Chapter 5 Lecture Notes

Reminder: Shoe-Size Lab

5.1  Probability Rules (objectives 1–3 only)

221 motivation for this chapter

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

Law of Large Numbers

probability of one = proportion of all — we’ll deal with that in depth on page 324

(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 defs: 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

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 values only

227 classical method gives exact values

229 SRS — every individual has equal chance of being chosen, ∴classical method computes prob of getting a particular sample

229–30 Ex 7; tree diagram for sample space — show a more compact alternative

233 try review #1–8

5.2  Addition Rule

238 def: disjoint a/k/a mutually exclusive

addition rule / Venn diagrams

239 Benford’s law (no need to memorize): P(D) = log(1 + 1/D)

241–2 general addition rule (issue of double counting)

Example 3

242–3 contingency table — be careful with your additions!

244–5 Complement Rule — Ex 6abc

245 try review 1–4

5.3  Multiplication Rule

250 def: independent events; difference from disjoint events

251 multiplication rule for independent events

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 the document)

254 try review 1–6, 24

5.4  Conditional Probability

256 notation P(F | E)

256–7 Example 1

257 learn formula: P(F | E) = P(E and F) / P(E)

With a 2-way table you don’t really need formulas. Just make sure to use the correct fraction denominators. Example: 80 cars on a lot, including large and compact, foreign and domestic. Suppose there are 38 foreign, 22 foreign compact, and 50 compact. Find these probabilities with symbols and numbers. Making a table first will help!

(a) ... of randomly selecting a large domestic car

(b) ... of randomly selecting a large or domestic car

(c) ... of the car being large if you know that it’s domestic

(d) ... of the car being domestic if you know that it’s large

(e) ... of randomly selecting a car that is large and domestic

(f) ... of a randomly selected domestic car being large

Answers are at the end of this document.

258 Example 3; general multiplication rule (transform of conditional formula)

Example: 8 molasses cookies and 12 chocolate chip in a bag. You draw two at random. What’s the probability you’ll get one of each? Hint: you have to consider molasses then choco-chip and choco-chip then molasses.

259–260 Examples 5–6

Usually “with replacement” means independence and “without replacement” means dependence. But assume independence, even without replacement, for samples <5% of a large population.

261 restates def of independence as formula

Example (from John Allen Paulos, A Mathematician plays the Stock Market):

Early in the fall 2002 Washington DC sniper case, ... police arrested a man who owned a white van, a number of rifles, and a manual for snipers. It was thought at the time that there was one sniper and that he owned all these items, so ... let’s assume that this turned out to be true. ...

There are about 4 million innocent people in the suburban Washington area and, we’re assuming, one guilty one. Let’s further estimate that ten people (including the guilty one) own all three of the items mentioned above.

Which is higher: the probability that an innocent man would own all these items, or the probability that a man who owned all these items would be innocent?

[answer at end of this document]

261 review 1–2

5.5  Counting Techniques (objective 1 only)

267 multiplication rule, Examples 2–3

268 factorials, Example 4

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.)

Answers to Cars Questions

When you have two attributes per individual, a table can often help you. It doesn’t matter which are rows and which are columns: if you do the table carefully you’ll get the same probabilities either way.

Pay attention to the bottoms of the fractions: with most probabilities the bottom will be the grand total (sample size), but with a conditional probability the bottom number is the row or column total for the “given-that” condition.

(a) P(Large and Domestic) = 14/80 or 7/40 (17.5%)

(b) P(Large or Domestic) = (30+42−14)/80 = 58/80 (72.5%)

(c) P(Large | Domestic) = 14/42 = 1/3 (about 33.3%)

(d) P(Domestic | Large) = 14/30 = 7/15 (about 46.7%)

(e) same as (a) — don’t be confused by different wording

(f) same as (c), not (d) — don’t be confused by different wording

Answer to Sniper Question

“The first probability — that an innocent man owns all these items — would be 9/4,000,000 or 1 in 400,000. The second probability — that a man owning all three of these items is innocent — would be 9/10. Whatever the actual numbers, these probabilities usually differ substantially. Confusing them is dangerous (to defendants).”

Yes, an innocent man is quite unlikely to own these items. But someone who owns them is still quite likely to be innocent. In statistical terms, P(owns | innocent) is very low, but P(innocent | owns) is high.

Another example: Medical False Positives and False Negatives