Probability Rules

When we want the probability that at least one of two events occurs, we must not double-count outcomes that belong to both.

Bounds on $P(A \cup B)$: Since $P(A \cap B) \geq 0$, we always have $P(A \cup B) \leq P(A) + P(B)$. Also, since $A \subseteq (A \cup B)$, we have $P(A \cup B) \geq P(A)$. The union can never be smaller than either individual probability — which answers our Think First question.

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$ — always subtract the intersection; $P(A \cup B) \geq \max(P(A), P(B))$ — the union is always at least as large as either event

Pause — copy the addition rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ and the reason you subtract the intersection (to avoid double-counting the overlap) into your book.

We just saw that $P(A \cup B)$ uses addition with a correction for double-counting. That raises a question: what rule do we use for the intersection $P(A \cap B)$ — the AND probability — especially when one event can influence the other? This card answers it → the multiplication rule: $P(A \cap B) = P(A) \times P(B \mid A)$, which simplifies to $P(A) \times P(B)$ only when events are independent.

When we want the probability that both events occur, we multiply probabilities — but we must account for how the first event affects the second.

Here $P(B \mid A)$ means "the probability of $B$ given that $A$ has occurred." This is conditional probability.

Example: A bag has 5 red and 5 blue marbles. You draw two without replacement. What is $P(\text{red then blue})$?

$P(\text{first red}) = \dfrac{5}{10} = \dfrac{1}{2}$. After removing one red, $P(\text{second blue} \mid \text{first red}) = \dfrac{5}{9}$.

So $P(\text{red then blue}) = \dfrac{1}{2} \times \dfrac{5}{9} = \dfrac{5}{18}$.

Independent events special case: If $A$ does not affect $B$, then $P(B \mid A) = P(B)$ and:

General: $P(A \cap B) = P(A) \times P(B \mid A)$ — always check whether events affect each other; Independent: $P(B \mid A) = P(B)$, so $P(A \cap B) = P(A) \times P(B)$

Pause — copy both multiplication forms: general $P(A \cap B) = P(A) \times P(B \mid A)$ and the simpler independent version $P(A \cap B) = P(A) \times P(B)$ — only use the simple version when independence is confirmed into your book.

We just saw that both the addition and multiplication rules require you to check whether events are mutually exclusive or independent before applying the simplified form. That raises a question: what is the most efficient way to confirm those conditions and avoid using the wrong simplification? This card answers it → use the definitions as tests: mutually exclusive means $P(A \cap B) = 0$; independent means $P(A \cap B) = P(A) \times P(B)$.

The complement rule is deceptively simple but extraordinarily powerful:

Its most common application is "at least one" problems. Directly counting "at least one success" requires summing probabilities for exactly 1, exactly 2, exactly 3, and so on. The complement needs only one calculation.

Example: Four people each guess a number from 1–10. What is $P(\text{at least one correct})$?

$P(\text{one person wrong}) = \dfrac{9}{10}$. Assuming independence: $P(\text{all wrong}) = \left(\dfrac{9}{10}\right)^4 = 0.6561$.

So $P(\text{at least one correct}) = 1 - 0.6561 = 0.3439$.

$P(A \cup B) = P(A)+P(B)$ only if mutually exclusive (given or proven); $P(A \cap B) = P(A) \times P(B)$ only if independent (given or proven)

Pause — copy both condition tests: mutually exclusive requires $P(A \cap B) = 0$ (given or proven); independent requires $P(A \cap B) = P(A) \times P(B)$ (given or proven — never assume!) into your book.