Tan S.T. Finite Mathematics for the Managerial, Life, and Social Sciences
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Solution
a. Each outcome of the experiment may be represented as a sequence of heads and
tails. Using the generalized multiplication principle, we see that the number of
outcomes of this experiment is given by 2
6
, or 64. Let E denote the event that the
coin lands heads exactly three times. Since there are C(6, 3) ways this can occur,
we see that the required probability is
b. Let F denote the event that the coin lands heads at most three times. Then n(F) is
given by the sum of the number of ways the coin lands heads zero times (no
heads!), the number of ways it lands heads exactly once, the number of ways it
lands heads exactly twice, and the number of ways it lands heads exactly three
times. That is,
Therefore, the required probability is
c. Let F denote the event that the coin lands heads on the first and the last toss. Then
n(F) 1
2
2
2
2
1 2
4
, so the probability that this event occurs is
EXAMPLE 2
Two cards are selected at random (without replacement) from a
well-shuffled deck of 52 playing cards. What is the probability that
a. They are both aces? b. Neither of them is an ace?
Solution
a. The experiment consists of selecting 2 cards from a pack of 52 playing cards.
Since the order in which the cards are selected is immaterial, the sample points
are combinations of 52 cards taken 2 at a time. Now there are C(52, 2) ways of
selecting 52 cards taken 2 at a time, so the number of elements in the sample
space S is given by C(52, 2). Next, we observe that there are C(4, 2) ways of
selecting 2 aces from the 4 in the deck. Therefore, if E denotes the event that the
cards selected are both aces, then
1
4
1
2
2
P1F 2
2
4
2
6
P1F2
n1F2
n1S2
42
64
21
32
⬇ .6563
1 6
6
#
5
2
6
#
5
#
4
3
#
2
42
6!
0! 6!
6!
1! 5!
6!
2! 4!
6!
3! 3!
n1F 2 C16, 02 C16, 12 C16, 22 C16, 32
6
#
5
#
4
3
#
2
64
20
64
5
16
.3125
P1E 2
n1E2
n1S2
C16, 32
64
6!
3! 3!
64
382 7 PROBABILITY
S
is a sample space of
the experiment.
87533_07_ch7_p353-416 1/30/08 10:05 AM Page 382
b. Let F denote the event that neither of the two cards selected is an ace. Since there are
C(48, 2) ways of selecting two cards neither of which is an ace, we find that
APPLIED EXAMPLE 3
Quality Control A bin in the hi-fi department
of Building 20, a bargain outlet, contains 100 blank cassette tapes, of which
10 are known to be defective. If a customer selects 6 of these cassette tapes at
random, determine the probability
a. That 2 of them are defective.
b. That at least 1 of them is defective.
Solution
a. There are C(100, 6) ways of selecting a set of 6 cassette tapes from the 100,
and this gives n(S), the number of outcomes in the sample space associated
with the experiment. Next, we observe that there are C(10, 2) ways of select-
ing a set of 2 defective cassette tapes from the 10 defective cassette tapes and
C(90, 4) ways of selecting a set of 4 nondefective cassette tapes from the
90 nondefective cassette tapes (Figure 13). Thus, by the multiplication princi-
ple, there are C(10, 2)
C(90, 4) ways of selecting 2 defective and 4 nonde-
fective cassette tapes. Therefore, the probability of selecting 6 cassette tapes
of which 2 are defective is given by
b. Let E denote the event that none of the cassette tapes selected is defective.
Then E
c
gives the event that at least 1 of the cassette tapes is defective. By the
rule of complements,
P(E
c
) 1 P(E)
⬇ .096
10
#
9
2
#
90
#
89
#
88
#
87
4
#
3
#
2
#
6
#
5
#
4
#
3
#
2
100
#
99
#
98
#
97
#
96
#
95
C110, 22
#
C190, 42
C1100, 62
10!
2! 8!
90!
4! 86!
100!
6! 94!
188
221
⬇ .8507
P1F 2
n1F2
n1S2
C148, 22
C152, 22
48!
2! 46!
52!
2! 50!
48
#
47
2
#
2
52
#
51
1
221
⬇ .0045
C14, 22
C152, 22
4!
2! 2
52!
2! 50!
