Introduction to Statistics (STAT 101) · Module 4 of 7
Probability & Random Variables
Probability rules, conditional probability and independence, expected value, and binomial distributions.
Probability Rules
- Complement: P(not A) = 1 − P(A). Addition: P(A or B) = P(A) + P(B) − P(A and B).
- Conditional probability: P(A | B) = P(A and B) / P(B). Events are independent if P(A | B) = P(A); then P(A and B) = P(A)·P(B).
- Mutually exclusive (disjoint) events can't happen together, so P(A and B) = 0. Disjoint events with nonzero probabilities are never independent.
P(A) = 0.5, P(B) = 0.4, and P(A and B) = 0.2. Find P(A or B), and decide whether A and B are independent.
Solution. P(A or B) = 0.5 + 0.4 − 0.2 = 0.7. P(A | B) = 0.2/0.4 = 0.5 = P(A), so A and B are independent.
Random Variables & Expected Value
- A random variable's expected value (mean) is E(X) = Σ x·P(x), its long-run average.
- For independent random variables, means add and VARIANCES add (never standard deviations): SD(X ± Y) = √(SD(X)² + SD(Y)²).
A game pays $10 with probability 0.2 and costs you $2 with probability 0.8. What is the expected value?
Solution. E = 10(0.2) + (−2)(0.8) = 2.0 − 1.6 = $0.40 per play, on average.
Binomial Distributions
- Binomial setting (BINS): Binary outcomes, Independent trials, a fixed Number of trials n, and the Same probability of success p.
- P(X = k) = C(n, k)·pᵏ(1 − p)ⁿ⁻ᵏ. Mean = np; SD = √(np(1 − p)).
A fair coin is flipped 10 times. What is the probability of exactly 3 heads?
Solution. C(10, 3)(0.5)³(0.5)⁷ = 120 × (0.5)¹⁰ = 120/1024 ≈ 0.117.
Common mistake: Standard deviations do NOT add. To combine independent random variables, add the variances, then take the square root.
Check yourself
2 of this module's 5 quiz questions. Pick your answers, then check them in the app, with an explanation for every question.
1. P(A) = 0.3 and P(B) = 0.5, and A and B are independent. P(A and B) is:
2. If A and B are mutually exclusive with P(A) = 0.2 and P(B) = 0.3, P(A or B) is: