Thread: 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.

Answers
1) 17/40
2) 8/23

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

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.

Answers
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:

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?

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 .....

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 .....

Sorry, my bad!
I got that point now! I just wrongly substituted 3/8 by 5/8 and didn't notice it was wrong substitution.