Comprehensive theory, key formulas, diagrams, and memory aids for Addition Theorem of Probability.
When calculating probabilities, we frequently encounter situations where we want to know the chance of at least one of two events occurring. In everyday language, this is usually phrased using the word "OR" (e.g., what is the probability of drawing a King or a Heart?).
In set theory, "OR" corresponds to the Union of sets, denoted by the symbol $\cup$. The Addition Theorem of Probability provides the mathematical framework for calculating the probability of the union of two or more events.
Before defining the theorem, we must understand two fundamental operations from set theory:
The General Addition Theorem applies to any two events, regardless of whether they can happen at the same time or not.
For any two events $A$ and $B$: $$ P(A \cup B) = P(A) + P(B) - P(A \cap B) $$
Imagine a Venn diagram with two overlapping circles representing Event $A$ and Event $B$. 1. When you add $P(A)$, you count the probability of everything inside circle $A$. 2. When you add $P(B)$, you count the probability of everything inside circle $B$. 3. The overlapping section (the intersection, $A \cap B$) was counted twice—once when adding $A$, and once when adding $B$. 4. To get the correct probability, we must subtract the intersection once so it is only counted a single time.
A card is drawn from a standard deck of 52 cards. What is the probability that the card is either a King ($K$) OR a Heart ($H$)? * $P(K) = \frac{4}{52}$ (There are 4 Kings) * $P(H) = \frac{13}{52}$ (There are 13 Hearts) * $P(K \cap H) = \frac{1}{52}$ (There is exactly 1 card that is both a King and a Heart: the King of Hearts)
Using the General Addition Theorem: $$ P(K \cup H) = P(K) + P(H) - P(K \cap H) $$ $$ P(K \cup H) = \frac{4}{52} + \frac{13}{52} - \frac{1}{52} = \frac{16}{52} = \frac{4}{13} $$
In the previous topic, we learned that mutually exclusive events cannot occur at the same time. If events $A$ and $B$ are mutually exclusive, their intersection is empty, meaning $P(A \cap B) = 0$.
If we plug this into the General Addition Theorem, the subtraction term vanishes. Therefore, for mutually exclusive events, the rule simplifies to:
$$ P(A \cup B) = P(A) + P(B) $$
A single six-sided die is rolled. What is the probability of rolling a 2 OR a 5? * Let $A$ be rolling a 2. $P(A) = \frac{1}{6}$. * Let $B$ be rolling a 5. $P(B) = \frac{1}{6}$. Since you cannot roll a 2 and a 5 on the same single die roll, the events are mutually exclusive. $$ P(A \cup B) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} $$
The Addition Theorem can be extended to three events ($A, B,$ and $C$). Following the logic of Venn diagrams, we add the individual probabilities, subtract the pairwise intersections (which were counted twice), and then add back the intersection of all three (which was subtracted one too many times).
$$ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) $$
If $A, B,$ and $C$ are all mutually exclusive with each other, the formula dramatically simplifies to: $$ P(A \cup B \cup C) = P(A) + P(B) + P(C) $$
When solving word problems, look for specific keywords to know when to use the Addition Theorem: * The word "OR" (e.g., "Find the probability of drawing a red ball or a blue ball"). * The phrase "At least one" (e.g., "Find the probability that at least one of the two students passes the exam"). Note: "At least one" can often be solved faster using the complement rule: $1 - P(\text{None})$.
A very common mistake for students is assuming events are mutually exclusive when they are not, and simply adding $P(A) + P(B)$ without checking for an intersection. Always ask yourself: "Can these two events happen at the exact same time?" If the answer is yes, you MUST use the General Addition Theorem and subtract $P(A \cap B)$.