Decomposing a compound experiment into individual experiments, we can multiply the number of outcomes in each experiment.

1. Solutions to Worksheet for Section 1.1–1.2
The Basic Principle of Counting
V63.0233, Theory of Probability
June 29, 2009
1. At Pat’s Cheese Steaks in Philadelphia you can order a cheesesteak with or without saut´ed
e
onions, and with your choice of provolone cheese, american cheese, or Cheez Whiz. (In the native
parlance, a typical order sounds like “Whiz, wit’.”)
How many different ways can you order your cheesesteak?
Solution. We can organize our choices into a tree:
with without
an
an
e
z
e
z
on
hi
on
hi
ic
ic
W
W
ol
ol
er
er
ov
ov
m
m
Pr
Pr
A
A
There are six paths in this tree, so six choices.
2. Consider flipping a coin two times.
(a) How many possible sequences of flips are there?
(b) What percentage of the possibilities have two heads?
(c) What percentage of the possibilities have an even number of heads?
Solution. Again a tree works to enumerate the possibilities.
1

2. H T
H T H T
(a) There are four: HH, HT, TH, TT
(b) One out of four has two heads, so the percentage is 25%.
(c) The even possibilities are zero or two heads. Two sequences have this property, so the total
percentage is 2/4 = 50%.
3. Consider flipping a coin four times.
(a) How many possible sequences of flips are there?
(b) What percentage of the possibilities have [exactly] two heads?
(c) What percentage of the possibilities have an even number of heads?
Solution. (a) We can create a tree with four levels, two branches at each node (it’s a little big to
put here). There are 16 paths, or leaves, on this tree.
(b) One possible way to see this is to use a part of the full tree. We’ll stop developing the tree
after two heads are on the path already.
H T
H T H T
H T H T H T
H T H T H T H T
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3. From there there’s only one way to continue the tree and still get two heads–the rest have to
be tails. There are therefore six sequences with two heads: HHTT, HTHT, HTTH, THHT,
THTH, TTHH. This means the probability of getting two heads is
6 3
= = 37.5%
16 8
If we mean “two heads” to mean “at least two heads” then we need to include five more
sequences with three or four heads: HHHT, HHTH, HTHH, THHH, HHHH. So the probability
11
of at least two heads is = 68.75%.
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(c) We need to add to the event of exactly two heads the event of getting zero heads or four heads.
In each of these there is only one possible outcome: TTTT or HHHH. Hence the possibility
of an even number of heads is
8
= 50%
16
4. In Indiana from 1963 through 2008, license plates were coded by the following scheme:
• The number of the county the licensee resides in, arranged alphabetically from 1 to 93
• a letter
• a four-digit number
An example might be “82 a5713”. How many license plates may be printed according to this
scheme?
Solution. The first choice is a number from 1 to 93 (93 alternatives), the second a letter (26
alternatives), and the third, fourth, fifth, and sixth a digit between 0 and 9. This makes 93 · 26 · 104
choices.
5. Since 2008, Indiana license plates are coded by a three-digit number and from one to three
letters. How many license plates are possible in this scheme?
Solution.
9 · 10 · 10 · (26 + 26 · 26 + 26 · 26 · 26) = 16, 450, 200
6. The current format of area codes is three digits, where
• The first digit can be any number but 0 or 1
• The second digit can be any number between 0 and 8
• The third digit can be any number at all, except that the last two digits cannot both be 1
So 781 is a valid area code but not 187 or 411. How many possible area codes are there?
3

4. Solution. We multiply the number of alternatives at each step, we get
8 × 9 × 10 = 720
anything but 0 or 1 0–8 0–9
But we need to eliminate the sequences 211, 311, 411, . . . , 911, which are either special or reserved.
So we subtract these eight and get 712 area codes.
If we want to count without subtraction, we count separately the cases where the middle digit
is or is not 1. If the middle digit is not 1, we have
8 × 8 × 10 = 640
2–9 0 or 2–8 0–9
choices. If the middle digit is 1, we have
8 × 1 × 9 = 72
2–9 1 0 or 2–9
Add these together and we get the same thing.
See http://en.wikipedia.org/wiki/North_American_Numbering_Plan
7. In the days of rotary telephones, area codes were restricted to a different format:
• The first digit could be any number but 0 or 1
• The second digit had to be 0 or 1
• The third digit could be from 1 to 9 if the second digit was 0, and from 2 to 9 if the second
digit was 1.
So 617 is a valid area code, but not 781 (back then), nor 411. This was to make dialing easier and
to give the most populous areas the smallest number of total clicks. How many possible area codes
were there then?
Solution. To count alternatives by multiplying, we have to break it up into two cases. If the second
digit is zero, then the choices are:
8 × 1 × 9 = 72
2–9 0 1–9
If the second digit is one, we have:
8 × 1 × 9 = 64
2–9 1 2–9
This makes altogether 136 area codes.
8. Packing his belongings to go to college, a student has to decide what to do about his CD
collection. In how many different ways can he take a long at least one of his 10 favorite CDs?
Solution. Each CD represents an independent “experiment:” the student can either take it or not.
That means there 210 = 1024 total ways the student can make a selection of CDs. We eliminate
the one selection of no CDs, so the total number of selections of at least one CD is 1023.
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