- #1

Kolmin

- 66

- 0

## Homework Statement

There are two boxes:

- in the first one there is 1 red ball,

- in the second one there are 2 balls, 1 red & 1 white.

**1st draw:**We take a ball from the second box without taking a look at the color and we put it in the first box.

**2nd draw:**We take a ball from the first box.

The ball is red.

What is the probability of having another red ball in the first box?

## Homework Equations

[itex]P(R)[/itex] is the probability of taking a red ball on first draw

[itex]P(r)[/itex] is the probability of having a red ball in the first box after the second draw

[itex]P(R|r)[/itex] is the probability of having taken a Red ball on the first draw, given that we have found a red ball in the first box after the second draw.

[itex]P(r|R)[/itex] is the probability of having a red ball in the first box, given that we took a Red ball on the first draw.

## The Attempt at a Solution

I can hardly find something more than Probability to which apply the sentence "it's not my cup of tea":

Actually I am not even sure about the way in which I framed the problem in

**[2.]**.

However, if that framework stands, then I would say that:

[itex]P(R)=\frac{1}{2}[/itex]

[itex]P(R|r)=1[/itex]

[itex]P(r|R)=1[/itex]

And here there is the mistake (I am almost sure about it, even if I dont' see why):

[tex]P(r)= \frac{P(R)P(r|R)}{P(R|r)}=\frac{1}{2}[/tex]

Right or wrong?