Answer
$P(E~or~F)=\frac{9}{19}\approx0.5263$
If we play the game 1900 times, we would expect that the metal ball falls into a green or red slot about 1000 times.
Work Step by Step
The sample space S = {0, 00, 1, 2, 3, ..., 34, 35, 36}. So, $N(S)=38$
Let E be the event "green slot" = {0, 00}. So, $N(E)=2$.
Let F be the event "red slot" = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35}. So, $N(F)=18$.
The events "green slot" and "red slot" are mutually exclusive (disjoint).
Using the Addition Rule for Disjoint Events (page 270):
$P(E~or~F)=P(E)+P(F)=\frac{N(E)}{N(S)}+\frac{N(F)}{N(S)}=\frac{2}{38}+\frac{18}{38}=\frac{20}{38}=\frac{9}{19}\approx0.5263$