4
#
3
2
#
2
52
#
51
P1E 2
n1E2
n1S2
7.4 USE OF COUNTING TECHNIQUES IN PROBABILITY 383
86
nondefective
tapes
8
defective
tapes
24
Sample
FIGURE 13
A sample of 6 tapes selected from
90 nondefective tapes and 10 defective
tapes
87533_07_ch7_p353-416 1/30/08 10:05 AM Page 383
To compute P(E), we observe that there are C(90, 6) ways of selecting a set
of 6 cassette tapes that are nondefective. Therefore,
The Birthday Problem
APPLIED EXAMPLE 4
The Birthday Problem A group of five people
is selected at random. What is the probability that at least two of them have
the same birthday?
Solution
For simplicity, we assume that none of the five people was born on
February 29 of a leap year. Since the five people were selected at random, we
also assume that each of them is equally likely to have any of the 365 days of a
year as his or her birthday. If we let A, B, C, D, and F represent the five people,
then an outcome of the experiment may be represented by (a, b, c, d, f ), where
the dates a, b, c, d, and f give the birthdays of A, B, C, D, and F, respectively.
We first observe that since there are 365 possibilities for each of the dates a,
b, c, d, and f, the multiplication principle implies that there are
abcdf
or 365
5
outcomes of the experiment. Therefore,
n(S) 365
5
where S denotes the sample space of the experiment.
Next, let E denote the event that two or more of the five people have the same
birthday. It is now necessary to compute P(E). However, a direct computation of
P(E) is relatively difficult. It is much easier to compute P(E
c
), where E
c
is the
event that no two of the five people have the same birthday, and then use the
relation
P(E) 1 P(E
c
)
To compute P(E
c
), observe that there are 365 ways (corresponding to the
365 dates) on which A’s birthday can occur, followed by 364 ways on which B’s
birthday could occur if B were not to have the same birthday as A, and so on.
Therefore, by the generalized multiplication principle,
n(E
c
) 365
364
363
362
361
A’s B’s C’s D’s F’s
birthday birthday birthday birthday birthday
365365365365365
⬇ .478
1
90
#
89
#
88
#
87
#
86
#
85
6
#
5
#
4
#
3
#
2
#
6
#
5
#
4
#
3
#
2
100
#
99
#
98
#
97
#
96
#
95
1
90!
6! 84!
100!
6! 94!
P1E
c
2 1
C190, 62
C1100, 62
P1E 2
C190, 62
C1100, 62
384 7 PROBABILITY
87533_07_ch7_p353-416 1/30/08 10:05 AM Page 384
Thus,
We can extend the result obtained in Example 4 to the general case involving r people.
In fact, if E denotes the event that at least two of the r people have the same birthday,
an argument similar to that used in Example 4 leads to the result
By letting r take on the values 5, 10, 15, 20, . . . , 50, in turn, we obtain the probabil-
ities that at least 2 of 5, 10, 15, 20, . . . , 50 people, respectively, have the same birth-
day. These results are summarized in Table 8.
The results show that, in a group of 23 randomly selected people, the chances are
greater than 50% that at least 2 of them will have the same birthday. In a group of
50 people, it is an excellent bet that at least 2 people in the group will have the same
birthday.
P1E2 1
365
#
364
#
363
#
p
#
1365 r 12
365
r
⬇ .027
1
365
#
364
#
363
#
362
#
361
365
5
P1E 2 1 P1E
c
2
365
#
364
#
363
#
362
#
361
365
5
P1E
c
2
n1E
c
2
n1S2
7.4 USE OF COUNTING TECHNIQUES IN PROBABILITY 385
1. Four balls are selected at random without replacement
from an urn containing 10 white balls and 8 red balls. What
is the probability that all the chosen balls are white?
2. A box contains 20 microchips, of which 4 are substandard.
If 2 of the chips are taken from the box, what is the proba-
bility that they are both substandard?
Solutions to Self-Check Exercises 7.4 can be found on
page 388.
