Probability Rules

Three rules answer most introductory event questions

For events A and B:

$P(A^c)=1-P(A)$ $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ $P(A\cap B)=P(A)P(B\mid A).$

The symbols ∪ and ∩ mean “A or B (or both)” and “both A and B.” The conditional probability P(B | A) describes the chance of B once A is known, and requires P(A) > 0 for the usual ratio definition. Independence is a special condition under which P(B | A) = P(B), allowing the simpler product P(A)P(B).

Worked example: use actual joint counts

Imagine 100 hypothetical orders. Of these, 40 use express shipping, 30 contain a gift, and 15 are both express and gift orders.

	Gift	Not gift	Total
Express	15	25	40
Not express	15	45	60
Total	30	70	100

The probability that a randomly selected order is express or a gift is (40 + 30 − 15)/100 = 0.55. Subtracting the intersection avoids counting its 15 orders twice.

The conditional probability of gift given express is 15/40 = 0.375. Multiplication gives P(express and gift) = 0.40 × 0.375 = 0.15. Multiplying 0.40 by the marginal gift probability 0.30 would instead give 0.12; that incorrectly assumes independence for this empirical distribution.

These proportions describe the 100 recorded orders. Differences between sample proportions do not by themselves prove dependence in a broader population; that requires uncertainty and design considerations.

Independence is different from mutually exclusive

Mutually exclusive events cannot both happen, so their intersection probability is 0. Independent events satisfy P(A ∩ B) = P(A)P(B). If two mutually exclusive events each have positive probability, they cannot be independent: their product is positive while their intersection is zero.

For a single fair die, “1” and “2” are mutually exclusive. For two independent die rolls, “first roll is 1” and “second roll is 2” can happen together with probability 1/36.

“At least one” is often a complement problem

Suppose four independent attempts each have a 20% success probability. No successes has probability 0.8⁴ = 0.4096. At least one success has probability 0.5904. Adding 0.2 four times would ignore overlapping successes.

The product model fails if attempts affect one another or the success chance changes. State those assumptions before using the expression.

Practice

If P(A) = 0.6, P(B) = 0.5 and P(A ∩ B) = 0.2, find P(A ∪ B) and P(B | A).

Show the explanation

P(A ∪ B) = 0.6 + 0.5 − 0.2 = 0.9. P(B | A) = 0.2/0.6 = 1/3. Since 0.2 differs from 0.6 × 0.5, A and B are not independent under these stated probabilities.

Continue

Use the probability calculator or study conditional probability. OpenStax explains the event distinctions.