Question 1
MATH40005 Probability and Statistics Resit Preparation
Question 1
1(a). A fair three-sector spinner labelled 1, 2, 3 is spun twice. Briefly describe the probability space (Ω, F, P). (3 marks)
1(b)(i). A game selects 3 distinct main numbers from {1, . . . ,25} and 2 distinct symbols from {1, . . . ,7}, with order irrele-
vant. How many outcomes are possible? (3 marks)
1(b)(ii). Given that a ticket matches exactly two main numbers, what is the probability that it also matches both symbols?
(3 marks)
1(b)(iii). Find the probability that a uniformly chosen ticket matches exactly two main numbers and both symbols. (4 marks)
1(b)(iv). Suppose the two symbols are instead drawn with replacement and order is ignored. How many symbol outcomes
are possible? (2 marks)
1(c). Let m ∈ Z and 0 < q < 1. A proposed pmf is
pX (x) = c(x − m + 1)qx−m, x = m, m + 1, . . . ,
and zero otherwise. Find c, verify validity, and give the CDF for all real x. (5 marks)
Total: 20 marks
2
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 1: solution and marking guide
1(a). Take Ω = {1, 2, 3}2, F = P(Ω), and P(A) = |A|/9 for each event A. The ordered pair records the first and second
spins.
3 marks; A:3
1(b)(i). There are 25
3
7
2
possible outcomes.
3 marks; A:3
1(b)(ii). The symbol selection is unaffected by the main-number information. Exactly one of the
7
2
symbol pairs matches,
so the probability is
17
2
= 1
21 .
3 marks; B:2, C:1
1(b)(iii). Choose two of the three drawn main numbers and one of the remaining 22 numbers. Therefore
P =
3
2
22
1
25
3
· 17
2
.
4 marks; A:1, D:3
1(b)(iv). This is the number of multisets of size two from seven symbols:
7 + 2 − 1
2
=
8
2
= 28.
2 marks; A:1, B:1
1(c). Put k = x − m. Since
∞X
k=0
(k + 1)qk = 1
(1 − q)2 ,
normalisation gives c = (1 − q)2. For x < m , FX (x) = 0 . For x ≥ m, put r = ⌊x⌋ − m and use the finite arithmetico-
geometric sum:
FX (x) = (1 − q)2
rX
k=0
(k + 1)qk = 1 − (r + 2)qr+1 + (r + 1)qr+2.
All probabilities are nonnegative, so the pmf is valid.
5 marks; A:2, B:1, C:1, D:1
Question total: 20 marks
2
Question 2
MATH40005 Probability and Statistics Resit Preparation
Question 2
2(a). Construct three Bernoulli random variables that are pairwise independent but not mutually independent. Justify your
construction. (3 marks)
2(b)(i). For x, y ≥ 0, let
FX,Y (x, y) = (1 − e−2x)(1 − e−3y),
and let it be zero otherwise. Find the marginal CDFs and densities and identify the distributions. (5 marks)
2(b)(ii). Are X and Y independent? (2 marks)
2(b)(iii). Find the MGF of X, including its domain. (3 marks)
2(b)(iv). Let Z = X 2. Find the CDF of Z. (3 marks)
2(c)(i). There are n independent sensors. Each alarms on a given day with probability 1/d. Let Yi indicate whether sensor i
alarms and let B = P
i Yi. Identify the distributions and parameters. (2 marks)
2(c)(ii). Find the probability that at least one sensor alarms. (2 marks)
Total: 20 marks
3
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 2: solution and marking guide
2(a). Let U, V be independent fair bits and let W = U ⊕ V . Each pair is uniform on {0, 1}2, hence pairwise independent.
But W is completely determined by U, V , so the three are not mutually independent.
3 marks; A:1, B:2
2(b)(i). The marginals are
FX (x) = 1 − e−2x, f X (x) = 2 e−2x,
and
FY (y) = 1 − e−3y, f Y (y) = 3 e−3y
for nonnegative arguments. Thus X ∼ Exp(2) and Y ∼ Exp(3).
5 marks; A:4, B:1
2(b)(ii). Y es, becauseFX,Y (x, y) = FX (x)FY (y) for all x, y.
2 marks; B:2
2(b)(iii). For t < 2,
MX (t) =
Z ∞
0
etx2e−2x dx = 2
2 − t .
3 marks; A:1, C:1, D:1
2(b)(iv). For z < 0, FZ(z) = 0 . For z ≥ 0, since X ≥ 0,
FZ(z) = P(X ≤ √z) = 1 − e−2√z.
3 marks; C:1, D:2
2(c)(i). Each Yi ∼ Bernoulli(1/d) and, by independence,
B ∼ Bin
n, 1
d
.
2 marks; A:2
2(c)(ii). Using the complement,
P(B ≥ 1) = 1 − P(B = 0) = 1 −
1 − 1
d
n
.
