# Probability and Bayes' formula

• September 23rd 2008, 09:15 PM
totalnewbie
[SOLVED] Probability and Bayes' formula
I need these solutions in 3 hours.

I've 2 boxes. THe first box has 2 new and 3 old balls. The second box has 3 new and 4 old balls. I took one ball from the first box and put it into the second box. Now I take the ball from the second box.
What's the probability that:
1) The ball taken from the second box is new.
2) The ball taken from the first box was new if I know that the ball taken from the second box was old.

1) 17/40
2) 8/23
• September 23rd 2008, 11:30 PM
jbpellerin
1)
2 cases where the ball is a new one
4 new 4 old -> 1/2
3 cases where it's an old one
3 new 5 old -> 3/8
5 cases total
2/5*1/2 + 3/5*3/8 = 17/40

too tired for the second one sorry
• September 24th 2008, 01:06 AM
mr fantastic
Quote:

Originally Posted by totalnewbie
I need these solutions in 3 hours.

I've 2 boxes. THe first box has 2 new and 3 old balls. The second box has 3 new and 4 old balls. I took one ball from the first box and put it into the second box. Now I take the ball from the second box.

[snip]

2) The ball taken from the first box was new if I know that the ball taken from the second box was old.

1) 17/40
2) 8/23

Using the result from 1), Pr(Old ball taken from box 2) = 1 - (17/40) = 23/40.

Using Bayes Theorem: $\Pr \left( \, \text{New ball taken from box 1} ~ | ~ \text{old ball taken from box 2} \, \right) = \frac{ \left( \frac{2}{5}\right) \, \left( \frac{1}{2}\right)}{\frac{23}{40}} = \, ....$
• September 28th 2008, 02:41 AM
totalnewbie
Quote:

Originally Posted by jbpellerin
1)
2 cases where the ball is a new one
4 new 4 old -> 1/2
3 cases where it's an old one
3 new 5 old -> 3/8
5 cases total
2/5*1/2 + 3/5*3/8 = 17/40

too tired for the second one sorry

I am solving different exercises and I understand how 2/5*1/2 + 3/5*3/8 comes but it's unclear how it equals to 17/40. It actually equals to 23/40. 17/40+23/40=1 Therefore these numbers are related, but how?
It seems to me that one of these is the old ball (23/40) taken and the second one of these is the new ball (17/40) taken? How to make difference?
• September 28th 2008, 04:11 AM
mr fantastic
Quote:

Originally Posted by totalnewbie
I am solving different exercises and I understand how 2/5*1/2 + 3/5*3/8 comes but it's unclear how it equals to 17/40. It actually equals to 23/40. 17/40+23/40=1 Therefore these numbers are related, but how?
It seems to me that one of these is the old ball (23/40) taken and the second one of these is the new ball (17/40) taken? How to make difference?

Pr(New ball taken from box 2) = 2/5*1/2 + 3/5*3/8 = 2/10 + 9/40 = 8/40 + 9/40 = 17/40 so I'm not sure what your problem is here .....
• September 28th 2008, 04:27 AM
totalnewbie
Quote:

Originally Posted by mr fantastic
Pr(New ball taken from box 2) = 2/5*1/2 + 3/5*3/8 = 2/10 + 9/40 = 8/40 + 9/40 = 17/40 so I'm not sure what your problem is here .....