2. A bag contains $x$ green and $y$ red sweets. A sweet is selected at random from the bag and its colour noted. It is then replaced into the bag together with 10 additional sweets of the same colour. A second sweet is next randomly drawn. Find the probability that the second sweet is red.
(A) $\frac{y+10}{x+y+10}$
(B) $\frac{y}{x+y+10}$
(C) $\frac{y}{x+y}$
(D) $\frac{x}{x+y}$
(E) $\frac{x+y}{x+y+10}$
2. (C)
Probability $=\left(\frac{x}{x+y} \times \frac{y}{x+y+10}\right)+\left(\frac{y}{x+y} \times \frac{y+10}{x+y+10}\right)$, which upon simplification yields $(\mathrm{C})$ as the answer.