Probability, conditioning and independence

What you’ll learn

• Use addition and multiplication rules.
• Distinguish independence from mutually exclusive events.
• Update probabilities when sampling without replacement.

Understand the idea

The union $A\cup B$ means at least one event occurs; the intersection $A\cap B$ means both occur. When adding $P(A)$ and $P(B)$, subtract the overlap once because it has otherwise been counted twice. A complement often makes an “at least one” question shorter.

A conditional probability restricts the reference population to the event after the vertical bar. In $P(A\mid B)$, only outcomes in $B$ are counted in the denominator. In a tree diagram multiply along a route and add probabilities of mutually exclusive routes to the required event.

Independent events do not change one another's probabilities. Mutually exclusive events cannot occur together; if both have positive probability, they cannot also be independent. Without replacement, change the available counts and the total after each draw.

Key formulae

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

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$$

$$P(A\cap B)=P(A)P(B\mid A)$$

$$P(A')=1-P(A)$$

Worked example

For two events, $P(A)=0.55$, $P(B)=0.40$ and $P(A\cap B)=0.25$. Find $P(A\cup B)$ and $P(A\mid B)$.

1. Add the marginal probabilities and subtract the overlap: $0.55+0.40-0.25$.
2. This gives $P(A\cup B)=0.70$.
3. For the conditional probability, divide the overlap by $P(B)$: $0.25/0.40=0.625$.

Answer: $P(A\cup B)=0.70$ and $P(A\mid B)=0.625$.