View Full Version : Probability Question
Jin314159
Sep9-04, 10:29 PM
It is known that the inhabitants of an island tell the truth 1/3 of the time and lie 2/3 of the time. On an occasion, one person made a statement and the person after him said the statement was true. What is the probability that the statement is actually true?
musky_ox
Sep9-04, 10:37 PM
I would think 1/9 but i dont actually remember how to do probablility. :wink:
Claude Bile
Sep10-04, 12:19 AM
Consider the 4 possibilities -
1st person truthful, 2nd person truthful
1st person truthful, 2nd person lying
1st person lying, second person truthful
1st person lying, second person lying.
Work out how probable each possibility is, then work out which possibilities actually result in the second person claiming the first person is truthful. The rest should be easy.
Claude.
Jin314159
Sep10-04, 02:08 AM
Here's what I did:
P (1st telling truth | second says first is telling truth) =
P (1st telling truth and second says first is telling truth) / [P(1st is telling truth and second says first is telling truth) + P(1st is lying and second says first is selling truth)]
= 1/3
Rogerio
Sep10-04, 04:49 PM
It is known that the inhabitants of an island tell the truth 1/3 of the time and lie 2/3 of the time. On an occasion, one person made a statement and the person after him said the statement was true. What is the probability that the statement is actually true?
Or they are telling true , or they are lying .
The answer is ( 1/3 * 1/3 ) / ( 1/3 * 1/3 + 2/3 * 2/3 ) = 1/5 .
Fredrik
Sep10-04, 06:42 PM
The answer is 1/3. A good way to see this is to do what Claude Bile suggested. Look at the four different "possibilities" one at a time. But you should also ask yourself if they're all really possible. (Big hint: They're not).
musky_ox
Sep10-04, 07:45 PM
I can not say something truthful, and then have my neighbor say something truthful also? theres a 1/3 chance that i would say something truthful, and a 1/3 chance that my neighbor would say something truthful.... no?
Fredrik
Sep11-04, 02:47 AM
Suppose that you say "it's raining", and your neighbor says "what musky ox is saying is true".
Which ones of the following alternatives are possible, and which ones aren't?
1. You're telling the truth and he's telling the truth.
2. You're telling the truth and he's lying.
3. You're lying and he's telling the truth.
4. You're lying and he's lying.
Once you have figured that out, the rest will be very easy.
Rogerio
Sep11-04, 03:30 PM
The answer is 1/3. A good way to see this is to do what Claude Bile suggested. Look at the four different "possibilities" one at a time. But you should also ask yourself if they're all really possible. (Big hint: They're not).
The answer is not 1/3 . Let's follow the Claude's suggestion:
Consider the 4 possibilities -
1st person truthful, 2nd person truthful
1st person truthful, 2nd person lying
1st person lying, second person truthful
1st person lying, second person lying.
The first pair of persons ocurrs with probability: 1/3 * 1/3 = 1/9
The second pair ocurrs with probability: 1/3 * 2/3 = 2/9
The third pair ocurrs with probability: 2/3 * 1/3 = 2/9
The fourth pair ocurrs with probability: 2/3 * 2/3 = 4/9
But, since the 2 persons agreed, we know only the first or the last case can be possible.
So, the probability we have a truthful pair of persons is (1/9) / (1/9 + 4/9) = 1/5.
This is classical Bayes. :smile:
Fredrik
Sep11-04, 04:50 PM
The answer is not 1/3 .
...
But, since the 2 persons agreed, we know only the first or the last case can be possible.
So, the probability we have a truthful pair of persons is (1/9) / (1/9 + 4/9) = 1/5.
This is classical Bayes. :smile:
Yes, you're right. I assigned probability 0 to the impossible "possibilities", but that's not the right way to do this. :redface:
Rogerio
Sep11-04, 05:03 PM
Well, in fact, you and Claude Bile did the job !
I just followed yours suggestion... :smile:
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