The Language of Argument

2 5 4

C HaP Te r 1 1 ■ C h a n c e s

includes the true positives plus the false positives, which are those who test

positive but do not have colon cancer. Given the stipulated probabilities,

in a normal population of 100,000 people, there will be 270 true positives

(100,000 3 0.003 3 0.9) and 2,991 false positives [(100,000 – 300) 3 0.03].

Thus, the probability that Wendy has colon cancer is about 270/(270 1

2,991). That is only about 8.3 percent, when most people estimate above

50 percent!

Why do people, including doctors, overestimate these probabilities so

badly? Part of the answer seems to be that they focus on the rate of true posi-

tives (90 percent) and forget that, because there are so many people without

colon cancer (99.7 percent of the total population), even a small rate of false

positives (3 percent) will yield a large number of false positives (2,991) that

swamps the much smaller number of true positives (270). (When the ques-

tion about probability was reformulated in terms of the number of people in

each group, most doctors come up with the correct answer.) For whatever

reason, people have a strong tendency to make mistakes in cases like these,

so we need to be careful, especially when so much is at stake.

One way to calculate probabilities like these uses a famous theorem that

was first presented by an English clergyman named Thomas Bayes (1702–

1761). A simple proof of this theorem applies the laws of probability from

the preceding section. We want to figure out Pr(h|e), that is, the probability

of the hypothesis h (for example, Wendy has colon cancer), given the evi-

dence e (for example, Wendy tested positive for colon cancer). To get there,

we start from Rule 2G:

1. Pr(e & h) 5 Pr(e) 3 Pr(h|e)
   Dividing both sides by Pr(e) gives us:


2. Pr 1 h^0 e 25

Pr 1 e & h 2

Pr 1 e 2

If two formulas are logically equivalent, they must have the same probabil-

ity. We can establish by truth tables (as in Chapter 6) that “e” is logically

equivalent to “(e & h) ∨ (e & ~h).“ Consequently, we may replace “e” in the

denominator of item 2 with “(e & h) ∨ (e & ~h)” to get:

3. Pr 1 h^0 e 25

Pr 1 e & h 2

Pr^31 e & h 2 ∨ 1 e & ~h 24

Since “e & h” and “e & ~h” are mutually exclusive, we can apply Rule 3 to

the denominator of item 3 to get:

4. Pr 1 h^0 e 25

Pr 1 e & h 2

Pr 1 e & h 21 Pr 1 e & ~h 2

97364_ch11_ptg01_239-262.indd 254 15/11/13 10:58 AM

some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materiallyCopyright 201^3 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights,

affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.