TABLE 8
Probability That at Least Two People
in a Randomly Selected Group of r
People Have the Same Birthday
rP(E)
5 .027
10 .117
15 .253
20 .411
22 .476
23 .507
25 .569
30 .706
40 .891
50 .970
Explore & Discuss
During an episode of the Tonight Show, a talk show host related “The Birthday Problem” to
the audience—noting that, in a group of 50 or more people, probabilists have calculated that
the probability of at least 2 people having the same birthday is very high. To illustrate this
point, he proceeded to conduct his own experiment. A person selected at random from the
audience was asked to state his birthday. The host then asked if anyone in the audience had
the same birthday. The response was negative. He repeated the experiment. Once again, the
response was negative. These results, observed the host, were contrary to expectations. In a
later episode of the show, the host explained why this experiment had been improperly con-
ducted. Explain why the host failed to illustrate the point he was trying to make in the ear-
lier episode.
7.4 Self-Check Exercises
87533_07_ch7_p353-416 1/30/08 10:05 AM Page 385
386 7 PROBABILITY
An unbiased coin is tossed five times. In Exercises 1–4,
find the probability of the given event.
1. The coin lands heads all five times.
2. The coin lands heads exactly once.
3. The coin lands heads at least once.
4. The coin lands heads more than once.
Two cards are selected at random without replacement
from a well-shuffled deck of 52 playing cards. In Exer-
cises 5–8, find the probability of the given event.
5. A pair is drawn.
6. A pair is not drawn.
7. Two black cards are drawn.
8. Two cards of the same suit are drawn.
Four balls are selected at random without replacement
from an urn containing three white balls and five blue
balls. In Exercises 9–12, find the probability of the given
event.
9. Two of the balls are white and two are blue.
10. All of the balls are blue.
11. Exactly three of the balls are blue.
12. Two or three of the balls are white.
Assume that the probability of a boy being born is the
same as the probability of a girl being born. In Exercises
13–16, find the probability that a family with three chil-
dren will have the given composition.
13. Two boys and one girl
14. At least one girl
15. No girls
16. The two oldest children are girls.
17. An exam consists of ten true-or-false questions. If a student
guesses at every answer, what is the probability that he or
she will answer exactly six questions correctly?
18. P
ERSONNEL
S
ELECTION
Jacobs & Johnson, an accounting
firm, employs 14 accountants, of whom 8 are CPAs. If a
delegation of 3 accountants is randomly selected from the
firm to attend a conference, what is the probability that
3 CPAs will be selected?
19. Q
UALITY
C
ONTROL
Two light bulbs are selected at random
from a lot of 24, of which 4 are defective. What is the prob-
ability that
a. Both of the light bulbs are defective?
b. At least 1 of the light bulbs is defective?
20. A customer at Cavallaro’s Fruit Stand picks a sample of
3 oranges at random from a crate containing 60 oranges, of
which 4 are rotten. What is the probability that the sample
contains 1 or more rotten oranges?
21. Q
UALITY
C
ONTROL
A shelf in the Metro Department Store
contains 80 colored ink cartridges for a popular ink-jet
printer. Six of the cartridges are defective. If a customer se-
lects 2 cartridges at random from the shelf, what is the
probability that
a. Both are defective?
b. At least 1 is defective?
22. Q
UALITY
C
ONTROL
Electronic baseball games manufactured
by Tempco Electronics are shipped in lots of 24. Before
shipping, a quality-control inspector randomly selects a
sample of 8 from each lot for testing. If the sample contains
any defective games, the entire lot is rejected. What is the
probability that a lot containing exactly 2 defective games
will still be shipped?
23. P
ERSONNEL
S
ELECTION
The City Transit Authority plans to
hire 12 new bus drivers. From a group of 100 qualified
applicants, of whom 60 are men and 40 are women,
12 names are to be selected by lot. Suppose that Mary and
John Lewis are among the 100 qualified applicants.
a. What is the probability that Mary’s name will be
selected? That both Mary’s and John’s names will be
selected?
b. If it is stipulated that an equal number of men and women
are to be selected (6 men from the group of 60 men and
6 women from the group of 40 women), what is the prob-
ability that Mary’s name will be selected? That Mary’s
and John’s names will be selected?
24. P
UBLIC
H
OUSING
The City Housing Authority has received
50 applications from qualified applicants for eight low-
7.4 Concept Questions
1. What is the probability of an event E in a uniform sample
space S?
2. Suppose we want to find the probability that at least two
people in a group of six randomly selected people have the
same birthday.
a. If S denotes the sample space of this experiment, what
is n(S)?
b. If E is the event that two or more of the six people in the
group have the same birthday, explain how you would
use P(E
c
) to determine P(E).
