Question

Using bayes rule
It's been a while since I've used bayes rule or even used probabilities.
Could someone tell me if I'm doing it right?

F := your house is on fire
B := neighbour tells you your house is on fire

P(F)=0.01
P(Neighbour is lying) = 0,1

Find
P(F | B)
P(~F | B)

My method:

P(B|F) = 0,9 AND P(B|~F) = 0,1

non-normalized P(F|B) = P(B|F) * P(F) = 0,009
P(F|B) = 1/a * non-normalized P(F|B)
a = 0,009 + 0,001 = 0,01
P(F|B) = 0,001 / 0,01 = 0,00009

in the same way

P(~F|B) = 1/a * non-normalized P(~F|B)
= 1/0,01 * 0,1 * 0,01 = 0,1

Answer 1

Can you post a picture of the original question?

Answer 2

It seems i've also made a typo.. P(F) = 0.001 instead of 0,01.

My calculations would then give P(F|B) = 0,09 and P(~F|B) = 0,01

Heres a picture

Answer 3

Those are wrong too.. as I would get a new a..

My calculations would be F|B = 0.9 and ~F|B = 0.1.

Which seem to be correct since they have to add up to 1

Answer 4

In order for me to properly give you an answer, I need to see the question. What you interpret my be wrong. If you have a link post it. That or a picture.

Answer 5

I attached the picture two posts above yours. Can you not see it?

Answer 6

I want to see the original question.

Answer 7

That is the question as I have recieved it...It's a screenshot from a video

Answer 8

In order for me to compute the posterior probability I must use bayes theorem. Bayes theorem relies on the conditional probabilities P(B|F) P(B|~F) and and the complements. Unless it is stated in the video, there is not enough information available to finish the problem.

Answer 9

Yes, I tried to reason the P(B|F) by saying:

Neighbour tells truth 9/10 times

P(B|F)
Given your house is burning,
The chances of your neighbour saying it's burning would be 0.9
Cause in this case there is a 0.1 chance of him lying.. (saying it's not burning)

P(B|~F)
Given your house is not burning
Telling the truth is again 0.9
The chances of your neighbour lying (saying it is burning) is 0.1

Is this faulty reasoning?