Simple yet confusing probability problem. A bag contains x green candies and y red candies. A candy is selected at random from the bag and it's color is noted. It is then replaced into the bag with an additional ten candies of the same color. A second candy is then randomly chosen. Find the probability that the second candy is red.
I think that the answer is y/(x+y+10) but am unsure.
Also where would you recommend to get more practice on probality problems
 A: We have x green candies and y red.

Case 1-

Green candy is picked in first attempt.
$\frac x{x+y}$
Now green candies x + 10.
Red candy is picked in second attempt.
$\frac {y}{x+y+10}$
Probability = $\frac {x}{x+y} × \frac {y}{x+y+10} $

Case 2-

Red candy is picked in first attempt.
Probability = $\frac y{x+y}$
Now red candies is y + 10.
Red candy is picked in second attempt.
$\frac {y+10}{x+y+10}$
Probability = $\frac {y}{x+y} × \frac {y+10}{x+y+10} $
Total Probability = $\frac {x}{x+y} × \frac {y}{x+y+10} +\frac {y}{x+y} × \frac {y+10}{x+y+10} $
A: P(red in 1st try) = $\frac{y}{x+y}$
P(red in second try) = $\frac{y}{x+y}. \frac{y+10}{x+y+10} + \frac{x}{x+y}. \frac{y}{x+y+10}$  (using mutual exclusivity of events)
$$= \frac{y}{x+y}$$
A: Also where would you recommend to get more practice on probality problems.
Play Dungeons and Dragons.  If that is too much of a commitment.  Play Backgammon.
As for your question.
Would you believe that it is exactly the same as the chance that the first candy is red?
The expected number of red candies added is proportional to the number of red candies.  Adding candies in this way doesn't change the probabilities.  
If that is too abstract:
The candies could be Green followed by red.
or red followed by red.
(Green)(red) + (red)(red) = $\frac {x}{x+y}\frac {y}{x+y+10} + \frac {y}{x+y}\frac {y+10}{x+y+10} = \frac {y(x+y+10)}{(x+y)(x+y+10)} =  \frac {y}{x+y} $
