TEST1_F22_STAT230_SOLUTIONS (A)
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School
University of Waterloo *
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Course
230
Subject
Mathematics
Date
Jan 9, 2024
Type
Pages
8
Uploaded by CommodoreFoxMaster973
{
Special instructions
•
Your final numerical answers should be given to 3 decimal places (e.g. 0.329). However, between
steps/parts you should carry more decimal places to avoid rounding errors.
•
If working is not specifically requested, a correct answer will receive full marks.
However, an
incorrect answer could still receive credit if working is provided. An incorrect answer with no
working will receive zero marks.
•
The exam consists of 4 questions for a total of 35 marks. The number of marks available per
question is indicated in [square brackets].
•
Answer the questions in the spaces provided.
You may use the last page of the test for any
additional work you would like to be graded. If you use this space, make it as clear as possible
which question(s) your work relates to.
Question 1
[8 marks]
Nine students (including James and George) have applied for a scholarship. Identical scholarships are
awarded at random to either 5, 6, or 7 of the applicants.
(a) [2 marks] How many possible ways are there to assign scholarships?
Since order does not matter, the number of possible ways is
(
9
5
)
+
(
9
6
)
+
(
9
7
)
= 126 + 84 + 36 = 246.
(b) [2 marks] Calculate the probability that James receives a scholarship.
If James receives a scholarship, we need to find the remaining students from the remaining 8 people.
So there are
(
8
4
)
+
(
8
5
)
+
(
8
6
)
= 70 + 56 + 28 = 154 possible ways for James to get a scholarship.
So the probability is 154
/
246 = 0
.
626
.
(c) [2 marks] Calculate the probability that either James or George (but not both) receives a scholar-
ship.
We found
P
(
J
) = 0
.
626 in b). By symmetry,
P
(
G
) = 0
.
626.
The probability both James and George have scholarships is
P
(
J
∩
G
) =
(
7
3
)
+
(
7
4
)
+
(
7
5
)
246
= 91
/
246 = 0
.
370
.
We want
P
(
J
∪
G
)
−
P
(
J
∩
G
) =
P
(
J
)+
P
(
G
)
−
P
(
J
∩
G
)
−
P
(
J
∩
G
) = 0
.
626+0
.
626
−
0
.
370
−
0
.
370 = 0
.
512
.
(d) [2 marks] Calculate the probability that neither James nor George receives a scholarship.
We want
P
(
J
c
∩
G
c
) =
P
(
J
∪
G
)
c
= 1
−
P
(
J
∪
G
) = 1
−
P
(
J
)
−
P
(
G
)+
P
(
J
∩
G
) = 1
−
0
.
626
−
0
.
626+
0
.
370 = 0
.
118
.
Alternate solution: exclude James and George and pick from the remaining 7 people:
(
7
5
)
+
(
7
6
)
+
(
7
7
)
246
=
29
/
246 = 0
.
118
.
Question 2
[8 marks]
Emily’s children are making a craft by placing beads on a string. One of the children is given 5 blue
beads, 2 yellow beads, and 4 green beads. The two ends of the string are not connected. Given the
child is a toddler, we assume the beads are placed randomly on the string.
(a) [2 marks] How many bead patterns can be created, if we consider beads of the same colour are
indistinguishable?
We have 11 beads total to arrange with 5 indistinguishable blue beads, 2 indistinguishable yellow
beads, and 4 indistinguishable green beads. Thus, the total number of arrangements is
11!
5!2!4!
= 6930.
(b) [2 marks] Find the probability that all 4 of the green beads are together.
Consider the 4 green beads as one object and then arrange the 8 items of GGGG Y Y B B B B B.
This can be done in
8!
1!2!5!
= 168 ways. So the probability the 4 G’s are together is
168
6930
= 0
.
0242.
(c) [2 marks] Find the probability that there is a green bead at each end of the string.
We fix a green bead at either end as G x x x x x x x x x G. The remaining 9 beads can be arranged
in the middle ‘x’ places in
9!
5!2!2!
= 756 ways. Thus, the probability is
756
6930
= 0
.
109.
(d) [2 marks] Find the probability that either all 4 of the green beads are together or there is one green
bead at either end of the string.
Our choice of wording for this question was ambiguous. Therefore we will accept both of the following
interpretations:
Solution 1
: This applies to the original intention of the question where the words “either end of the
string” are to be interpreted as both ends of the string. I.e. we find the probability that either all 4
of the green beads are together or there is one green bead at EACH end of the string.
Since these are two disjoint events, we can add the number of outcomes from parts (b) and (c) to-
gether by the addition rule. Thus, there are 168 + 756 = 924 ways in which the green beads are either
together or at each end of the string. Thus, the probability is
924
6930
= 0
.
133.
Solution 2
: This solution is also acceptable.
We want to count the number of ways we can have (1) GGGG together, or (2) there is a G at both ends
of the string, or (3) there is a G at the beginning of the string but not at the end, or (4) there is a G
the end of the string but not at the beginning. For case 1 and 2, this is still from (b) and (c), so there
are 168 + 756 = 924 ways in which the green beads are either together or at each end of the string.
For counting case 3 we have to be careful not to double count the arrangements that have already been
counted in case 1 and 2. We only want to count the outstanding arrangements that have G at the
beginning but haven’t been counted yet. First, fix the first G as G x x x x x x x x x x. The remaining
beads can be arranged in
10!
5!3!2!
