The Multiplication Rule and Tree Diagrams
Apply the general multiplication rule P(A∩B) = P(A)·P(B|A). Use tree diagrams to organize sequential probability problems. Apply Bayes's Theorem in simple two-stage settings.
A lesson of AP Statistics, in Probability, about 10 minutes.
The General Multiplication Rule and Bayes' Theorem
By the end of this lesson, you will be able to apply the general multiplication rule, use Bayes' Theorem to reverse conditional probabilities, and solve multi-stage probability problems.
The General Multiplication Rule
The probability that both A and B occur:
$$P(A \cap B) = P(A) \cdot P(B \mid A)$$
Equivalently:
$$P(A \cap B) = P(B) \cdot P(A \mid B)$$
For independent events only: $P(A \cap B) = P(A) \cdot P(B)$ (since $P(B \mid A) = P(B)$)
Example (without replacement): A deck has 52 cards. What is the probability of drawing two Aces in a row?
$P(\text{1st Ace}) = 4/52$
$P(\text{2nd Ace} \mid \text{1st Ace}) = 3/51$ (one Ace is gone, 51 cards remain)
$P(\text{both Aces}) = \frac{4}{52} \cdot \frac{3}{51} = \frac{12}{2652} = \frac{1}{221} \approx 0.0045$
Total Probability and Partitions
When event B can only occur alongside one of several mutually exclusive, exhaustive events $A_1, A_2, \ldots, A_k$ (a partition of the sample space):
$$P(B) = P(B \mid A_1)P(A_1) + P(B \mid A_2)P(A_2) + \cdots + P(B \mid A_k)P(A_k)$$
This is the Law of Total Probability.
Example: A factory has two machines. Machine 1 produces 60% of output; 5% defective. Machine 2 produces 40% of output; 10% defective. What fraction of all items are defective?
$P(\text{defective}) = P(\text{def} \mid M_1) \cdot P(M_1) + P(\text{def} \mid M_2) \cdot P(M_2)$
$= 0.05 \times 0.60 + 0.10 \times 0.40 = 0.03 + 0.04 = 0.07$
7% of all items are defective.
Bayes' Theorem
Bayes' Theorem "reverses" a conditional probability. Given $P(B \mid A)$, it finds $P(A \mid B)$.
$$P(A \mid B) = \frac{P(A) \cdot P(B \mid A)}{P(B)}$$
Using the law of total probability to expand P(B):
$$P(A \mid B) = \frac{P(A) \cdot P(B \mid A)}{P(A) \cdot P(B \mid A) + P(A^c) \cdot P(B \mid A^c)}$$
Example (continuing the factory problem): A randomly selected item is defective. What is the probability it came from Machine 1?
$P(M_1 \mid \text{defective}) = \frac{P(M_1) \cdot P(\text{def} \mid M_1)}{P(\text{defective})} = \frac{0.60 \times 0.05}{0.07} = \frac{0.03}{0.07} \approx 0.429$
Despite Machine 1 producing more of the output, it accounts for only 43% of defective items because its defect rate is lower.
Medical Testing Example
A disease affects 1% of the population. A test is 95% sensitive (P(positive | disease) = 0.95) and 90% specific (P(negative | no disease) = 0.90, so P(positive | no disease) = 0.10).