Q:

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} $

