Bilkent University Fall 2022
MATH 255 - Probability and Statistics
Solutions to Midterm Exam I
Problem 1. [6pt] Suppose A, B, and C are events for a probability experiment such that A and
B are mutually independent, P (A) = P (B) = P (C) = 0.5, P (A ∩ C) = P (B ∩ C) = 0.3, and
P (A ∩ B ∩ C) = 0.1. Fill in the probabilities of all events in the Karnaugh map below. Show
your work.
Due to mutual independence, we have P (A ∩ B) = 0.25.
P (A ∩ B ∩ C) = 0.1
P (Ac ∩ B ∩ C) = 0.2 (+0.5 pt)
P (A ∩ B c ∩ C) = 0.2 (+0.5 pt)
P (Ac ∩ B c ∩ C) = 0 (+1 pt)
P (A ∩ B ∩ C c ) = 0.15 (+1 pt)
P (Ac ∩ B ∩ C c ) = 0.05 (+0.5 pt)
P (A ∩ B c ∩ C c ) = 0.05 (+0.5 pt)
P (Ac ∩ B c ∩ C c ) = 0.25 (+2 pt)
1 09-25-2025 13:26:12 GMT -05:00
This study source was downloaded by 100000899606396 from CourseHero.com on
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, Problem 2. [8pt] For a given graph, two vertices, i and j, are selected at random, with all
possible values of (i, j) having equal probability, including the cases with i = j. Let D denote
the distance between i and j, which is the minimum number of edges that must be crossed
to walk in the graph from i to j. If i = j, then D = 0. Find and sketch the PMF of D, and
find its expected value and variance for each of the three undirected graphs below. There is no
designated space for the final answer. Hint: For (b) and (c), by symmetry, it can be assumed that i = 1
and only j is selected at random.
(a) The PMF can be sketched as (+2 pt)
6 10 8 6 4 2
E[D] = ×0+ ×1+ ×2+ ×3+ ×4+ ×5
36 36 36 36 36 36
10 + 16 + 18 + 16 + 10 70 35
= = = (+0.5pt)
36 36 18
To compute the variance, we can use the second definition by computing
6 10 8 6 4 2
E[D2 ] = × 02 + × 12 + × 22 + × 32 + × 42 + × 52
36 36 36 36 36 36
10 + 32 + 54 + 64 + 50 210 105 35
= = = = .
36 36 18 6
Hence, we obtain
2
105 35 105 × 18 − 35 × 35 1890 − 1225 665
var(D) = − = 2
= = (+1pt).
18 18 18 324 324
(b) The PMF can be sketched as (+1 pt)
Based on the center of gravity interpretation, the expected value is
3
given by (+0.5 pt).
2
Note that the expected value is well-defined since the random variable
takes finitely many values.
2 09-25-2025 13:26:12 GMT -05:00
This study source was downloaded by 100000899606396 from CourseHero.com on
https://www.coursehero.com/file/243107133/2022Fall-Midterm1Solutionspdf/
MATH 255 - Probability and Statistics
Solutions to Midterm Exam I
Problem 1. [6pt] Suppose A, B, and C are events for a probability experiment such that A and
B are mutually independent, P (A) = P (B) = P (C) = 0.5, P (A ∩ C) = P (B ∩ C) = 0.3, and
P (A ∩ B ∩ C) = 0.1. Fill in the probabilities of all events in the Karnaugh map below. Show
your work.
Due to mutual independence, we have P (A ∩ B) = 0.25.
P (A ∩ B ∩ C) = 0.1
P (Ac ∩ B ∩ C) = 0.2 (+0.5 pt)
P (A ∩ B c ∩ C) = 0.2 (+0.5 pt)
P (Ac ∩ B c ∩ C) = 0 (+1 pt)
P (A ∩ B ∩ C c ) = 0.15 (+1 pt)
P (Ac ∩ B ∩ C c ) = 0.05 (+0.5 pt)
P (A ∩ B c ∩ C c ) = 0.05 (+0.5 pt)
P (Ac ∩ B c ∩ C c ) = 0.25 (+2 pt)
1 09-25-2025 13:26:12 GMT -05:00
This study source was downloaded by 100000899606396 from CourseHero.com on
https://www.coursehero.com/file/243107133/2022Fall-Midterm1Solutionspdf/
, Problem 2. [8pt] For a given graph, two vertices, i and j, are selected at random, with all
possible values of (i, j) having equal probability, including the cases with i = j. Let D denote
the distance between i and j, which is the minimum number of edges that must be crossed
to walk in the graph from i to j. If i = j, then D = 0. Find and sketch the PMF of D, and
find its expected value and variance for each of the three undirected graphs below. There is no
designated space for the final answer. Hint: For (b) and (c), by symmetry, it can be assumed that i = 1
and only j is selected at random.
(a) The PMF can be sketched as (+2 pt)
6 10 8 6 4 2
E[D] = ×0+ ×1+ ×2+ ×3+ ×4+ ×5
36 36 36 36 36 36
10 + 16 + 18 + 16 + 10 70 35
= = = (+0.5pt)
36 36 18
To compute the variance, we can use the second definition by computing
6 10 8 6 4 2
E[D2 ] = × 02 + × 12 + × 22 + × 32 + × 42 + × 52
36 36 36 36 36 36
10 + 32 + 54 + 64 + 50 210 105 35
= = = = .
36 36 18 6
Hence, we obtain
2
105 35 105 × 18 − 35 × 35 1890 − 1225 665
var(D) = − = 2
= = (+1pt).
18 18 18 324 324
(b) The PMF can be sketched as (+1 pt)
Based on the center of gravity interpretation, the expected value is
3
given by (+0.5 pt).
2
Note that the expected value is well-defined since the random variable
takes finitely many values.
2 09-25-2025 13:26:12 GMT -05:00
This study source was downloaded by 100000899606396 from CourseHero.com on
https://www.coursehero.com/file/243107133/2022Fall-Midterm1Solutionspdf/
