Probability foundations · Definition

Addition rule

probabilityfoundations

Before you start you need to know

Computing the probability of «this or that» is the most common question when combining events; this rule avoids double counting.

Definition

The addition rule gives the probability \(P\) that at least one of two events \(A\) and \(B\) occurs: add their probabilities and subtract that of their intersection.

\[P(A\cup B)=P(A)+P(B)-P(A\cap B)\]

When we add \(P(A)\) and \(P(B)\), outcomes lying in both events enter twice. That is why \(P(A\cap B)\) is subtracted. On a die, with \(A\) = «even» and \(B\) = «multiple of 3», the 6 is in both: \(\frac{3}{6}+\frac{2}{6}-\frac{1}{6}=\frac{4}{6}\).

If \(A\) and \(B\) are mutually exclusive, \(P(A\cap B)=0\) and the rule reduces to \(P(A\cup B)=P(A)+P(B)\), the third Kolmogorov axiom for two events. The general rule also yields the bound \(P(A\cup B)\le P(A)+P(B)\), useful for bounding probabilities without knowing the intersection.

For three events the correction continues:\[P(A\cup B\cup C)=P(A)+P(B)+P(C)-P(A\cap B)-P(A\cap C)-P(B\cap C)+P(A\cap B\cap C).\]This is the inclusion–exclusion principle, which extends to any number of events by alternating signs.

A typical mistake is adding without subtracting when events overlap, which can give «probabilities» above 1. Another is using the product \(P(A)P(B)\) instead of the sum: the product answers «both», and only under independence. The rule also works backwards: if \(P(A\cup B)\) is known, solve for \(P(A\cap B)\).

Think about it

If \(P(A)=0.6\) and \(P(B)=0.7\), can they be exclusive?

No: it would give \(0.6+0.7=1.3>1\). Necessarily \(P(A\cap B)\ge 0.3\).

Exercises 10

References

  • Joseph K. Blitzstein, Jessica Hwang, Introduction to Probability, 2nd edition, CRC Press, 2019. §1.6
  • Sheldon Ross, A First Course in Probability, 10th edition, Pearson, 2019. §2.4
  • Larry Wasserman, All of Statistics, 1st edition, Springer, 2004. §1.3