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Unformatted text preview: WEEK 3 — Mon, Sept 13 Review of main idea from Section 25. For all events A1, A2 ,..., An P(A1UA2U...UAn)=1~P(EﬂZQm...mZn) Thus, for independent events we have the textbook’s P(A1k_JA2 U...UAn)=1"P(E)P(ZQ)"'P(Z«) (211) Exercise 236. Ans. 1  0.355 = 0.995 Exercise 238. Students work on this in class. Exercise 2—42. Ans. 1 — (0.10)(0.25)(0.40) = 0.99 Exercise 248. Ans. 1 — (0.04)(0.09)(0.20) = 0.99928 Exercise 2—50. Students work on this in class. I Expected New Coverage for Today. Ideas in Chapter 2, Sections 26 and 2—7. Section 26. COUNTING — PERMUTATIONS AND COMBINATIONS ' Consider sets 81 and 32 of cardinality m and 112, respectively. There are nlnz possible
ordered pairs (x1, x2) with x1 8 $1 and x2 a 8;. The product 111112 gives the number of
ordered arrangements of symbols with the ﬁrst symbol taken from set 81 and the second
taken from set Sz. This multiplication rule extends to more than two sets. It is the basis
for the following counting rules. The multiplication rule implies that in creating an ordered arrangement of 7' items selected
with replacement from a single set of n distinct elements, there are nr possibilities. It also
implies that in creating an ordered arrangement (permutation) of r items selected without replacement from a single set of n distinct items, there are 11!
HP, =n(n—l)(n—2)...(n—r+l)= (Fl—r)! possibilities. This is a product with n factors; it is referred to as the number of
permutations of :1 things taken 11 at a time. If r = n, “P, 2 n!, which is the number of ordered arrangements of n distinct objects. In creating an unordered arrangement (combination) of r items selected without
replacement from a single set of n distinct items, there are n P n! r ﬂ —rKn—rﬂ (2—13) possibilities. The second equality follows from the fact that “Pr = VIC,  r! , itself a consequence of the fact that each unordered arrangement of r elements gives rise to r! 0f
the permutations. Permutations. In Excel, typing =PERMUT(14,5) into a cell returns 14P5 = 240,240.
Using the TI, ﬁrst enter 14, then KP, from the MATH menu, then 5 and the evaluation 14P5 = 240,240 returns. This is the number of choices that Coach 1220 has to ﬁll the starting team in basketball
from a 14 person roster. There are 240,240 possibilities for a starting lineup (where a
person is assigned to be center, a person is assigned to be small forward, a person is
assigned to be power forward, a person is assigned to be shooting guard, and a person is
assigned to be point guard). The starting line—up (Roe, Sherman, Summers, Lucious,
Lucas) is different from the starting line—up (Lucious, Lucas, Roe, Summers, Sherman).
Order is relevant. Combinations. In Excel, typing =COMBIN(14,5) into a cell returns 14C5 = 2,002. Using
the TI, ﬁrst enter 14, then nC, from the MATH menu, then 5 and the evaluation 14C5 =
2,002 returns. Coach 1220 has this many choices of groups of 5 players to start the game Without regard
to order, here positions played. T here are MC, = 2,002 possibilities for a group of players to start the game without regard to the position on the ﬂoor. Here Coach 1220 may be
saying, “you ﬁve guys go play;” he has 2,002 possible groups of ﬁve to choose from for
the group that starts the game. The group (set) {Roe, Sherman, Summers, Lucious, Lucas}
is the same as the group {Lucious, Lucas, Roe, Summers, Sherman}. Order is not
relevant. I! Example 1. An engineer must select 5 sites to test from 12 sites and then go to them one
after the other. One possibility is (S4, Sg, 31, 87, S”) which denotes a trip that takes the
engineer first to S4, from there to Sg, from there to SI, from there to S7, from there to S“.
