1. ## Another ball problem

Bowl A contains 4 red balls, 3 blue balls and 2 green balls. Bowl B contains 2 red balls, 3 blue balls and 4 green balls. One ball is drawn from bowl A and then put into bowl B. After that one ball is drawn from bowl B. What is the probability that the ball that is drawn from bowl B is red?

2. Originally Posted by Ond
Bowl A contains 4 red balls, 3 blue balls and 2 green balls. Bowl B contains 2 red balls, 3 blue balls and 4 green balls. One ball is drawn from bowl A and then put into bowl B. After that one ball is drawn from bowl B. What is the probability that the ball that is drawn from bowl B is red?

Well, think I've got it, but I'm not quite sure. Is it: 4/9 * 3/10 + 5/9 * 2/10

3. Originally Posted by Ond
Bowl A contains 4 red balls, 3 blue balls and 2 green balls. Bowl B contains 2 red balls, 3 blue balls and 4 green balls. One ball is drawn from bowl A and then put into bowl B. After that one ball is drawn from bowl B. What is the probability that the ball that is drawn from bowl B is red?
$\displaystyle \mathcal{P}\left(R_2\right)=\mathcal{P}\left(R_1R_ 2\right)+\mathcal{P}\left(B_1R_2\right)+\mathcal{P }\left(G_1R_2\right)$

4. Well, I'll throw in my 2 cents' worth:

Prob that 1st ball is red = 4/9

So we have 2 + 4/9 reds = 22/9 reds

Prob that next draw is red: (22/9) / (22/9 + 3 + 4) = 22/85

Anything wrong with this "1st cup of coffee" approach?