The addition theorem

What you will be able to do

• use $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ and its special case for mutually exclusive events;
• find the probability of exactly one, of neither and of "$A$ but not $B$";
• use the rule for three events.

A card that is a heart or a king

A card is drawn from a pack. What is the chance that it is a heart or a king? There are $13$ hearts and $4$ kings, so adding gives $\dfrac{13}{52} + \dfrac{4}{52}$. But the king of hearts was counted twice. Take it away once:

The addition theorem

For any two events $A$ and $B$ of the same experiment,

Why. The union is made of three mutually exclusive parts: $A - B$, $A \cap B$ and $B - A$. So $P(A \cup B)$ is the sum of their probabilities. Also $P(A) = P(A - B) + P(A \cap B)$ and $P(B) = P(B - A) + P(A \cap B)$. Adding $P(A)$ and $P(B)$ counts the middle part twice, so we take it away once.

For mutually exclusive events $P(A \cap B) = 0$, and the theorem becomes $P(A \cup B) = P(A) + P(B)$. This is the third axiom of the last lesson. Many Statistics texts, such as AP Statistics, call the general form the general addition rule.

Four more probabilities from the same data

Given $P(A)$, $P(B)$ and $P(A \cap B)$, all of these follow:

• $A$ but not $B$: $P(A - B) = P(A) - P(A \cap B)$.
• Exactly one of $A$ and $B$: $P(A) + P(B) - 2P(A \cap B)$.
• Neither: $P(A' \cap B') = 1 - P(A \cup B)$.
• Not both: $P((A \cap B)') = 1 - P(A \cap B)$.

In the figure, $P(A) = 0.5$, $P(B) = 0.4$ and $P(A \cap B) = 0.2$. Then $P(A \cup B) = 0.7$, exactly one has probability $0.5 + 0.4 - 0.4 = 0.5$, and neither has probability $0.3$.

How large can the overlap be?

Because $P(A \cup B) \le 1$, we get $P(A \cap B) \ge P(A) + P(B) - 1$. Because $A \cap B$ lies inside both $A$ and $B$, we get $P(A \cap B) \le \min\{P(A), P(B)\}$. So

Also $P(A \cup B) \le P(A) + P(B)$. A set of numbers that breaks these limits cannot be a set of probabilities.

Three events

For three events,

Add the singles, take away the pairs, and put back the triple. The same pattern, with plus and minus signs taking turns, works for four events and more. This is the inclusion and exclusion rule.

A number is chosen at random from $1$ to $100$. Let $A$, $B$ and $C$ be "divisible by $2$", "divisible by $3$" and "divisible by $5$". Then $n(A) = 50$, $n(B) = 33$, $n(C) = 20$, $n(A \cap B) = 16$, $n(B \cap C) = 6$, $n(A \cap C) = 10$ and $n(A \cap B \cap C) = 3$. So

A worked example

Two events have $P(A) = 0.6$, $P(B) = 0.5$ and $P(A \cup B) = 0.8$. Find $P(A \cap B)$, the probability of exactly one, and the probability of neither.

1. The overlap. From the theorem, $0.8 = 0.6 + 0.5 - P(A \cap B)$, so $P(A \cap B) = 0.3$.
2. Exactly one. $0.6 + 0.5 - 2 \times 0.3 = 0.5$.
3. Neither. $1 - 0.8 = 0.2$.
4. Check the limits. The overlap must lie between $\max\{0, 0.1\} = 0.1$ and $\min\{0.6, 0.5\} = 0.5$. The value $0.3$ does.

Practise with new numbers

Try it in Python

The program uses the numbers from $1$ to $100$ as the sample space. It checks the addition theorem, the three-event rule and the formula for exactly one.

Common mistakes

• Adding without taking away the overlap. $P(A) + P(B)$ counts the overlap twice unless the events are mutually exclusive.
• Using the mutually exclusive form by habit. Check that $A \cap B$ really is empty first.
• Confusing "or" with "exactly one". "$A$ or $B$" includes both. Exactly one does not.
• Stopping the three-event rule early. The last term, plus $P(A \cap B \cap C)$, is easy to forget.

Remember these

• $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, and without the last term for mutually exclusive events.
• $P(A - B) = P(A) - P(A \cap B)$. Exactly one: $P(A) + P(B) - 2P(A \cap B)$. Neither: $1 - P(A \cup B)$.
• $\max\{0, P(A) + P(B) - 1\} \le P(A \cap B) \le \min\{P(A), P(B)\}$.
• Three events: singles, minus pairs, plus triple.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. $P(A) = 0.45$, $P(B) = 0.3$ and $P(A \cap B) = 0.1$. Find $P(A \cup B)$.

2. A card is drawn from a pack. Find the probability that it is a spade or a face card.

3. For mutually exclusive events, $P(A) = 0.3$ and $P(B) = 0.45$. Find the probability that neither happens.

4. Is $P(A) = 0.7$, $P(B) = 0.5$, $P(A \cap B) = 0.1$ possible?

5. Why is $P(A \cap B) \le P(A)$?

References

• NCERT, Mathematics for Class XI (earlier edition), Probability, section on the addition rule.
• OpenStax, Introductory Statistics, Chapter 3, section on two basic rules of probability.