7.4 Exercises
87533_07_ch7_p353-416 1/30/08 10:05 AM Page 386
income apartments. Three of the apartments are on the
north side of town, and five are on the south side. If the
apartments are to be assigned by means of a lottery, what
is the probability that
a. A specific qualified applicant will be selected for one of
these apartments?
b. Two specific qualified applicants will be selected for
apartments on the same side of town?
25. A student studying for a vocabulary test knows the mean-
ings of 12 words from a list of 20 words. If the test contains
10 words from the study list, what is the probability that at
least 8 of the words on the test are words that the student
knows?
26. D
RIVING
T
ESTS
Four different written driving tests are
administered by the Motor Vehicle Department. One of
these four tests is selected at random for each applicant for
a driver’s license. If a group consisting of two women and
three men apply for a license, what is the probability that
a. Exactly two of the five will take the same test?
b. The two women will take the same test?
27. B
RAND
S
ELECTION
A druggist wishes to select three brands
of aspirin to sell in his store. He has five major brands to
choose from: A, B, C, D, and E. If he selects the three
brands at random, what is the probability that he will select
a. Brand B?
b. Brands B and C?
c. At least one of the two brands B and C?
28. B
LACKJACK
In the game of blackjack, a 2-card hand con-
sisting of an ace and a face card or a 10 is called a black-
jack.
a. If a player is dealt 2 cards from a standard deck of
52 well-shuffled cards, what is the probability that the
player will receive a blackjack?
b. If a player is dealt 2 cards from 2 well-shuffled standard
decks, what is the probability that the player will
receive a blackjack?
29. S
LOT
M
ACHINES
Refer to Exercise 25, Section 6.3, where
the “lucky dollar” slot machine was described. What is the
probability that the three “lucky dollar” symbols will
appear in the window of the slot machine?
30. R
OULETTE
In 1959 a world record was set for the longest
run on an ungaffed (fair) roulette wheel at the El San Juan
Hotel in Puerto Rico. The number 10 appeared six times in
a row. What is the probability of the occurrence of this
event? (Assume that there are 38 equally likely outcomes
consisting of the numbers 1–36, 0, and 00.)
In “The Numbers Game,” a state lottery, four numbers
are drawn with replacement from an urn containing balls
numbered 0–9, inclusive. In Exercises 31–34, find the
probability that a ticket holder has the indicated winning
ticket.
31. All four digits in exact order (the grand prize)
32. Two specified, consecutive digits in exact order (the first
two digits, the middle two digits, or the last two digits)
33. One digit (the first, second, third, or fourth digit)
34. Three digits in exact order
A list of poker hands, ranked in order from the highest to
the lowest, is shown in the accompanying table along
with a description and example of each hand. Use the
table to answer Exercises 35–40.
7.4 USE OF COUNTING TECHNIQUES IN PROBABILITY 387
Hand Description Example
Straight 5 cards in sequence
flush in the same suit
Four of 4 cards of the same
a kind rank and any other
card
Full 3 of a kind and a
house pair
Flush 5 cards of the same
suit that are not
all in sequence
Straight 5 cards in sequence
but not all of the
same suit
Three 3 cards of the
of a same rank and 2
kind unmatched cards
Two 2 cards of the same
pair rank and 2 cards
of any other rank
with an unmatched
card
One 2 cards of the
pair same rank and 3
unmatched cards
A2345
K KKK 2
333
77
569J K
10 J Q K A
KKK24
KK224
KK524
If a 5-card poker hand is dealt from a well-shuffled deck
of 52 cards, what is the probability of being dealt the
given hand?
35. A straight flush (Note that an ace may be played as either
a high or low card in a straight sequence—that is, A, 2, 3,
4, 5 or 10, J, Q, K, A. Hence, there are ten possible
sequences for a straight in one suit.)
36. A straight (but not a straight flush)
37. A flush (but not a straight flush)
38. Four of a kind
39. A full house
40. Two pairs
87533_07_ch7_p353-416 1/30/08 10:05 AM Page 387
41. Z
ODIAC
S
IGNS
There are 12 signs of the Zodiac: Aries,
Taurus, Gemini, Cancer, Leo, Virgo, Libra, Scorpio, Sagit-
tarius, Capricorn, Aquarius, and Pisces. Each sign corre-
sponds to a different calendar period of approximately
1 month. Assuming that a person is just as likely to be born
under one sign as another, what is the probability that in a
group of five people at least two of them
a. Have the same sign?
b. Were born under the sign of Aries?
