The Language of Argument

2 4 9

S o m e ru l e s o f P r o b a b i l i t y

half the class is female and exactly half the class is over nineteen and the age

distribution is the same for females and males. What is the probability that a

randomly selected student will be either a female or over nineteen? If we simply

add the probabilities (1/2 1 1/2 5 1), we would get the result that we are cer-

tain to pick someone who is either female or over nineteen. But that answer is

wrong, because a quarter of the class is male and not over nineteen, and one of

them might have been randomly selected. The correct answer is that the chances

are 3/4 of randomly selecting someone who is either female or over nineteen.

We can see that this is the correct answer by examining the following

table:

Over Nineteen Not over Nineteen

Female 25% 25%

Male 25% 25%

It is easy to see that in 75 percent of the cases, a randomly selected student

will be either female or over nineteen. The table also shows what went

wrong with our initial calculation. The top row shows that 50 percent of

the students are female. The left column shows that 50 percent of the stu-

dents are over nineteen. But we cannot simply add these figures to get the

probability of a randomly selected student being either female or over nine-

teen. Why? Because that would double-count the females over nineteen. We

would count them once in the top row and then again in the left column. To

compensate for such double-counting, we need to subtract the students who

are both female and over nineteen. The upper left figure shows that this is

25%. So the correct way to calculate the answer is 50% 1 50% 2 25% 5 75%.

This pattern is reflected in the general rule governing the calculation of

disjunctive probabilities:

Rule 3G: Disjunction in General. The probability that at least one of

two events will occur is the sum of the probabilities that each

of them will occur, minus the probability that they will both

occur. Symbolically:

Pr(h 1 or h 2 ) 5 Pr(h 1 ) 1 Pr(h 2 ) 2 Pr(h 1 & h 2 )

If h 1 and h 2 are mutually exclusive, then Pr(h 1 & h 2 ) 5 0, and Rule 3G reduces

to Rule 3. Thus, as with Rules 2 and 2G, Rule 3 is simply a special case of the

more general Rule 3G.

Probabilities in a Series

Before stating Rule 4, we can think about a particular example. What is the

probability of tossing heads at least once in eight tosses of a coin? Here it

is tempting to reason in the following way: There is a 50 percent chance

of getting heads on the first toss and a 50 percent chance of getting heads

on the second toss, so after two tosses it is already certain that we will toss

97364_ch11_ptg01_239-262.indd 249 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.