2 marks; C:1, D:1
Question total: 20 marks
3
Question 3
MATH40005 Probability and Statistics Resit Preparation
Question 3
3(a). Let X have finite first and second moments. State the minimiser and minimum of E[(X − a)2], and state a minimiser
of E|X − b|. (3 marks)
3(b). Prove the bias–variance decomposition for an estimator bθ. (3 marks)
3(c). Name a suitable plot for (i) the relationship between weekly study hours and examination mark, and (ii) the median,
spread and outliers of examination marks in each of four degree programmes. (2 marks)
3(d)(i). Eleven independent observations are assumed normal with unknown mean and variance. The sample mean is54.6 and
sample variance is 6.25. Compute a 90% confidence interval for σ2. Y ou may use χ2
10,0.05 = 3 .940 and χ2
10,0.95 = 18 .307.
(4 marks)
3(d)(ii). Using the same data, test H0 : µ = 56 against H1 : µ ̸= 56. Compute the test statistic and p-value and conclude at
level 0.05. (5 marks)
3(e). Define a p-value and explain how it should be interpreted. (3 marks)
Total: 20 marks
4
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 3: solution and marking guide
3(a). The squared-loss minimiser is a = EX, with minimum Var (X). Any median of X minimises E|X − b|.
3 marks; A:2, B:1
3(b). Add and subtract Ebθ and expand:
E(bθ − θ)2 = E(bθ − Ebθ)2 + (Ebθ − θ)2,
because the cross term has expectation zero. Thus MSE equals variance plus squared bias.
3 marks; A:1, B:1, D:1
3(c). Use (i) a scatterplot and (ii) side-by-side boxplots.
2 marks; A:2
3(d)(i). The interval is 10(6.25)
18.307 , 10(6.25)
3.940
≈ (3.41, 15.86).
4 marks; A:2, B:1, C:1
3(d)(ii). The standard error is 2.5/
√
11. Hence
t = 54.6 − 56
2.5/
√
11 ≈ −1.857
with 10 degrees of freedom. The two-sided p-value is approximately 0.0929. Since it exceeds 0.05, fail to reject H0.
5 marks; A:1, B:1, C:1, D:2
3(e). It is the probability, calculated under H0, of a test statistic at least as extreme as the observed value in the direction(s)
specified by the alternative. A small p-value is evidence that the observed result is unusual under H0; it is not the probability
that H0 is true.
3 marks; B:2, D:1
Question total: 20 marks
4
Question 4
MATH40005 Probability and Statistics Resit Preparation
Question 4
4(a). Suppose X ∼ N(6, 4) and Y ∼ N(3, 9) with Corr(X, Y ) = −1/3. Let Z = 3X + Y . Compute Corr (Y, Z). (4 marks)
4(b). A simple linear regression has R2 = 0.97 and the residual plot below. Does the model fit well? Suggest an improvement.
(3 marks)
4(c). Give a real-world example in which linear regression can investigate a relationship between two quantities. (2 marks)
4(d). T wo unrelated annual series have sample correlation 0.94. Can one conclude that one causes the other? (2 marks)
4(e)(i). The data are {9, 14, 18, 22, 31}. Is {31, 14, 14, 9, 22} a bootstrap sample? (2 marks)
4(e)(ii). Is {9, 14, 18, 22} a bootstrap sample of the same data? (2 marks)
4(f). Independent observations are uniformly distributed on [θ − 1, θ + 2]. Find the full set of MLEs of θ. (5 marks)
0 5 10 15 20
−1
−0.5
0
0.5
xi
Residual bεi
Total: 20 marks
5
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 4: solution and marking guide
4(a). Since sd(X) = 2 and sd(Y ) = 3 ,
Cov(X, Y ) = − 1
3 (2)(3) = −2.
Then
Cov(Y, Z) = 3 Cov(Y, X ) + Var(Y ) = 3( −2) + 9 = 3 ,
and
Var(Z) = 9 Var(X) + Var(Y ) + 6 Cov(X, Y ) = 36 + 9 − 12 = 33 .
Therefore
Corr(Y, Z) = 3
3
√
33 = 1√
33 .
4 marks; A:1, C:1, D:2
4(b). No. The residuals show systematic curvature despite the high R2. Include a nonlinear term such as x2 or use an
appropriate transformation, and then reassess the residuals.
3 marks; B:2, D:1
4(c). For example, model crop yield (response) as a function of fertiliser amount (explanatory variable). The slope estimates
the change in expected yield per unit fertiliser within the range and subject to the model assumptions.
2 marks; A:1, B:1
4(d). No. Shared trends, confounding, reverse causation or coincidence can produce high correlation. The correlation alone
does not identify a causal effect.
2 marks; B:1, C:1
4(e)(i). Y es. It has the original sample size five and every value is drawn from the observed data with replacement.
2 marks; A:2
4(e)(ii). No. A standard nonparametric bootstrap sample must have the same size as the original sample, but this sample has
size four.
2 marks; A:1, B:1
4(f). The likelihood is 3−n when all observations satisfy θ − 1 ≤ xi ≤ θ + 2, and zero otherwise. Equivalently,
θ ≤ xmin + 1, θ ≥ xmax − 2.
Under the stated model, the sample range is at most 3 almost surely, and the full set of MLEs is
bθ ∈ [xmax − 2, x min + 1].
An externally supplied sample with range greater than 3 would be incompatible with this model.
5 marks; A:1, C:3, D:1
Question total: 20 marks
5