A brief note on Independence, Trees and Streaks . Read it before a test. Try each question at the end before you open its answer.
Independent events Meaning: one event does not change the chance of the other. Multiplication rule: P ( A ∩ B ) = P ( A ) × P ( B ) P(A \cap B) = P(A) \times P(B) P ( A ∩ B ) = P ( A ) × P ( B ) for independent events. At least one: P ( at least one ) = 1 − P ( none ) P(\text{at least one}) = 1 - P(\text{none}) P ( at least one ) = 1 − P ( none ) . Not independent: weather on two days, two cards from one pack, two facts about one die. Trees and replacement Tree: branches from a point add up to 1 1 1 ; multiply along a path; add the paths. Without replacement: the second branches use the counts that are left. Big population: draws without replacement are almost independent. Streaks Gambler's fallacy: after a streak, the next independent result is not "due". Every sequence: n n n fair tosses give each particular sequence the probability ( 1 2 ) n \left(\dfrac{1}{2}\right)^n ( 2 1 ) n . Long runs are normal: in 100 100 100 fair tosses, a run of 5 5 5 or more equal faces is very likely. A worked example Two made-up machines work independently, with probabilities 0.8 0.8 0.8 and 0.6 0.6 0.6 .
Both work. 0.8 × 0.6 = 0.48 0.8 \times 0.6 = 0.48 0.8 × 0.6 = 0.48 . At least one works. Both fail: 0.2 × 0.4 = 0.08 0.2 \times 0.4 = 0.08 0.2 × 0.4 = 0.08 . So 1 − 0.08 = 0.92 1 - 0.08 = 0.92 1 − 0.08 = 0.92 . Exactly one works. 0.92 − 0.48 = 0.44 0.92 - 0.48 = 0.44 0.92 − 0.48 = 0.44 . Practice with new numbers
A program to try The program checks the worked example, and then finds the chance that two cards from one pack are both red.
Run it. Why is the chance for the cards not exactly 1 4 \dfrac{1}{4} 4 1 ?
1. Change b = Fraction(6, 10) to b = Fraction(1, 2). What is the chance that at least one machine works now?
Mistakes to avoid Multiplying linked events. Check independence first. Adding along a path. Multiply along it. Thinking a result is due. Independent tries have no memory. Forgetting an order. "Exactly one" often has two paths. Check yourself Two independent events have probabilities 0.3 0.3 0.3 and 0.6 0.6 0.6 . Find the probability that both happen. Stuck? Show a hint Show the answerA bag holds 3 3 3 blue and 5 5 5 green counters. Two are taken without replacement. Find the probability that both are blue. Stuck? Show a hint Show the answerA fair die is rolled 3 3 3 times. Find the probability of at least one six. Stuck? Show a hint Show the answerAfter 4 4 4 heads in a row, what is the chance that a fair coin lands tails? Stuck? Show a hint Show the answerSources OpenStax, Introductory Statistics , Chapter 3, the sections on independent events and on tree diagrams.