STEP-BY-STEP LESSON · §9.9
Conditional Probability, Bayes’ Formula, and Independent Events
Condition on information and distinguish reversed probabilities.
Before you begin
Intersection
A∩B contains objects in both A and B. In probability this is the event that both conditions occur. Count each common outcome once.
Events and their probability
An event is a set of outcomes. In a finite probability model, its probability is the sum of the probabilities of its outcomes. Two events can overlap; their intersection describes simultaneous occurrence.
Symbols
- P(A|B)
- probability of A given B
- Aᶜ
- Complement of event A in the sample space
Definitions and notation for this topic
Conditional probability
P(A|B)=P(A∩B)/P(B), P(B)>0
The probability of A given that B occurs, provided P(B)>0.
Here the vertical bar means “given.”
The bar reads given, not divides or set-builder such that. Independence and mutually exclusive events are different conditions.
Separate glossary example
P(A|B)=P(A∩B)/P(B)
Independent events
P(A∩B)=P(A)P(B)
Events A and B are independent when P(A∩B)=P(A)P(B). If P(B)>0, this is equivalent to P(A|B)=P(A).
Disjoint events with positive probabilities are not independent: their intersection has probability 0 but the product is positive. Independence is a property of the probability model, not just different event names.
Separate glossary example
For two independent fair coin tosses, let A be a head on the first and B a head on the second: P(A∩B)=1/4=(1/2)(1/2).
Bayes’ formula
P(B_j|E)=P(E|B_j)P(B_j)/∑_i P(E|B_i)P(B_i)
Let B₁,…,B_m form a finite partition of the sample space, each with positive probability, and let P(E)>0. Bayes’ formula reverses conditioning using these prior weights and the likelihoods of E.
The denominator is the total probability of E, summed over exhaustive disjoint cases. P(E|A)=2/10 here, which differs from P(A|E)=2/11.
Separate glossary example
Of 100 items, 10 are type A with 2 defective; the other 90 contain 9 defective. Select one item uniformly. Given it is defective, P(A|defective)=2/11.
Step by step
Step 1 / 6
Restrict attention
If B is known, outcomes outside B are excluded. We require P(B)>0.
The information changes the reference population.
Worked example
Of 100 items, 20 are type A. Four A items and eight other items are defective. One item is selected uniformly at random from all 100 items. Given that the selected item is defective, what is the probability that it is type A?
- Conditioning leaves 4+8=12 defective items.
- Four are type A: P(A|defective)=4/12=1/3.
- But P(defective|A)=4/20=1/5 is a different probability.
Optional self-check
A fair die gives A={1,2} and B={2,3,4}. Compute P(A|B) and decide whether A,B are independent.
Show answer
The overlap is {2}: P(A|B)=(1/6)/(3/6)=1/3. This equals P(A)=2/6; equivalently 1/6=(1/3)(1/2), so they are independent.