Ch. 14 Resources

Chapter 14: From Randomness to Probability

The computations in Chapter 14 are fairly basic; in fact, you have already been doing very similar calculations, but without the new notation and terminology related to probability. Be sure to read the chapter carefully, and read the problems carefully. Make pictures (e.g. Venn diagrams) whenever you can and create tables listing the probabilities mentioned in a problem (if one isn't already given to you). The computations won't take a lot of time; figuring out which computations to perform will, so reserve plenty of time and be sure to work as many problems as possible.

Since there are no specialized calculator or Data Desk commands, we'll just work through a few brief examples here.

Web browsers

Many different Web browsers are available to computer users, but the two most commonly used are Internet Explorer and Firefox. When a Internet surfer visits a Web site, the Web server records various information about the browser being used, including the browser name and the computer's operating system (Windows XP, Mac, Linux, etc.). Web designers find it useful to know which browsers visitors to their sites are using, since different browsers render HTML and CSS differently (especially Internet Explorer, which does not comply with published Web standards).

Summary statistics for browser usage during December 2006 at one Web site can be found at http://www.w3schools.com/browsers/browsers_stats.asp and are also displayed in the following table:

browser	relative frequency
IE7	10.7%
IE6	45.3%
IE5	3.4%
Firefox	30.3%
other	10.3%

Of course, browser usage varies from site to site, as tech sites might be more likely to have visitors who use open source browsers like Firefox and open source operating systems like Linux. Let's use the w3schools.com information to answer some basic probability questions.

What's the probability that a randomly selected visit to this Web site used Firefox?

`P(mbox{Firefox}) = 0.303 = 30.3%`

What's the probability that a randomly selected visit did not use Firefox?

`P(mbox(not Firefox)) = P(mbox(IE7 or IE6 or IE5 or other)) = P(mbox(IE7)) + P(mbox(IE6)) + P(mbox(IE5))+ P(mbox(other)) = 0.107+0.453+0.034+0.103 = 0.697`

Notice that since the outcomes "IE7" and "IE6" and "IE5" and "other" are disjoint (someone can't be using more than one type of browser on a single visit to a Web site) we can employ the Addition Rule.

Note also that there's another way to solve this problem, using the Complement Rule:

`P(mbox{not Firefox}) = 1 - P(mbox{Firefox}) = 1-0.303 = 0.697`

If we randomly select three visits to this Web site, what's the probability that all of them used Firefox?

`P(mbox(FF and FF and FF)) = P(mbox(FF)) times P(mbox(FF)) times P(mbox(FF)) = 0.303 times 0.303 times 0.303 approx 0.028`

Notice here that there are three events: the first visit, the second visit and the third visit, so we rephrase the original statement using "and"; since the browser used by the first visitor is unlikely to affect the browser used by the second randomly selected visitor, the events appear to be independent, so we can use the Multiplication Rule. (We do have to be a bit careful here: if the visits were not randomly selected and the three visits in question all happened on a day when the Web site was mentioned on a blog frequented by Firefox users, the visits would not be independent.)

If we select four visits to this Web site at random, what's the probability that at least one of them used Firefox?

`P(mbox(at least one FF)) = 1 - P(mbox(no FF)) = 1- (P(mbox(not FF)))^4 = 1 - (1-0.303)^4 = 1 - (0.697)^4 approx 0.764`

Whenever we see the phrase "at least one" we should think of the complement ("none"). For example, in this case "at least one used Firefox" means "1 used Firefox or 2 used Firefox or 3 used Firefox or 4 used Firefox": the only alternative unaccounted for is "0 used Firefox" so the complement of "at least one" is "none"; we thus employ the Complement Rule in the first step. After that we note that there are four events (one for each randomly selected visit) and rephrase "none used Firefox" using "and"; then we employ the Multiplication Rule since (as before) it seems reasonable that the browser used during one of these four randomly selected visits does not affect the browser used during any of the other three. We then use the fact that P("not Firefox") = 0.697 (since the probability someone does use Firefox is 0.303) and complete the probability computation using straightforward arithmetic.

Homework

Work the following exercises in Chapter 14: 11, 13, 17–27 odd, 31 and 35–41 odd.

Errata

On page 372, the second equation in the For Example is missing a parenthesis: `P(mbox{red}) = P(mbox{not(green OR yellow)})`

The margin note on page 374 should read "eggs in that carton" (not "cartoon").

Part c of Exercise 1 should read: Flip a coin until you get a head or 3 consecutive tails; record the order of heads and/or tails that result.

The last sentence in the first paragraph of Exercise 17 should read "must be about 0.9%" (not "0.09%").

ActivStats

Work the activities on pages 14-1 and 14-2 in the ActivStats lesson book, as time permits.

Additional Resources

What Is Probability?

Episode 15 from Against All Odds features a discussion of some of these same topics, although some of the terminology used may be different.

Carnegie Mellon: Introduction to Statistics

Modules 5 and 6 of Carnegie Mellon's open source Introduction to Statistics course cover many of the same ideas.

Sofia: Elementary Statistics

Lessons 3.1, 3.2 and 3.3 of the Sofia Open Content Initiative's Elementary Statistics course include a discussion of probability terminology and rules.

Probabilities in the Game of Monopoly

An interesting exercise in computer simulations—plus, more than you ever wanted to know about the game of Monopoly!

M&M's

Color distribution of M&M's plain chocolate candies, as mentioned in Exercise 31.