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## ? regarding probability

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• I am stuck on a probability question .... Can anyone help ? =) The Question is ... you have 2 boxes ... one contains 1 white marble and 2 black marbles the
Message 1 of 4 , Dec 1, 2003
I am stuck on a probability question ....

Can anyone help ? =)

The Question is ...

you have 2 boxes ...
one contains 1 white marble and 2 black marbles
the other box contains 3 black marbles and 4 white marbles
without looking at the color of the marble - 1 marble is randomly
drawn from one of the boxes (also randomly selected) and placed into
the remaining box. If you now draw a marble out of the box to which
the marble was added ... what is the probability that it is white ?

My best guess after considerable work is 157/336

I hope someone can confirm this - or educate me about where I went
wrong ! thanks
Mark
• ... Mike, I agree with 157/336. The challenge is to come up with a clear and convincing proof... I am gonna think about it. Noel.
Message 2 of 4 , Dec 1, 2003
> My best guess after considerable work is 157/336

Mike,

I agree with 157/336. The challenge is to come up with a clear and
convincing proof... I am gonna think about it.

Noel.
• Hi, using a three-level probability tree, I find 349/672... Pedro ... Van: travelguy027 [mailto:travelguy027@aol.com] Verzonden: maandag 1 december 2003 18:15
Message 3 of 4 , Dec 1, 2003
Hi,

using a three-level probability tree, I find 349/672...

Pedro

-----Oorspronkelijk bericht-----
Van: travelguy027 [mailto:travelguy027@...]
Verzonden: maandag 1 december 2003 18:15
Aan: probability@yahoogroups.com
Onderwerp: [probability] ? regarding probability

I am stuck on a probability question ....

Can anyone help ? =)

The Question is ...

you have 2 boxes ...
one contains 1 white marble and 2 black marbles
the other box contains 3 black marbles and 4 white marbles
without looking at the color of the marble - 1 marble is randomly
drawn from one of the boxes (also randomly selected) and placed into
the remaining box. If you now draw a marble out of the box to which
the marble was added ... what is the probability that it is white ?

My best guess after considerable work is 157/336

I hope someone can confirm this - or educate me about where I went
wrong ! thanks
Mark

[Non-text portions of this message have been removed]
• ... into ... which ... Mike, I am not too happy with what follows, as I don t find it very elegant, but I cannot think of anything better for the time being.
Message 4 of 4 , Dec 2, 2003
> you have 2 boxes ...
> one contains 1 white marble and 2 black marbles
> the other box contains 3 black marbles and 4 white marbles
> without looking at the color of the marble - 1 marble is randomly
> drawn from one of the boxes (also randomly selected) and placed
into
> the remaining box. If you now draw a marble out of the box to
which
> the marble was added ... what is the probability that it is white ?
>
> My best guess after considerable work is 157/336

Mike,

I am not too happy with what follows, as I don't find it very
elegant, but I cannot think of anything better for the time being.

Let X0 be a radom variable with values in {0,1}, modelling the draw
of one marble in box 0, i.e. with distribution:

P(X0=0)= 1/3 and P(X0=1)=2/3

where it is understood that '0' means 'white' and '1' means 'black'.
Likewise, let X1 be a random variable with values in {0,1}, modelling
the draw of one marble in box 1, i.e. with distribution:

P(X1=0)=4/7 and P(X1=1)=3/7

Let e be a random variable with values in {0,1}, modelling the radom
choice between box 0 and box 1, i.e. with distribution:

P(e=0)=1/2 and P(e=1)=1/2

where it is understood that 'e=0' means that box 0 has been selected,
whereas 'e=1' means that box 1 has been selected.

One marble will be drawn from one box and added to the other. If box
0 is initially selected (e=0), then box 1 will contain 8 marbles. If
box 1 is initially selected then box 0 will contain 4 marbles.

Let e0 be a random variable with values in {0,1} modelling whether a
draw from "box 0 with one additional marble" will result in a draw of
a marble from box 0, or a draw of the additional marble, i.e. with
distribution:

P(e0=0)=3/4 and P(e0=1)=1/4

where it is understood that if e0=0 then the draw from "box 0 with
one additional marble" brings a marble from box 0, whereas e0=1 means
that the additonal marble has been drawn.

Let e1 be a random variable with values in {0,1} modelling whether a
draw from "box 1 with one additional marble" will result in a draw of
a marble from box 1, or a draw of the additional marble, i.e. with
distribution:

P(e1=0)=7/8 and P(e0=1)=1/8

where it is understood that if e1=0 then the draw from "box 1 with
one additional marble" brings a marble from box 1, whereas e1=1 means
that the additonal marble has been drawn.

Let X be the radom variable modeling our experiment (i.e. the draw
from the second box). Then X is defined as:

If e=0 and e1=0 then X=X1
If e=0 and e1=1 then X=X0

If e=1 and e0=0 then X=X0
If e=1 and e0=1 then X=X1

Choices in box 0 and 1 (i.e. X0 and X1) , choice between box 0 and
box 1 (i.e. e), choices wether additional marble or initial marbles
(i.e. e0 and e1), should all be independent in our model, i.e.

X0,X1,e,e0,e1 are independent.

Our question is to compute P(X=0). This goes as follows:

P(X=0)

=P(e=0,e1=0,X=0)
+P(e=0,e1=1,X=0)
+P(e=1,e0=0,X=0)
+P(e=1,e0=1,X=0)

=P(e=0,e1=0,X1=0)
+P(e=0,e1=1,X0=0)
+P(e=1,e0=0,X0=0)
+P(e=1,e0=1,X1=0)

=P(e=0)P(e1=0)P(X1=0)
+P(e=0)P(e1=1)P(X0=0)
+P(e=1)P(e0=0)P(X0=0)
+P(e=1)P(e0=1)P(X1=0)

=(1/2)(7/8)(4/7)
+(1/2)(1/8)(1/3)
+(1/2)(3/4)(1/3)
+(1/2)(1/4)(4/7)

=(1/2)(1/8)[4+(1/3)+2+(8/7)]
=(1/2)(1/8)(1/3)(1/7)[84+7+42+24]
=157/336

Noel.
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