The last term P(E and F) is there regardless of discrete or continuous probability. If events E and F are mutually exclusive, then P(E and F) is 0. Then P(E or F) = P(E) + P(F);

For example, if there are 3 green balls, 4 white balls and 2 red balls, if your draw 1 ball what is the probability that it is green or red?

Here the ball cannot be red and green, so P(E and F) = 0. So answer = p(green) + p(red) - 0 = 3/9+2/9 = 5/9.

An example where P(E and F) is non zero is as follows. What is the probability that you get an even number or a number greater than 3 when you roll a die?

P(even number)=3/6 = 1/2
P(number>3)=3/6 = 1/2
P(even number and number>3) = 2/6 = 1/3.
P(even number or number > 3) = 1/2+1/2-1/3=2/3

Think of the same problem another way:{2, 4, 5, 6} is the set of required events, so answer = 4/6 = 2/3.

great explanation, thanks

Another way of looking at it is that you have avoid counting twice the overlapping probabilities; in the above example:
Prob (even) = {2,4,6} = 3/6 = 1/2
Prob (>3) = {4,5,6} = 3/6 = 1/2
Prob (even) + Prob (>3) = {2,4,6,4,5,6} = 6/6 = 1 ?

You have to eliminate the overlap:
Prob (even AND >3) = {4,6}
Prob (even OR >3) = {2,4,6,4,5,6} - {4,6} = {2,4,5,6} = 4/6/ = 2/3