Initially, there are 5 red beans and 9 black beans in a bag. Then, 3 red beans and 3 black beans are added to the bag. If one bean is randomly chosen from the bag, what is the probability that this bean is red?
(A) $\frac{3}{8}$
(B) $\frac{2}{5}$
(C) $\frac{4}{5}$
(D) $\frac{5}{8}$
(E) $\frac{8}{17}$
When 3 red beans are added to the bag, the number of red beans in the bag is $5+3=8$.

When 3 black beans are added to the bag, the number of black beans in the bag is $9+3=12$. The number of beans now in the bag is $8+12=20$.

If one bean is randomly chosen from the bag, the probability that the bean is red is $\frac{8}{20}=\frac{2}{5}$.

ANswer: (B)