The complement rule handles a single event and its negation. The addition theorem handles the probability that at least one of two events occurs — the event $A \cup B$, read "$A$ or $B$".
For any two events $A$ and $B$ of the same sample space:
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
The reason for the subtraction is geometric. If you simply add $P(A)$ and $P(B)$, the outcomes lying in both events — the overlap $A \cap B$ — get counted twice, once inside each event. Subtracting $P(A \cap B)$ removes the double count and restores the correct total. On a Venn diagram, you are adding two overlapping discs and then peeling off the lens-shaped middle that you covered twice.
When the two events are mutually exclusive, they have no common outcome, so $A \cap B = \varnothing$ and $P(A \cap B) = 0$. The theorem then collapses to the simple additive form, which is exactly the third axiom:
$$P(A \cup B) = P(A) + P(B) \qquad (\text{when } A \cap B = \varnothing)$$
The result extends to three events, again alternately adding and subtracting overlaps:
$$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$$
A pairing with the complement rule is worth memorising: "neither $A$ nor $B$" is the complement of "$A$ or $B$", which by De Morgan's law equals $\overline{A} \cap \overline{B}$ and gives a handy shortcut.
Rearranged forms of the theorem solve "find the missing piece" questions. From the master formula you can read off any one quantity given the other three:
$$P(A \cap B) = P(A) + P(B) - P(A \cup B), \qquad P(A) = P(A \cup B) + P(A \cap B) - P(B)$$
Mixing counting with probability. Many board and competitive questions wrap the addition theorem around a permutations / combinations count. The recipe is always the same: use $\binom{n}{r}$ to count favourable and total selections, form $P = \dfrac{\text{favourable}}{\text{total}}$, then combine events with the addition theorem or the complement rule. For choosing a committee of $r$ from $n$ people, the total number of ways is $n(S) = \binom{n}{r}$; "at least one woman" is fastest as $1 - P(\text{no woman})$. For arrangements, $n(S) = n!$ and favourable cases are counted by treating blocks together or fixing positions. The key judgement is whether order matters: a committee or a hand of cards is a selection (use $\binom{n}{r}$), whereas seating in a row or forming a number is an arrangement (use $n!$ or $^{n}P_{r}$).
Deeper Insight — the addition theorem is inclusion–exclusion in probability dress: If the subtraction in $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ feels familiar, that is because you have already met its set-counting twin: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$ from the chapter on Sets. The two are the same identity. Divide every term of the counting version by $n(S)$ and each count becomes a probability, turning the inclusion–exclusion principle for sizes into the addition theorem for likelihoods. This is the clearest illustration of the chapter's central message: probability is built directly on set theory, and the overlap must be subtracted for exactly the same reason a Venn diagram cannot count its shared region twice. Recognising this lets you reuse one mental tool in two settings, and it explains why the three-event formula alternates signs in the inclusion–exclusion pattern. When combinations enter, nothing changes conceptually — $\binom{n}{r}$ is merely how you count the favourable and total outcomes before the same ratio and the same addition rule take over.
A card is drawn from a pack of $52$. Find the probability that it is a king or a heart.
Solution- Let $A$ = "king" and $B$ = "heart". Then $P(A) = \dfrac{4}{52}$ and $P(B) = \dfrac{13}{52}$.
- The overlap is the king of hearts, a single card: $P(A \cap B) = \dfrac{1}{52}$.
- Addition theorem: $P(A \cup B) = \dfrac{4}{52} + \dfrac{13}{52} - \dfrac{1}{52} = \dfrac{16}{52} = \dfrac{4}{13}$.
Answer: $\dfrac{4}{13}$.
A die is rolled. Find the probability of getting an even number or a number greater than $4$.
Solution- $A$ = even $= \{2,4,6\}$, so $P(A) = \dfrac{3}{6}$. $B$ = greater than $4$ $= \{5,6\}$, so $P(B) = \dfrac{2}{6}$.
- Overlap $A \cap B = \{6\}$, so $P(A \cap B) = \dfrac{1}{6}$.
- $P(A \cup B) = \dfrac{3}{6} + \dfrac{2}{6} - \dfrac{1}{6} = \dfrac{4}{6} = \dfrac{2}{3}$.
Answer: $\dfrac{2}{3}$.
If $P(A) = 0.5$, $P(B) = 0.3$ and $A$, $B$ are mutually exclusive, find $P(A \cup B)$.
Solution- Mutually exclusive means $A \cap B = \varnothing$, so $P(A \cap B) = 0$.
- The theorem reduces to $P(A \cup B) = P(A) + P(B)$.
- $P(A \cup B) = 0.5 + 0.3 = 0.8$.
Answer: $P(A \cup B) = 0.8$.
Two dice are thrown. Find the probability that the sum is $7$ or the sum is $11$.
Solution- $n(S) = 36$. Sum $= 7$: $6$ outcomes, so $P(A) = \dfrac{6}{36}$. Sum $= 11$: $(5,6),(6,5)$, so $P(B) = \dfrac{2}{36}$.
- A roll cannot total both $7$ and $11$, so the events are mutually exclusive: $P(A \cap B) = 0$.
