Addition rule and applications

The addition rule of probability is the single most useful identity for combining events. It is the probabilistic version of the inclusion-exclusion principle from set theory.

The addition rule

For any two events $A, B$: $\boxed{P(A \cup B) = P(A) + P(B) - P(A \cap B).}$

Derivation: $A \cup B = A \cup (B - A)$ with disjoint pieces. So $P(A \cup B) = P(A) + P(B - A)$. Also, $B = (A \cap B) \cup (B - A)$ disjoint, so $P(B - A) = P(B) - P(A \cap B)$. Substitute.

The three-event rule

$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C).$

Complement rule

$P(A') = 1 - P(A).$

This is invaluable when computing "$A$ does not happen" is easier than "$A$ happens".

"At least one" trick

For events $A_1, A_2, \dots, A_n$, the probability of "at least one occurs" is $P(A_1 \cup A_2 \cup \dots \cup A_n) = 1 - P(\text{none occurs}) = 1 - P(A_1' \cap A_2' \cap \dots \cap A_n').$

When the $A_i$ are "independent" (Class XII), this becomes very useful.

Worked examples

Example 1. Roll a die. Probability of getting "even" OR "more than $3$".

$P(\text{even}) = 3/6$. $P(>3) = 3/6$. $P(\text{even and} > 3) = P(\{4, 6\}) = 2/6$.

$P(\text{even or} > 3) = 3/6 + 3/6 - 2/6 = 4/6 = 2/3$.

Example 2. Draw one card. Probability of "ace" OR "spade".

$P(\text{ace}) = 4/52$. $P(\text{spade}) = 13/52$. $P(\text{ace and spade}) = P(\text{ace of spades}) = 1/52$.

$P = 4/52 + 13/52 - 1/52 = 16/52 = 4/13$.

Example 3. Two dice rolled. Probability sum is $7$ OR doubles.

$P(\text{sum 7}) = 6/36 = 1/6$. $P(\text{doubles}) = 6/36 = 1/6$. $P(\text{both})$? Doubles have sums $2, 4, 6, 8, 10, 12$. So no doubles have sum $7$. Intersection is empty.

$P = 1/6 + 1/6 - 0 = 1/3$.

Example 4. A bag has $3$ red, $4$ blue, $5$ green balls. Two balls drawn. Probability they are of the same colour.

$|S| = \binom{12}{2} = 66$. Favourable: same colour. $\binom{3}{2} + \binom{4}{2} + \binom{5}{2} = 3 + 6 + 10 = 19$.

$P = 19/66$.

Example 5. In a class of $100$ students, $40$ play cricket, $25$ play football, $15$ play both. A student is picked at random.

Probability of playing at least one sport: $(40 + 25 - 15)/100 = 50/100 = 0.5$. Probability of playing neither: $1 - 0.5 = 0.5$.

Try it yourself

1. $P(A) = 0.6, P(B) = 0.4, P(A \cap B) = 0.2$. Find $P(A \cup B)$.
2. Draw a card. Probability of "heart" OR "king".
3. Two dice rolled. Probability of sum $\le 4$ OR doubles.
4. From a deck, draw a card. Probability "red" OR "face card".
5. A die rolled. Probability "even" OR "prime".
6. Three coins tossed. Probability "at least one head".
7. $20$ students, $12$ play cricket, $8$ play football, $5$ play both. Probability a random student plays at least one game?
8. $P(A) = 1/3, P(B) = 1/2, P(A \cup B) = 2/3$. Find $P(A \cap B)$.
9. In a survey, $60\%$ read English newspapers, $40\%$ read Hindi, $20\%$ both. Probability a person reads at least one.
10. Probability of not getting $6$ in two rolls of a die.
11. Two dice. Probability sum is $9$ OR doubles.
12. From $\{1, 2, \dots, 20\}$, pick a number. Probability it is divisible by $2$ or $3$ or $5$.

Pitfalls / Tricks

• The addition rule subtracts the intersection only once.
• For mutually exclusive events, $P(A \cup B) = P(A) + P(B)$ (the intersection is empty).
• Use complement when "at least one" appears , it is often simpler.
• Insight. The addition rule is just inclusion-exclusion translated into probability. Master inclusion-exclusion (Chapter 1) and probability becomes much friendlier.