42. B
IRTHDAY
P
ROBLEM
What is the probability that at least
two of the nine justices of the U.S. Supreme Court have the
same birthday?
43. B
IRTHDAY
P
ROBLEM
Fifty people are selected at random.
What is the probability that none of the people in this group
have the same birthday?
44. B
IRTHDAY
P
ROBLEM
There were 42 different presidents of
the United States from 1789 through 2000. What is the
probability that at least two of them had the same birthday?
Compare your calculation with the facts by checking an
almanac or some other source.
388
7 PROBABILITY
1. The probability that all 4 balls selected are white is given
by
⬇ .069
10
#
9
#
8
#
7
4
#
3
#
2
#
4
#
3
#
2
18
#
17
#
16
#
15
10!
4! 6!
18!
4! 14!
C110, 42
C118, 42
The number of ways of selecting 4 white
balls from the 10 in the urn
The number of ways of selecting any
4 balls from the 18 balls in the urn
2. The probability that both chips are substandard is given by
⬇ .032
4
#
3
2
#
2
20
#
19
4!
2! 2!
20!
2! 18!
C14, 22
C120, 22
The number of ways of choosing any
2 of the 4 substandard chips
The number of ways of choosing any
2 of the 20 chips
Conditional Probability
Suppose that three cities, A, B, and C, are vying to play host to the Summer Olympic
Games in 2016. If each city has the same chance of winning the right to host the Games,
then the probability of city A hosting the Games is . Now suppose city B decides to pull
out of contention because of fiscal problems. Then it would seem that city A’s chances
of playing host will increase. In fact, if each of the two remaining cities have equal
chances of winning, then the probability of city A playing host to the Games is .
In general, the probability of an event is affected by the occurrence of other events
and/or by the knowledge of information relevant to the event. Basically, the injection
of conditions into a problem modifies the underlying sample space of the original
problem. This in turn leads to a change in the probability of the event.
EXAMPLE 1
Two cards are drawn without replacement from a well-shuffled deck
of 52 playing cards.
a. What is the probability that the first card drawn is an ace?
b. What is the probability that the second card drawn is an ace given that the first
card drawn was not an ace?
1
2
1
3
7.5 Conditional Probability and Independent Events
7.4 Solutions to Self-Check Exercises
87533_07_ch7_p353-416 1/30/08 10:05 AM Page 388
c. What is the probability that the second card drawn is an ace given that the first
card drawn was an ace?
Solution
a. The sample space here consists of 52 equally likely outcomes, 4 of which are
aces. Therefore, the probability that the first card drawn is an ace is , or .
b. Having drawn the first card, there are 51 cards left in the deck. In other words,
for the second phase of the experiment, we are working in a reduced sample
space. If the first card drawn was not an ace, then this modified sample space of
51 points contains 4 “favorable” outcomes (the 4 aces), so the probability that the
second card drawn is an ace is given by .
c. If the first card drawn was an ace, then there are 3 aces left in the deck of 51
playing cards, so the probability that the second card drawn is an ace is
given by , or .
Observe that in Example 1 the occurrence of the first event reduces the size of the
original sample space. The information concerning the first card drawn also leads us
to the consideration of modified sample spaces: In part (b) the deck contained 4 aces,
and in part (c) the deck contained 3 aces.
The probability found in part (b) or part (c) of Example 1 is known as a condi-
tional probability, since it is the probability of an event occurring given that another
event has already occurred. For example, in part (b) we computed the probability of
the event that the second card drawn is an ace given that the first card drawn was not
an ace. In general, given two events A and B of an experiment, under certain circum-
stances one may compute the probability of the event B given that the event A has
already occurred. This probability, denoted by P(B 兩 A), is called the conditional
probability of B given A.
A formula for computing the conditional probability of B given A may be discov-
ered with the aid of a Venn diagram. Consider an experiment with a uniform sample
space S, and suppose A and B are two events of the experiment (Figure 14).
