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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

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

"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

"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

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.

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 half-joking, of course)
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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 1-1/16 = 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.

"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. B-cos the solution assumes that his lie is restricted to numbers 1-6. His lies can actually be limitless...

"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

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...

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.

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 1-x 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 * (1-x) * 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.