Download Chapter 7: Section 7-5 Applications of Counting Principles

Chapter 7: Section 7-5
Applications of Counting Principles
D. S. Malik
Creighton University, Omaha, NE
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Example
An experiment consists of forming license plates consisting of three letters
followed by three digits, with repetition allowed. Let S be the sample
space of the experiment. Then
n (S ) = 26 26 26 10 10 10 = 17, 576, 000.
Let E be the event that a license plate begins with the letter A. Let us
determine the probability of E . To accomplish this …rst we …nd the
number of elements in E . Now the number of license plates that begin
with the letter A are
1 26 26 10 10 10 = 676, 000.
Thus, n (E ) = 676, 000. Hence,
Pr[E ] =
1 26 26 10 10 10
1
n (E )
=
= .
n (S )
26 26 26 10 10 10
26
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Example
A bag contains 5 red, 6 blue, and 4 brown marbles. An experiment
consists of drawing three marbles at random.
The total number of marbles is 5 + 6 + 4 = 15. So the number of ways to
select 3 marbles from 15 marbles is
C (15, 3) =
15!
15!
=
= 455.
3! (15 3)!
3! 12!
Hence, n (S ) = 455.
We …nd the probability that 2 are red and 1 is blue. There are 5 red
marbles. The number of ways to select 2 red marbles is C (5, 2). Similarly,
the number of ways to select 1 blue marble is C (6, 1). Thus, number of
ways to select 2 red and 1 blue marbles is: C (5, 2) C (6, 1) = 10 6 = 60.
So the probability of drawing 3 marbles so that 2 are red and 1 is blue is
C (5, 2) C (6, 1)
60
12
=
= .
C (15, 3)
455
91
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Example
Two cards are drawn from a well-shu- ed deck of 52 cards. Note that the
number of elements in the sample space is
C (52, 2) = 1326.
We …nd the probability that exactly one of the cards is a king.
There 4 kings and 48 cards are not kings. So we must select 1 card from 4
kings and 1 card from the remaining 48 non-king cards. Thus, the number
of ways to select two cards so that one is a king is
C (4, 1) C (48, 1) = 4 48 = 192.
Hence, the probability that exactly one of the cards is a king is
192
32
C (4, 1) C (48, 1)
=
=
.
C (52, 2)
1326
221
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Probabilities and Tree Diagrams
Two cards are drawn at random from a deck of 52 cards. What is the
probability that exactly one of the cards is a king?
Drawing two cards from a deck of 52 cards can be thought of as
drawing two cards in sequence without replacement. So this can be
considered a two step experiment. This means that we can draw a
tree diagram of this experiment as shown in the following diagram:
Step 2
Step 1
Outcome
K
K
K = KK
K’
K
K’ = KK’
K
K’ K = K’K
K’
K’
K
K’
K’ = K’K’
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Note that, in the previous …gure, at the end of each branch we write
the outcome of the branch.
After step 2, we draw straight lines and at the end of each straight
line, we write the outcome of the path.
For example, in the top row, the path of the two branches leads to
the outcome K \ K , which we sometimes write as KK . That is, the
…rst outcome is a king and the second outcome is also a king.
At the end of the second straight line (the second row), the outcome
is K 0 \ K = K 0 K , i.e., the …rst card is not a king and the second card
is a king.
Similar conventions apply for the remaining branches.
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Next in the tree diagram, on each branch we write the probability of
that outcome, see the following …gure:
Step 2
Step 1
4/52
48/52
Outcome
3/51
K
K
K = KK
48/51
K’
K
K’ = KK’
4/51
K
K’ K = K’K
47/51
K’
K’
K
K’
K’ = K’K’
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Note that the probability of the second king, if the …rst is also a king,
3
is 51
because after drawing the …rst king, 3 kings are left and a total
of 51 cards are left.
Finally, to complete the tree diagram, on each straight line, we write
the multiplication of all of the probabilities of the branches leading to
that straight line.
For example, in the very top row, the probabilities of the branches are
3
4
52 and 51 . So on the top straight line, we write
4 3
1
=
.
52 51
221
This is the probability that the …rst card is a king and the second card
is also a king, i.e., both of the cards are kings. That is,
Pr[Both kings] = Pr[KK ] =
1
.
221
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After writing all of the probabilities on the straight line, we obtain the
following tree diagram:
Outcome
Step 2
Step 1 3/51
4/52
K
(4/52)(3/51) = 1/221
K
K = KK
K
K’ = KK’
K
48/52
48/51
K’ (4/52)(48/51) = 16/221
4/51
K
47/51
K’
(48/52)(4/51) = 16/221
K’ K = K’K
K’
(48/52)(47/51) = 188/221
K’
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K’ = K’K’
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Exercise: At a local university, 80% of the …rst year students live on
campus. Of those who live on campus, 75% eat at the
student center cafeteria. Of those who do not live on
campus, 60% eat at the student center cafeteria. Find the
following probabilities of a randomly selected …rst year
student at the university.
