What this chapter is about
This chapter builds on the probability ideas you learned in earlier classes and introduces conditional probability, the multiplication rule, independence of events, and Bayes' theorem. These tools let you handle situations where one event affects the chance of another, which is how most real-world uncertainty works.
You will also study probability distributions for discrete random variables, focusing on the mean (expectation) and variance. The chapter ends with the binomial distribution, which models repeated independent trials where each trial has exactly two outcomes (success or failure). After working through this material, you should be able to compute probabilities in multi-stage experiments, apply Bayes' theorem to update beliefs with new information, and analyse random variables quantitatively.
Key ideas
- Conditional probability: P(A|B) is the probability of event A given that event B has already occurred. It equals P(A ∩ B) / P(B), provided P(B) > 0.
- Multiplication rule: For any two events, P(A ∩ B) = P(B) × P(A|B) = P(A) × P(B|A). This lets you find the probability of both events happening together.
- Independent events: Two events A and B are independent if P(A ∩ B) = P(A) × P(B). Knowing one occurred does not change the probability of the other.
- Total probability theorem: If B₁, B₂, …, Bₙ form a partition of the sample space, then P(A) = Σ P(Bᵢ) × P(A|Bᵢ).
- Bayes' theorem: Given prior probabilities of causes B₁, B₂, …, Bₙ and a new observation A, the updated probability is P(Bᵢ|A) = [P(Bᵢ) × P(A|Bᵢ)] / P(A).
- Random variable: A function that assigns a real number to each outcome of an experiment. Its probability distribution lists every value with its probability.
- Mean and variance: For a discrete random variable X with values xᵢ and probabilities pᵢ, the mean is E(X) = Σ xᵢ pᵢ and the variance is Var(X) = Σ pᵢ (xᵢ − E(X))² = E(X²) − [E(X)]².
- Binomial distribution: If an experiment has n independent trials, each with success probability p, the probability of exactly r successes is P(X = r) = C(n, r) × p^r × (1 − p)^(n − r), where C(n, r) is "n choose r".
Formulas and facts to remember
- Formula: P(A\|B) = P(A ∩ B) / P(B) · Meaning: Conditional probability of A given B.
- Formula: P(A ∩ B) = P(A) × P(B\|A) · Meaning: Multiplication rule for joint probability.
- Formula: P(A ∩ B) = P(A) × P(B) · Meaning: Holds only when A and B are independent.
- Formula: P(A) = Σ P(Bᵢ) × P(A\|Bᵢ) · Meaning: Total probability over a partition.
- Formula: P(Bₖ\|A) = [P(Bₖ) × P(A\|Bₖ)] / P(A) · Meaning: Bayes' theorem for updating probabilities.
- Formula: E(X) = Σ xᵢ pᵢ · Meaning: Mean (expectation) of a discrete random variable.
- Formula: Var(X) = E(X²) − [E(X)]² · Meaning: Variance in shortcut form.
- Formula: P(X = r) = C(n, r) p^r q^(n − r), q = 1 − p · Meaning: Binomial probability for r successes in n trials.
- Formula: Mean of binomial = np; Variance = npq · Meaning: Quick results for the binomial distribution.
Worked examples
Example 1: Conditional probability
A box contains 5 red and 3 green balls. Two balls are drawn one after the other without replacement. Find the probability that the second ball is green given that the first ball was red.
Solution
After the first red ball is removed, the box has 4 red and 3 green balls, totalling 7 balls.
P(second is green | first is red) = 3 / 7.
Example 2: Bayes' theorem
A factory has two machines, M₁ and M₂. Machine M₁ produces 60 % of the items and M₂ produces 40 %. The defective rates are 2 % for M₁ and 5 % for M₂. An item is picked at random and found defective. What is the probability it came from M₂?
Solution
Let D denote "item is defective".
P(M₁) = 0.60, P(M₂) = 0.40. P(D|M₁) = 0.02, P(D|M₂) = 0.05.
Total probability of defect: P(D) = P(M₁) × P(D|M₁) + P(M₂) × P(D|M₂) = 0.60 × 0.02 + 0.40 × 0.05 = 0.012 + 0.020 = 0.032.
By Bayes' theorem: P(M₂|D) = [P(M₂) × P(D|M₂)] / P(D) = 0.020 / 0.032 = 5/8 = 0.625.
So the probability is 0.625 or 62.5 %.
Example 3: Binomial distribution
A fair coin is tossed 6 times. Find the probability of getting exactly 4 heads, and find the mean number of heads.
Solution
Here n = 6, p = 1/2, q = 1/2.
P(X = 4) = C(6, 4) × (1/2)⁴ × (1/2)² = 15 × 1/16 × 1/4 = 15 / 64.
Mean = np = 6 × (1/2) = 3 heads.
Common mistakes
- Using P(A) × P(B) when A and B are not independent → first check independence; if unsure, use the multiplication rule with conditional probability.
- Confusing P(A|B) with P(B|A) → read the condition after the vertical bar carefully; they are generally different.
- Forgetting that probabilities in a distribution must sum to 1 → always verify Σ pᵢ = 1 before computing mean or variance.
- Using the binomial formula when trials are not independent or when the probability changes each trial → binomial requires identical, independent trials.
- Computing variance as Σ xᵢ² pᵢ without subtracting [E(X)]² → remember Var(X) = E(X²) − [E(X)]².
Quick revision
- Conditional probability adjusts the sample space to the given condition.
- Bayes' theorem reverses conditioning: from P(effect|cause) to P(cause|effect).
- Independent events satisfy P(A ∩ B) = P(A) × P(B).
- Mean of a random variable is a weighted average of its values, weights being probabilities.
- Binomial distribution: n trials, constant p, P(X = r) = C(n, r) p^r (1 − p)^(n − r), mean = np, variance = np(1 − p).