Complements and independence

Idea: Count what you do not want

Some events are easy to count from the other side. The complement $A^c$ is everything outside $A$.

• Flip on "at least one", "not all", "some pair matches"
• Find the chance of none, then subtract it from $1$

Formula: The complement rule

$P(A)=1-P(A^c)$

An event and its complement cover the list exactly once.

Example: At least one 6 in 3 rolls

Roll a fair die $3$ times. What is the chance of at least one $6$?

1. No $6$: $5\times 5\times 5=125$ of $6^3=216$ outcomes
2. At least one: $1-\frac{125}{216}=\frac{91}{216}$

Formula: At least one success in n tries

$P(\ge 1\text{success})=1-(1-p{)}^n$

n independent tries, each with chance p. For 5 coin tosses, at least one head: 31/32.

Idea: Exclusive is not independent

Exclusive events cannot both happen. Independent events tell nothing about each other.

• Exclusive: add, $\frac{1}{6}+\frac{1}{6}=\frac{1}{3}$
• Independent: multiply, $\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$
• Exclusive events with positive chances are never independent

Formula: Inclusion and exclusion

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

When events overlap, subtract the part counted twice. ∪ is either or both, ∩ is both.

Common mistake: Adding chances across tries

$3\times \frac{1}{6}=\frac{1}{2}$ for at least one $6$ in $3$ rolls counts rolls with several sixes more than once.

Instead Use the complement: $1-{\left(\frac{5}{6}\right)}^3\approx 0.42$. With $7$ rolls the habit gives $\frac{7}{6}>1$.

Quick check: Some pair matches

Roll $3$ fair dice. What is the chance that at least $2$ of them match?

Idea: Say it out loud

Say "I will find the chance of none and subtract" before you compute.

• Before you multiply, say why the events are independent
• Before you add, say they cannot happen together
• State any independence the question did not give