Question

A bag contains x green and y red sweets. A sweet is selected at random and its colour noted. It is then replaced into the bag together with 10 additional sweets of the same color. A second sweet is next randomly drawn. Find the probability that the second sweet is red.

Answer 1

OK, there's an x/(x+y) chance that the first sweet was green, in which case there's a y/(x+y+10) chance that the second is red. There's also a y/(x+y) chance that the first sweet was red, which leads to a (y+10)/(x+y+10) chance that the second is red. Dependent events must have probabilities multiplied, and we've described the only two scenarios which lead to the result we want. That means we can add them:
\[\frac{y^2+10y+yx}{(x+y+10)(x+y)}\]
Someone please check my work.

Answer 2

Thank you Sir