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Chapter 16: Introducing Probability
The world, of course, isn’t that simple. In fact, probability problems that
live in statistics textbooks aren’t even that simple. Most of the time, sample
spaces are large and it’s not convenient to list every elementary outcome.
Take, for example, rolling a die twice. How many elementary outcomes are in
the sample space consisting of both tosses? You can sit down and list them,
but it’s better to reason it out: Six possibilities for the first toss, and each
of those six can pair up with six possibilities on the second. So the sample
space has 6 × 6 = 36 possible elementary outcomes. (This is similar to the
coin-and-die sample space in Table 16-1, where the sample space consists of
2 × 6 = 12 elementary outcomes. With 12 outcomes, it was easy to list them all
in a table. With 36 outcomes, it starts to get . . . well . . . dicey.)
Events often require some thought, too. What’s the probability of rolling a die
twice and totaling five? You have to count the number of ways the two tosses
can total five, and then divide by the number of elementary outcomes in the
sample space (36). You total a five by getting any of these pairs of tosses: 1
and 4, 2 and 3, 3 and 2, or 4 and 1. That’s four ways and they don’t overlap
(excuse me, intersect), so
Listing all the elementary outcomes for the sample space is often a night-
mare. Fortunately, shortcuts are available, as I show in the upcoming sub-
sections. Because each shortcut quickly helps you count a number of items,
another name for a shortcut is a counting rule.
Believe it or not, I just slipped one counting rule past you. A couple of para-
graphs ago, I say that in two tosses of a die you have a sample space of 6 × 6
= 36 possible outcomes. This is the product rule: If N
1
outcomes are possible
on the first trial of an experiment, and N
2
outcomes on the second trial, the
number of possible outcomes is N
1
N
2
. Each possible outcome on the first trial
can associate with all possible outcomes on the second. What about three
trials? That’s N
1
N
2
N
3
.
Now for a couple more counting rules.
Permutations
Suppose you have to arrange five objects into a sequence. How many ways
can you do that? For the first position in the sequence, you have five choices.
After you make that choice, you have four choices for the second position.
Then you have three choices for the third, two for the fourth, and one for the
fifth. The number of ways is (5)(4)(3)(2)(1) = 120.
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