- $P(A \cup B) = \dfrac{6}{36} + \dfrac{2}{36} = \dfrac{8}{36} = \dfrac{2}{9}$.
Answer: $\dfrac{2}{9}$.
In a class, $P(\text{a student plays cricket}) = 0.6$, $P(\text{plays football}) = 0.5$, and $P(\text{plays both}) = 0.3$. Find the probability that a student plays neither game.
Solution- First find "cricket or football": $P(A \cup B) = 0.6 + 0.5 - 0.3 = 0.8$.
- "Neither" is the complement of "at least one": $P(\overline{A \cup B}) = 1 - P(A \cup B)$.
- $P(\text{neither}) = 1 - 0.8 = 0.2$.
Answer: $0.2$.
A card is drawn from a pack of $52$. Find the probability that it is a spade or a face card.
Solution- $A$ = spade: $13$ cards, $P(A) = \dfrac{13}{52}$. $B$ = face card (J, Q, K of each suit): $12$ cards, $P(B) = \dfrac{12}{52}$.
- Overlap = face cards that are spades (J, Q, K of spades): $3$ cards, $P(A \cap B) = \dfrac{3}{52}$.
- $P(A \cup B) = \dfrac{13}{52} + \dfrac{12}{52} - \dfrac{3}{52} = \dfrac{22}{52} = \dfrac{11}{26}$.
Answer: $\dfrac{11}{26}$.
Given $P(A) = 0.25$, $P(B) = 0.5$ and $P(A \cap B) = 0.14$, find (a) $P(A \cup B)$, (b) $P(\text{neither } A \text{ nor } B)$.
Solution- (a) Addition theorem: $P(A \cup B) = 0.25 + 0.5 - 0.14 = 0.61$.
- (b) "Neither" $= 1 - P(A \cup B)$ (De Morgan: $\overline{A} \cap \overline{B} = \overline{A \cup B}$).
- $= 1 - 0.61 = 0.39$.
Answer: (a) $0.61$; (b) $0.39$.
For two events, $P(A \cup B) = 0.7$, $P(A) = 0.4$ and $P(B) = 0.5$. Find $P(A \cap B)$, and hence the probability of "exactly one of $A$, $B$".
Solution- Rearrange: $P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.4 + 0.5 - 0.7 = 0.2$.
- "Exactly one" $= P(A \cup B) - P(A \cap B)$.
- $= 0.7 - 0.2 = 0.5$.
Answer: $P(A \cap B) = 0.2$ and $P(\text{exactly one}) = 0.5$.
A committee of $3$ is chosen at random from $4$ men and $3$ women. Find the probability that it has (a) all men, (b) at least one woman.
Solution- Total ways to choose $3$ from $7$: $n(S) = \binom{7}{3} = 35$.
- (a) All men: $\binom{4}{3} = 4$, so $P(\text{all men}) = \dfrac{4}{35}$.
- (b) "At least one woman" is the complement of "all men": $P = 1 - \dfrac{4}{35} = \dfrac{31}{35}$.
Answer: (a) $\dfrac{4}{35}$; (b) $\dfrac{31}{35}$.
Two cards are drawn together from a pack of $52$. Find the probability that (a) both are kings, (b) both are of the same colour.
Solution- Total ways: $n(S) = \binom{52}{2} = 1326$.
- (a) Both kings: $\binom{4}{2} = 6$, so $P = \dfrac{6}{1326} = \dfrac{1}{221}$.
- (b) Same colour = both red or both black: $\binom{26}{2} + \binom{26}{2} = 325 + 325 = 650$, so $P = \dfrac{650}{1326} = \dfrac{25}{51}$.
Answer: (a) $\dfrac{1}{221}$; (b) $\dfrac{25}{51}$.
The four distinct letters of the word "MATH" are arranged in a row at random. Find the probability that a random arrangement (a) begins with M, (b) has the two consonants T and H together.
Solution- The $4$ distinct letters give $n(S) = 4! = 24$ arrangements.
- (a) Fix M first; arrange the other $3$ in $3! = 6$ ways: $P = \dfrac{6}{24} = \dfrac{1}{4}$.
- (b) Treat (TH) as one block with M, A: that is $3!$ block-orders, and the block has $2!$ internal orders, giving $3! \times 2! = 12$. So $P = \dfrac{12}{24} = \dfrac{1}{2}$.
Answer: (a) $\dfrac{1}{4}$; (b) $\dfrac{1}{2}$.
A number is selected at random from $1$ to $100$. Find the probability that it is divisible by $2$ or by $5$.
Solution- $n(S) = 100$. $A$ = divisible by $2$: $50$ numbers. $B$ = divisible by $5$: $20$ numbers.
- Overlap = divisible by $10$: $10$ numbers, so $P(A \cap B) = \dfrac{10}{100}$.
- $P(A \cup B) = \dfrac{50}{100} + \dfrac{20}{100} - \dfrac{10}{100} = \dfrac{60}{100} = \dfrac{3}{5}$.
Answer: $\dfrac{3}{5}$.