How many possible trips are there? Solution: HP, = 12  1 1  109  8 = 95,040 possible trips. If the engineer has to choose 5 sites
for future testing without regard to order, there are12 C5 2 12  1 1  10  9  8/51: 792 possible
combinations. C] Example 2. Parts are to be labeled with 3 letters chosen from a 26 letter alphabet
followed by 4 digits chosen from the digits 1 through 9. How many unique labels are there
if letters can be repeated and digits can be repeated in the labeling? How many unique
labels are there if letters cannot be repeated but digits can be repeated in the labeling? Solution: In the ﬁrst case, 26 26  26  9  9  99 = 115,316,136 possibilities, and, in the second
case, 26  25  24  9  9  9 ~ 9 = 102,351,600 possibilities. E It is easy to remember the formula fornﬂ. Think this way: there are n ways to choose the ﬁrst item to go into the permutation; given that choice, there are (n — 1) ways to select the
second item; given the two choices so far, there are (n — 2) ways to select the third item;
continuing to create the permutation of n ordered items we ﬁnd the total number of
possibilities to be “P, = n(n — 1)‘(n — 2)°(n — r + 1), which is a product with r factors. This
is an application of the multiplication rule for counting. The elements in a combination of r from n can be ordered in r! ways to produce a permutation, and it follows that CznPr_ n! n r r! —rl(n—r)l' Example 3. A syndicate gets lucky. A syndicate in Australia was organized to place
wagers in the lotteries around the world when situations became favorable. The payout
from a state lottery in the United States had grown to $7,000,000. This was a pick 6 from
41 lottery. There are only 41C6 = 4,496,388 possible combinations so covering them all at
a cost of $4,496,3 88 would guarantee a share of the jackpot. Since the State would not
accept a single payment of $4,496,388, the Syndicate used other means to buy as many
combinations as it could before the drawing. It succeeded in covering somewhat more
than half of the combinations. The Syndicate was “lucky” in that it had purchased the
winning combination and was the sole winner. E1 Exercise 252. Students work on this in class.
Exercise 254. Students work on this in class.
Exercise 256. Students work on this in class. Exercise 258. Ans. —6  i  :5  3 3 ' i — I (366) = 36C6 = 1,946,792. 36 35 34 33 32 31—1,946,792 Some fun examples. Example 7. The probability model of equally likely outcomes for the Michigan ThreeDigit
Lottery, results in P(all 3 digits are different) = 0.720. P(at least one repeated digit) = l 
0.720 = 0.280. Example 8. Consider the probability model of equally likely birthdays of the year for an
individual and independence of birthdays for individuals. Taking 365 days as a year: P(In a random sample of 23 individuals, all have different birthdays) = 0.4927. P(ln a random sample of 40 individuals, all have different birthdays) : 0.1088.
Complements : P(In a random sample of 23 individuals, at least two share the same birthday) I 0.5073. P(In a random sample of 40 individuals, at least two share the same birthday) = 0.8912. Tree Diagrams Tree diagrams are a useful way to depict What is going on in many situations. Example 4. Consider counting the number of permutations of 4 letters {a, b, c, d} taken 2 at
a time, that is, determining the value of 4P2 . Second Letter Permutation l l First Letter There are 43=12 paths through the tree starting at the left and moving to the right. 4P3 = 12. Example 5. Consider the tree shown below. Suppose a rat starts at the left and moves to the
right through the tree. Suppose that when confronted with choices of paths, the rat chooses
with equal probability. We work out the probabilities for each trip through the tree using the
multiplication rule. With this probability model, what is the probability that the rat gets some
cheese? I] Ans. (l/4)(1/2) + (1/4)(1/4) = 3/16 = 0.1875. First Choice Second Choice Trip 1 cl
c/—//_2 .02
\3 nc3
4 c4 d
\1 d1
\2 d2
3 d3 The trips are not egually likely! Example 6. On December 30, 1981, the Michigan Lottery selected ﬁve ﬁnalists from among
11,561 winners of $50 in the instant lottery game "Three Aces” for a $1,000,000 drawing to
be held later. I will describe the method used by the Lottery that evening to pick a single
ﬁnalist "at random". You will be instructed to determine the probabilities of certain
outcomes. The lottery used as a sampling frame the set of 11,561 ﬁvedigit numbers , s: {00001, 00002, 00003,...,11561} aﬁer having assigned one of these numbers to each of the 11,561 contestants. (Here I am
giving a simpliﬁed version of what in fact was the case to keep the analysis simple. In fact,
there were fewer than 11,561 winners in the pool for the selection of ﬁnalists, but S was used
as a frame. By using nowfamiliar devices that blow numbered ping pong balls around and allow that
one be drawn off at the push of a button, an element was selected from the frame S. The
Lottery ofﬁcials used the following procedure. First, a device with two balls labeled 0 and l
was used to select the leading digit in the ﬁve—digit number. If the outcome were 0, then a
device with ten balls labeled 0, 1,...,9 was used to select the second digit; if the ﬁrst outcome
were 1, then a device with two balls labeled 0 and 1 was used to select the second digit. mote
that there are no elements of S that start with the digits 12, 13, 14, 15, 16, 17, 18 or 19.) The
partial tree diagram shown gives some of the probability structure for the method used by the
Lottery to select a single ﬁnalist. Mr. X was assigned the frame element 00400. He was the person who raised the issue of
unequal probabilities of selection. Mr. Y was assigned the ﬂame element 11560. (a) Use the
multiplication law for probabilities to show that Mr. X is selected with probability 1/20,000
and Mr. Y is selected with probability 1/336. (b) Determine the probability that 00001 is
selected. (0) Determine the probability that 11098 is selected. D Em: F... .
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 Spring '08
 Anderson
 probability model

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