- 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 > 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

-----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] > you have 2 boxes ...

into

> 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

> 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 ?

Mike,

>

> My best guess after considerable work is 157/336

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.