Probability of Either of Two Teams Winning the World Cup I'm just trying to get my head around some basic probability theory. Say we have a World Cup sweepstake and each better has been drawn two random teams from a hat.
Let's assume that I have been given Argentina and Honduras. The probability of Argentina winning is $0.2$ and the probability of Honduras winning is $0.00039$. To calculate the probability of either team winning I would calculate the union of their probabilities?
$$P(A~\text{or}~B)=P(A\cup B)=P(A)+P(B)$$
$$P(\text{Argentina}\cup\text{Honduras}) = P(\text{Argentina}) + P(\text{Honduras}) = 0.2 + 0.00039 = 0.20039$$
 A: This is correct, but only because the events $\{\text{Argentina wins the World Cup}\}$ and $\{\text{Honduras wins the World Cup}\}$ are mutually exclusive. In other words, if $A$ is the former event and $H$ is the latter, then $\mathbb{P}(A \cap H) = 0$. Instead, suppose we have two events $A$ and $B$, which are not necessarily mutually exclusive.  Then the probability of their union (either event occurring) is given by
$$\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cap B)\,\,,$$
where $A \cap B$ is the intersection of these events (the event that both events occur).  This is a special case of what is called the inclusion-exclusion principle in probability theory. 
A: $\newcommand{\P}{\operatorname{\mathcal{P}}}$
Yes, that is correct, as long as the events are mutually exclusive; that is to say only one team can win (ties are broken) and the intersection of the events has zero probability of occurring. $(\P(A\cap B)=0)$
$$\P(A\cup B) = \P(A)+\P(B) \quad \text{iff }A,B\text{ are mutually exclusive}$$
If a tie were a possible outcome ($\P(A\cap B)>0$) then it would be need to be excluded from the result (because it would be double counted).  $$\P(A\cup B) = \P(A)+\P(B)-\P(A\cap B) \quad\text{otherwise}$$
