TC3 → Stan Brown → Statistics → Ch 5 Guide
revised 2 Oct 2011 (What’s New?)

Guide to Chapter 5

Copyright © 2008–2012 by Stan Brown, Oak Road Systems

This is your guide to what’s important in the chapter, with comments on some things that the chapter leaves out or doesn’t explain well. Page numbers refer to Sullivan, Michael, Fundamentals of Statistics 3/e (Pearson Prentice Hall, 2011), which is equivalent to the “second custom edition” for TC3.

Always check Corrections to Sullivan’s Fundamentals of Statistics, 3rd Edition for known mistakes in the textbook.

Agenda: 

Pictures:  page 228: rolls of two dice; page 240: 52 cards in the standard deck

Announce:  Shoe-Size Lab is due next week and will use material from Chapter 4 only. Staple it before you come to class.

5.1  Probability Rules

223 2 definitions of probability: likelihood, and long-term relative frequency

Your book gives examples, but doesn’t really make interpretation of probability explicit until page 330. It’s not hard: just remember probability of one = proportion of all. If an event has probability 0.28, then you can say it has a 28% likelihood of occurring on any one trial, or you can say that it will occur in the long run on about 28% of possible occasions. Which of those you select depends on the particular situation.

Example: say the probability that your bus is late is 80%. Interpretation: there’s an 80% chance the bus will be late on any particular occasion (probability of one), or the bus is late about 80% of the time in the long run (proportion of all). In this case the second interpretation is probably more natural unless you’re standaing at the bus stop waiting to get to work.

223–4The Law of Large Numbers ties into the idea of probability as long-term relative frequency. (Some people call it the law of averages.)

(not in book) The Gambler’s Fallacy is the wrong idea that things are more likely to even out on the next try.

Example: An honest roulette wheel once got twenty-six black in a row (1913, Monaco).

Example: a fair coin is not “due for a tail” after a run of heads. “The surprising length and frequency of consecutive runs of heads or tails is yet another lesson of penny flipping. If Henry and Thomas were to continue flipping pennies once a day, then there's a better-than-even chance that within about two months Henry will have won at least five flips in a row, as will Tommy. If they continue flipping for six years, there’s a better-than-even chance that each will have won at least ten flips in a row.”   —John Allen Paulos, A Mathematician Plays the Stock Market

224 definitions: experiment, outcome, event, sample space

225 2 rules: probabilities are between 0 and 1, add up to 1

what does it mean if P(A) is closer to 1 than P(B)? closer to 0?

225 def: probability model (We’ll do much more with them in Chapter 6.)

p = n% means the event happens about n times out of 100

def: unusual event (usually p < 0.05) — can you give examples?

226 empirical a/k/a experimental probability gives approx probabilities only

227 classical method requires equally likely outcomes but gives exact probabilities

229–30 Example 7 shows using a tree diagram for to construct sample space — more compact alternatives exist

231–2 Example 8 is simplified but realistic. In business and science, simulations are now often used to answer questions about probabilities. They require much more sophisticated random-number generators than you’ll find in your calculator.

233 try review #1–8

5.2  Addition Rule and Complements

238 def: disjoint a/k/a mutually exclusive

addition rule / Venn diagrams

239 Benford’s Law is just an illustration; you won’t be working problems with it in our course.

interesting factoid (no need to memorize): P(D) = log(1 + 1/D)

If interested in how it’s used to catch cheating, see optional extras for Chapter 5 on Web page.

240 Work Example 2.

Skip objective 2 (general addition rule). Just know when you can and cannot use the addition rule for disjoint events.

242–3 Get used to contingency tables a/k/a 2-way tables. Sometimes you’ll be given row and column totals; sometimes you’ll have to compute them. Make sure row totals and column totals add to the same grand total.

Work Example 4, but skip (d).

243–5 The Complement Rule  is your friend when computations seem too laborious, especially for “at least”-type problems. Work Example 6 carefully. (Caution: the next-to-last number should be 3891, not 3981.)

245 try review 1–4, and page 246 #29

5.3  Multiplication Rule

250 def: independent events; difference from disjoint events

251 multiplication rule for independent events

Work Example 2.

not in book: If sample size is very small relative to the population, you can assume independence and your probability will be approximately correct.

253 “at least” — be on the lookout for complements here and elsewhere

The Challenger space shuttle was secured by six gaskets. If any one failed, the shuttle would explode (as actually happened). Assuming each gasket had a 97% reliability rate and that they were independent, what is the probability of failure for the mission?   (answer at end of document)

254–5 try review 1–6, 22; plus page 263 #29 and page 264 #28

Skip sections 5.4–5.6

Caution: I want you always to think about when a technique that you have learned does not apply. Therefore, the homework includes one or more problems that we have not learned to solve, because the solution techniques are in parts of the chapter that we skipped. For such problems, just write down briefly the reason you can’t solve it: probability “or” for non-disjoint events, or probability “and” for non-independent events.

Answer to Shuttle Question

The mission fails if one or more gaskets fail — this is a typical “at least” problem. Rather than compute the probability of one failing, two failing, and so forth, compute the complement of what you want. In other words, compute the probability that all six gaskets hold, and the mission succeeds.

If each gasket has a 97% reliability rate, and they’re independent, then the probability that all six hold is (.97)(.97)(.97)(.97)(.97)(.97) = (.97)^6 = about 0.83. The probability that one or more fail is therefore about 1−0.83 = about 0.17 or about 17%. (This is almost exactly the odds of taking a bullet in Russian roulette.)

What’s New


This page is used in instruction at Tompkins Cortland Community College in Dryden, New York; it’s not an official statement of the College. Please visit www.tc3.edu/instruct/sbrown/ to report errors or ask to copy it.

For updates and new info, go to http://www.tc3.edu/instruct/sbrown/stat/