implementation of Bayes law(probabilities)

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dionys 101 Oct 17, 2004 at 12:52

Hi guys
i know that i post a problem without my own answer but this problem is
very difficult for me and i’m very comfused.
The exercise says that we must implement the Bayes’ Law to solve the exercise.My problem is that i cant distinguish and write down the events ):

Problem:In my kitchen i put my 10 forks in the left drawer and my 10 knifes in the right drawer.**
before my room mate change something:
k=knife,f=forks
left drawer : (10f,0k)
right drawer: (0f,10k))
**
When my room-mate came he took 2 forks from the left drawer and he put them in the right drawer.
**
after my room mate’s alteration:
left drawer : (8f,0k)
right drawer: (2f,10k)
**

After this action he chooses one piece(knife or fork) from the right drawer and he puts the piece in the left drawer.
**
left drawer : (9f,0k) or (8f,1k)
right drawer: (1f,10k) or (2f,9k)
**
After all these changes i come and choose one piece from 1 of the 2 drawers.
IF Is given that i hold a knife,what is the probability that i opened the left drawer?

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Baaecfb3a09127d5bc5045f89025e49e
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UnknownStranger 101 Oct 17, 2004 at 18:05

Ok, my guess is on:

Left drawer: 0.1
Right drawer: 0.9

Explanation follows…

Baaecfb3a09127d5bc5045f89025e49e
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UnknownStranger 101 Oct 17, 2004 at 18:14

Well, I don’t have much experience with causality, but anyway, here’s what I think:

K … got knife

D1 … from Drawer 1
D2 … from Drawer 2

P( D1 K ) = ( P( D1 ) * P( K D1 ) ) / P( K )
P( D2 K ) = ( P( D2 ) * P( K D2 ) ) / P( K )

P( D1 ) = 0.5
P( D2 ) = 0.5
P( K ) = 0.5

So…

P( D1 | K ) = P( K | D1 )
and
P( D2 | K ) = P( K | D2 )

Now…

mK … Mate took knife
mF … Mate took fork

P( K | D1 ) = P( mK ) * (1 / 9) + P( mF ) * (0 / 9) (law of total probability)
P( K | D2 ) = P( mK ) * (9 / 11) + P( mF ) * (10 / 11) (law of total probability)

P( mK ) = 10 / 12
P( mF ) = 2 / 12

==>

P( D1 K ) = P( K D1 ) = 0.925…
P( D2 K ) = P( K D2 ) = 0.833…

==>

P( D1 K ) = 0.1
P( D2 K ) = 0.9