he claims (lies) he rolled a 6 is incorrect as I understand. A plausible lie should have multiplied the total outcomes by 1/5 and an improbable assumption should have by 1/infinity

the answer should be 3/4. and you need to multiply 1/5 the way Ian said. A simple way to check this is - If he tells it is a 6, the probability of this being 6 = probability that this guys tells the truth = 3/4. This makes sense. Right? At least it does to me. So per Bayes theorem P(B/A) = P(B int A)/P(A) = [3/4*1/6]/[(3/4*1/6)+(1/4*5/6*1/5)] =3/4

To check this change the probability of telling truth to 2/3. Then per my theory the probably that it is actually a 6 should be 2/3. check it out using formula [2/3*1/6]/[(2/3*1/6)+(1/3*5/6*1/5)] =2/3

A simple way to check this is - If he tells it is a 6, the probability of this being 6 = probability that this guys tells the truth = 3/4. This makes sense. Right? At least it does to me.

I think in this case, this does actually hold through but you should be very careful in making generalizations like this one above. They often depend on assumptions which may not be obvious.

For instance, if let's say you were to consider that the dice is loaded and that the probability of it showing a 6 is 1/2 and any other number is 1/10. Then you will realize the answer to this problem becomes 15/16 instead of 3/4. So it is clearly hinged on a uniform probability of each outcome.
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I also think that the correct answer is 3/4. The number of faces on the die has no significance to the question.

Compare our 3/4 truth speaker these other 3 speakers. They each have a die in their pockets which they roll. They each say (speaking about their own roll) that they have rolled a 6:

A always speaks the truth. A says it is a 6. The probability that it is a 6 is 1.

B always lies. B says it is a 6. The probability that it is a 6 is 0.

C tells the truth 1/2 the time. C says it's a 6. Half the time it is a 6, and half the time it's not a 6. So the probability that it is a 6 when C reports is .5 (50%).

bunuel...this question is strikingly similar to ones solved by Bayes theorem...is the knowledge of bayes theorem required for gmat? btw +1 for solving the cheeky question!
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confused here by the favorable outcomes/ possible outcomes, usually i think of it as possibilities, positive integers not fractions

why are we using probabilities in place of the outcomes? and with that said, shouldn't the total possible outcomes add up to 1 anyway if it were a probability?

A simple way to check this is - If he tells it is a 6, the probability of this being 6 = probability that this guys tells the truth = 3/4. This makes sense. Right? At least it does to me.

I think in this case, this does actually hold through but you should be very careful in making generalizations like this one above. They often depend on assumptions which may not be obvious.

For instance, if let's say you were to consider that the dice is loaded and that the probability of it showing a 6 is 1/2 and any other number is 1/10. Then you will realize the answer to this problem becomes 15/16 instead of 3/4. So it is clearly hinged on a uniform probability of each outcome.

can you show your work for the loaded dice variant

In my opinion the answer 3/8 is wrong. Consider this way. If the man speaks the truth always then the probability that it is a six when he says it is a six is 1. So when he speaks truth 3/4 times the probability that it is a six when he says it is a six is, 3/4.
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A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually six is 1. 1/8 2. 2/8 3. 3/8 4. 1/2

Probability of speaking truth is 3/4. And, the probability that a thrown die reports 6 is 1/6. So, the probability that it is actually six is (3/4)*(1/6) = 1/8.

As discussed above, the question is not very clear. We are not sure how he lies. If we have to make an assumption, we might assume that every time he lies, he calls it a 6 or we may assume that he lies consistently i.e. he will use every number equally while lying. So say if he throws a 1 twenty times, he will lie 5 times. When he lies, he will call it a 2 once, a 3 once, a 4 once, a 5 once and a 6 once.

Assuming he lies consistently: So consider a case when he throws the die 24 times. He gets a 1 four times, 2 four times, 3 four times and so on since all are equally likely. Of the 4 times he throws a 1, he lies once. Say he calls it a 2. Of the 4 times he throws a 2, he lies once. Say he calls it a 3. Of the 4 times he throws a 3, he lies once. Say he calls it a 4. Of the 4 times he throws a 4, he lies once. Say he calls it a 5. Of the 4 times he throws a 5, he lies once. Say he calls it a 6. Of the 4 times he throws a 6, he lies once. Say he calls it a 1. He says the truth in other 3 instances and calls it a 6.

We have to find the probability that when he says it is 6, it actually is a 6. So 4 times he says it is a 6. Out of these 4 times, it is actually 6 in 3 instances. So the probability that it actually is 6 (when he says it is 6) is 3/4.

Assuming he calls it a 6 every time he lies

Of the 4 times he throws a 1, he lies once. Say he calls it a 6. Of the 4 times he throws a 2, he lies once. Say he calls it a 6. Of the 4 times he throws a 3, he lies once. Say he calls it a 6. Of the 4 times he throws a 4, he lies once. Say he calls it a 6. Of the 4 times he throws a 5, he lies once. Say he calls it a 6. Of the 4 times he throws a 6, he lies once. Say he calls it something else. He says the truth in other 3 instances and calls it a 6. So 8 times he says that it is a 6. Out of these 8 times, it is actually 6 in 3 instances. So the probability that it actually is 6 (when he says it is 6) is 3/8. You can calculate this as given by Bunuel above.

Where did you go wrong: You are finding the probability that he throws a 6 AND then calls it a 6 (P(Throws a 6) - 1/6; P(Calls it 6) - 3/4; Does both = 1/6 * 3/4 = 1/8). But that it not the question. You have to find the probability that it is actually 6 (after he has said that it is a 6). So we need to find P(No of time it is a 6 when he calls it a 6)/(Total number of time he calls it a 6) It is already given that he calls it a 6. Now we have to figure the probability that it actually is a 6.
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