= 2520 ways. However we have to subtract off the ways in which there
is also a green bead at the other end or all greens are together at the beginning. There are 756 ways
where there is a green bead at each end from (c), so we will deduct that. Second, having GGGG as
the first four beads can be done in
7!
2!5!
= 21 ways. So for case 3, we have 2520
−
756
−
21 = 1743 ways.
By symmetry, we also have 1743 ways for case 4. Finally, the probability is
1743+1743+924
6930
= 0
.
636
.
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Question 3
[12 marks]
A PIN on a loyalty card must be 4 digits long (each digit from 0-9) and it cannot have more than two
of the same digit (e.g. 3434 is OK but 3343 is not.)
(a) [3 marks] How many possible PINs are there?
Case 1: All digits are different. We have 10
(4)
options.
Case 2: Two digits are the same, two are different. There are
(
4
2
)
possible locations for the repeated
digits, 10 choices for the repeated digit and 9 and 8 choices for the other two digits.
Case 3: Two pairs of identical digits. We have
(
10
2
)
options for the two repeated digits and they can
be permuted in 6 ways, e.g. 1122, 1212, 1221, 2211, 2121, 2112.
Answer : 10
(4)
+
(
4
2
)
10
(3)
+
(
10
2
)
(6) = 9630
.
(b) [3 marks] Calculate the probability that a randomly selected PIN contains at least one “9”.
The complement is the event that it contains no “9”, and similarly to a), we have 9
(4)
+
(
4
2
)
9
(3)
+
(
9
2
)
(6) possibilities for the complement since we have 9 digits to choose from instead of 10. Answer :
1
−
9
(4)
+
(
4
2
)
9
(3)
+
(
9
2
)
(6)
9630
= 1
−
6264
9630
= 0
.
350
.
(c) [3 marks] Calculate the probability that a randomly selected PIN contains three even digits. (Recall
that 0 is even.)
Case 1: All digits are different. We have 5
(3)
permutations for the even digits and 4 places to insert
the odd digit, with 5 possible odd digits.
Case 2:
One even digit is repeated twice.
We have
(
5
2
)
ways to choose the two even digits.
The
permutations of the even digits can take one of the 6 following forms 224, 242, 422, 442, 424, 244. We
have 4 places to insert the odd digit, with 5 possible odd digits.
Answer :
5
(3)
(4)(5)+
(
5
2
)
(6)(4)(5)
9630
=
2400
9630
= 0
.
249
.
(d) [3 marks] Calculate the probability that a randomly selected PIN contains the pattern “89”
Case 1: “89” and two distinct digits from 0-7.
We have 8
(2)
permutations for the 0-7 digits and 3
possible placements around “89”.
Case 2: “89” and one of 0-7 repeated twice. We have 8 choices for the repeated digit and 3 possible
placements around “89”.
Case 3: “889” or “898” or “899” or “989” and one of 0-7. We have 8 choices for the digit and 3 possible
placements without breaking “89”.
Case 4: 8899 or 8989 or 9898 or 8998 or 9889
Answer :
8
(2)
(3)+8(3)+4(8)(3)+5
9630
=
293
9630
= 0
.
030
.
Question 4
[7 marks]
Holly loves ice cream. Every time Holly passes an ice cream shop, she has to stop to buy herself an
ice cream cone.
Let C be the event that Holly gets chocolate ice cream, V the event that she gets
vanilla ice cream, and S the event that she gets strawberry ice cream. Note that
P
(
V
) = 1
, P
(
C
C
) =
0
.
3
, P
(
C
∩
V
) =
P
(
V
∩
S
) and
P
(
C
∩
V
∩
S
) = 0
.
5.
(a) [2 marks] Draw a Venn diagram of the events C, V, and S, indicating the respective probabilities
of each of the regions formed by the intersections of the various events.
0
1
−
0
.
5
−
2
x
0
0
C
V
S
0
.
5
x
0
x
Solving
P
(
C
C
) = 0
.
3 = 1
−
(
x
+ 0
.
5 + 0 + 0) for
x
, we get
x
= 0
.
2.
(b) [2 marks] Find the odds in favour of the event that Holly gets at least two different flavours of ice
cream.
P
((
C
∩
V
∩
S
C
)
∪
(
S
∩
V
∩
C
C
)
∪
(
C
∩
S
∩
V
C
)
∪
(
C
∩
S
∩
V
)) =
x
+
x
+ 0 + 0
.
5 = 0
.
9. The odds in
favor are 0
.
9
/
0
.
1 = 9 : 1 = 9.
(c) [2 marks] Find the probability distribution of all possible ice cream flavour combinations.
Based on the Venn diagram we have
Flavours
no flavour
C
V
S
C & V, not S
V & S, not C
C & S, not V
C & V & S
Probability
0
0
0.1
0
0.2
0.2
0
0.5
Note: the columns with 0 probability may be omitted.
(d) [1 mark] Find
P
(
C
C
∩
S
)
C
.
P
(
C
C
∩
S
)
C
=
P
(
C
∪
S
C
) = 1
−
(
x
+ 0) = 0
.
8, using De Morgan’s law.
END OF EXAMINATION
Use this page for any additional work you would like to be graded. If you use this
space, make it as clear as possible which question(s) your work relates to. If there is
any ambiguity only your work on the previous question pages will be graded.
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Use this page for any additional work you would like to be graded. If you use this
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any ambiguity only your work on the previous question pages will be graded.
Use this page for any additional work you would like to be graded. If you use this
space, make it as clear as possible which question(s) your work relates to. If there is
any ambiguity only your work on the previous question pages will be graded.