(a) The …rst year student lives on campus and does not eat
at the cafeteria.
(b) The …rst year student eats at the cafeteria.
Solution: Let L be the set of …rst year students who live on cam
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Solution: Let L be the set of …rst year students who live on campus
and E be the set of students who eat at the cafeteria. The
tree diagram of this problem is:
Outcome
Step 2
0.80
0.20
Step 1 0.75
L
0.25
E’
0.60
E
0.40
E’
E
L’
(0.80)(0.75) = 0.60
L
E = LE
L
E ’= LE’
(0.20)(0.60) = 0.12
L’
E = L’E
(0.20)(0.40) = 0.08
L’
E’ = L’E’
(0.80)(0.25) = 0.20
(a) The probability that a randomly selected …rst year
student lives on campus and does not eat at the cafeteria is
Pr[L \ E 0 ] = (0.80)(0.25) = 0.20.
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(b) The probability that a randomly selected …rst year
student eats at the cafeteria is Pr[E ]. From the …gure,
Pr[E ] = Pr[fLE , L0 E g] = Pr[LE ] + Pr[L0 E ]
= 0.60 + 0.12 = 0.72.
Note that Pr[E ] is the probability of students who live on
campus and eat at the cafeteria or students who do not live
on campus but eat at the cafeteria.
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Exercise: Find the probability of randomly drawing 5 cards from a
well-shu- ed deck of 52 cards so that 3 are diamonds and 2
are hearts.
Solution: The number of ways to select 5 cards such that 3 are
diamonds and 2 are hearts is
C (13, 3) C (13, 2).
Also, the number of ways to select 5 cards is C (52, 5).
Hence, the probability of randomly drawing 5 cards such that
3 are diamonds and 2 are hearts is
C (13, 3) C (13, 2)
286 78
1859
=
=
.
C (52, 5)
2598960
216, 580
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Exercise: Find the probability of randomly selecting 4 marbles from a
bag of 5 red and 3 blue marbles such that all are blue.
Solution: Note that we are selecting 4 marbles and the bag has only 3
blue marbles. Thus, the number of ways to select 4 marbles
so that all are blue is 0. Hence, the probability of randomly
selecting 4 marbles from a bag of 5 red and 3 blue marbles
such that all are blue is 0.
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Exercise: Consider an experiment in which 2 marbles are drawn at
random from an urn containing 8 red, 6 blue, 4 green, and 2
white marbles. Determine the probability that both marbles
are red.
Solution: The total number of marbles is 8 + 6 + 4 + 2 = 20. The
number of ways to select 2 marbles from 20 marbles is
C (20, 2) = 190.
The number of ways to select 2 marbles such that both are
red is C (8, 2) = 28. Hence, the probability of selecting two
marbles such that both are red is
14
28
= .
190
95
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Exercise: Two fair dice are rolled. Find the probability such that the
sum of the numbers rolled is 9.
Solution: Note that the number of elements in the sample space is 36.
Let E be the event that the sum of the numbers rolled is 9.
Then E = f(3, 6), (4, 5), (5, 4), (6, 3)g. So n (E ) = 4.
Hence, the probability that the sum of the numbers rolled is
9 is
1
4
= .
36
9
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Exercise: If four distinct coins are tossed, …nd the probability of
getting exactly three heads.
Solution: Because the coin is tossed 4 times, the total number of
outcomes is 24 = 16. The outcomes that contains exactly 3
heads is HHHT , HTHH, HHTH, and THHH. Hence, the
probability is
1
4
= .
16
4
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Exercise: Samantha’s piggy bank contains at most $100.00. What is
the probability that the piggy bank contains an exact dollar
amount, such as $15.00 and $68.00?
Solution: Let S be the sample space. Then S can contain any number
between 0.00 and 100.00. Hence n (S ) = 10001. (Note that
an amount such as 25.68 is an element of S.) Let E be the
event that piggy bank contains an exact dollar amount.
Then n (E ) = 101. Hence, the probability that the piggy
bank contains an exact dollar amount is
101
.
10001
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Exercise: A bag contains 5 dimes, and 3 nickel. Find the probability of
selecting 3 coins so that the total value of the coins is 25
cents.
Solution: We are selecting 3 coins and the total value of the coins is
25 cents. So we must select 2 dime from 5 dimes, and 1
nickel from 3 nickels. This can be done in
C (5, 2) C (3, 1) = 10 3 = 30 ways.
Also the number of ways to select 3 coins from 5 + 3 = 8
coins is C (8, 3) = 56. Hence, the probability of selecting 3
coins so that the total value of the coins is 25 cents is
15
30
= .
56
28
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