The condition that the event A has occurred tells us that the possible outcomes
of the experiment in the second phase are restricted to those outcomes (elements)
in the set A. In other words, we may work with the reduced sample space A instead
of the original sample space S in the experiment. Next we observe that, with respect
to the reduced sample space A, the outcomes in the event B are precisely those
elements in the set A 傽 B. Consequently, the conditional probability of B given
A is
n1A 艚 B2
n1A2
P1B 0A2
Number of elements in A 艚 B
Number of elements in A
1
17
3
51
4
51
1
13
4
52
7.5 CONDITIONAL PROBABILITY AND INDEPENDENT EVENTS 389
A
S
B A
(a) Original sample space
(b) Reduced sample space A. The
shaded area is A 傽 B.
FIGURE
14
n
(
A
) 0
87533_07_ch7_p353-416 1/30/08 10:05 AM Page 389
Dividing the numerator and the denominator by n(S), the number of elements in S, we
have
which is equivalent to the following formula.
EXAMPLE 2
A pair of fair dice is rolled. What is the probability that the sum of
the numbers falling uppermost is 7 if it is known that one of the numbers is a 5?
Solution
Let A denote the event that the sum of the numbers falling uppermost is
7 and let B denote the event that one of the numbers is a 5. From the results of
Example 4, Section 7.1, we find that
A {(6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6)}
B {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),
(1, 5), (2, 5), (3, 5), (4, 5), (6, 5)}
so that
A 傽 B {(5, 2), (2, 5)}
(Figure 15). Since the dice are fair, each outcome of the experiment is equally
likely; therefore,
Recall that
n
(
S
) 36.
Thus, the probability that the sum of the numbers falling uppermost is 7 given that
one of the numbers is a 5 is, by virtue of Equation (2),
APPLIED EXAMPLE 3
Color Blindness In a test conducted by the
U.S. Army, it was found that of 1000 new recruits (600 men and 400
women), 50 of the men and 4 of the women were red-green color-blind. Given that
a recruit selected at random from this group is red-green color-blind, what is the
probability that the recruit is a male?
Solution
Let C denote the event that a randomly selected subject is red-green
color-blind and let M denote the event that the subject is a male recruit. Since 54
out of the 1000 subjects are color-blind, we have
P1A 0B2
2
36
11
36
2
11
P1A 艚 B2
2
36
and
P1B2
11
36
Conditional Probability of an Event
If A and B are events in an experiment and P(A) 0, then the conditional prob-
ability that the event B will occur given that the event A has already occurred is
(2)
P1B 0A2
n1A 艚 B2
n1S2
n1A2
n1S2
390 7 PROBABILITY
FIGURE 15
A 傽 B
{
(5, 2), (2, 5)
}
P1B 0A2
P1A 艚 B2
P1A2
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Therefore, by Equation (2), the probability that a subject is male given that the
subject is red-green color-blind is
In certain problems, the probability of an event B occurring given that A has
occurred, written P(B 兩 A), is known, and we wish to find the probability of A and B
occurring. The solution to such a problem is facilitated by the use of the following
formula:
This formula is obtained from Equation (2) by multiplying both sides of the equation
by P(A). We illustrate the use of the product rule in the next several examples.
APPLIED EXAMPLE 4
Seniors with Driver’s Licenses There are
300 seniors in Jefferson High School, of which 140 are males. It is known
that 80% of the males and 60% of the females have their driver’s license. If a
student is selected at random from this senior class, what is the probability that the
student is
a. A male and has a driver’s license?
b. A female who does not have a driver’s license?
Product Rule
(3)
.05
.054
.926
P1M 0C 2
P1M 艚 C2
P1C2
P1C2
54
1000
.054
7.5 CONDITIONAL PROBABILITY AND INDEPENDENT EVENTS 391
Explore & Discuss
Let A and B be events in an experiment and suppose P(A) 0. In n trials, the event A
occurs m times, the event B occurs k times, and the events A and B occur together l times.
1. Explain why it makes good sense to call the ratio l/m the conditional relative frequency
of the event B given the event A.
2. Show that the relative frequencies l/m, m/n, and l/n satisfy the equation
3. Explain why the result of part 2 suggests that Equation (2)
is plausible.
l
m
l
n
m
n
P1B 0A 2
P1A 艚 B 2
P1A2
3P1A 2 04
P1A 艚 B2 P1A2
#
P1B 0A2
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