A fair die is rolled once and a fair coin is flipped once. What is the

Because 1/6 is basically 1/6*1: the die landing on 3 and the coin landing on any side. The same way 1/2 is basically 1/2*1: the coin landing on heads and the die landing on any side. So, both 1/6 and 1/2 count (include) the probability of the case when the die lands on 3 and the coin lands on heads, hence we should subtract P(3 on the die and heads on the coin)=1/6*1/2 once, to avoid double counting.

P=1/6+1/2-1/6*1/2=7/12.

A fair die is rolled once and a fair coin is flipped once. What is the probaility that either the die will land on 3 or that the coin will land on heads?

This question can be solved with an easier approach: P(3 on a die OR heads on a coin)=1-P(neither 3 on a die nor heads on a coin)=1-5/6*1/2=7/12.

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.

Hope it helps.