Two dice are tossed once. The probability of getting an even number at

OR probability:
If Events A and B are independent, the probability that Event A OR Event B occurs is equal to the probability that Event A occurs plus the probability that Event B occurs minus the probability that both Events A and B occur: \(P(A \ or \ B) = P(A) + P(B) - P(A \ and \ B)\).

This is basically the same as 2 overlapping sets formula:
{total # of items in groups A or B} = {# of items in group A} + {# of items in group B} - {# of items in A and B}.

Note that if event are mutually exclusive then \(P(A \ and \ B)=0\) and the formula simplifies to: \(P(A \ or \ B) = P(A) + P(B)\).

Also note that when we say "A or B occurs" we include three possibilities:
A occurs and B does not occur;
B occurs and A does not occur;
Both A and B occur.

AND probability:
When two events are independent, the probability of both occurring is the product of the probabilities of the individual events: \(P(A \ and \ B) = P(A)*P(B)\).

This is basically the same as Principle of Multiplication: if one event can occur in \(m\) ways and a second can occur independently of the first in \(n\) ways, then the two events can occur in \(mn\) ways.

BACK TO THE ORIGINAL QUESTION:
Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is
A. 1/36
B. 3/36
C. 11/36
D. 20/36
E. 23/36

1. The probability of getting an even number at the first die is 1/2 (as the probability of even = the probability of odd = 1/2);
2. The probability of getting a total of 8 is 5/6^2, as there are 5 different favorable scenarios: (2,6), (6,2), (3,5), (5,3) and (4,4);
3. The probability of getting an even number at the first die AND a total of 8 is 3/6^2 (from above case);

Hence, The probability of getting an even number at the first die OR a total of 8 is 1/2+5/36-3/36=20/36.

Answer: D.