Probability space, combinatorics and geometric pmf
MATH40005 Probability and Statistics Resit Preparation
Question 1
1(a). A box contains cards labelled 1, 2, . . . ,8. One card is drawn, replaced, and a second card is drawn.
Briefly describe a probability space (Ω, F, P) for this experiment when all cards are equally likely. (3
marks)
1(b)(i). A prize draw asks a player to choose 4 distinct main numbers from {1, . . . ,30} and 2 distinct
bonus symbols from {1, . . . ,9}. Order is irrelevant. 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 bonus symbols? (3 marks)
1(b)(iii). Find the probability that a uniformly chosen ticket matches exactly two main numbers and
both bonus symbols. (4 marks)
1(b)(iv). Suppose the two bonus symbols are drawn with replacement and their order is ignored. How
many bonus outcomes are possible? (2 marks)
1(c). Let m ∈ Z and 0 < q < 1. A random variable has proposed pmf pX (x) = cqx−m for x = m, m+1, . . .,
and 0 otherwise. Find c, verify validity, and obtain the CDF for all real x. (5 marks)
T otal: 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, . . . ,8}2, with ordered outcomes (i, j). Let F = P(Ω) and, for A ⊆ Ω, define
P(A) = |A|/64. Each singleton has probability 1/64.
3 marks; A:3
1(b)(i). There are
30
4
main-number choices and
9
2
bonus choices, hence
30
4
9
2
outcomes.
3 marks; A:3
1(b)(ii). The main-number information does not alter the bonus choice. Exactly one of the
9
2
unordered
bonus pairs matches, so the probability is 1/
9
2
= 1/36.
3 marks; B:3
1(b)(iii). Choose the two correct main numbers in
4
2
ways and the two incorrect main numbers from
the remaining 26 in
26
2
ways. Therefore
P =
4
2
26
2
30
4
· 1 9
2
.
4 marks; D:4
1(b)(iv). This is the number of multisets of size 2 from 9 symbols:
9+2−1
2
=
10
2
= 45.
2 marks; B:2
1(c). Non-negativity holds when c ≥ 0. Normalisation gives
1 =
∞X
x=m
cqx−m = c
∞X
k=0
qk = c
1 − q ,
so c = 1 − q. For x < m , FX (x) = 0 . For x ≥ m,
FX (x) =
⌊x⌋−mX
k=0
(1 − q)qk = 1 − q⌊x⌋−m+1.
5 marks; A:2, C:3
Question total: 20 marks
2
Independence, joint distributions and moments
MATH40005 Probability and Statistics Resit Preparation
Question 2
2(a). Construct three Bernoulli random variables that are pairwise independent but not mutually inde-
pendent, and justify the claim. (3
marks)
2(b)(i). For x, y ≥ 0, let FX,Y (x, y) = (1 − e−2x)(1 − e−y), and let it be 0 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 moment generating function of X, including its domain. (3 marks)
2(b)(iv). Let Z =
√
X. Find the CDF of Z. (3 marks)
2(c)(i). There are n independent components, each failing on a specified day with probability 1/d. Let Yi
indicate failure of component i on that day and let B = P
i Yi. Identify the distributions and parameters.
(2 marks)
2(c)(ii). Find the probability that at least one component fails on that day. (2 marks)
T otal: 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 Bernoulli variables and set W = U ⊕ V (addition modulo 2). Each
variable is Bernoulli (1/2). Every pair is uniformly distributed on {0, 1}2, hence pairwise independent.
But W = U ⊕ V always, so, for example, P(U = 0 , V = 0 , W = 1) = 0 ̸= 1 /8; they are not mutually
independent.
3 marks; A:3
2(b)(i). Taking the other variable to infinity gives FX (x) = 1 − e−2x for x ≥ 0 and FY (y) = 1 − e−y for
y ≥ 0. Hence fX (x) = 2 e−2x and fY (y) = e−y on [0, ∞). Thus X ∼ Exp(2) and Y ∼ Exp(1).
5 marks; A:5
2(b)(ii). Yes. For all x, y, FX,Y (x, y) = FX (x)FY (y).
2 marks; B:2
2(b)(iii).
MX (t) =
Z ∞
0
etx2e−2x dx = 2
2 − t , t < 2.
3 marks; C:3
2(b)(iv). For z < 0, FZ(z) = 0 . For z ≥ 0,
FZ(z) = P(X ≤ z2) = 1 − e−2z2
.
3 marks; B:3
2(c)(i). Yi ∼ Bernoulli(1/d) and, by independence, B ∼ Binomial(n, 1/d).
2 marks; D:2
2(c)(ii).
P(B ≥ 1) = 1 − P(B = 0) = 1 − (1 − 1/d)n.
2 marks; D:2
Question total: 20 marks
3
Estimation, intervals and Poisson comparison
MATH40005 Probability and Statistics Resit Preparation
Question 3
3(a). Let X have finite first and second moments. State the minimiser of E[(X − a)2], the minimum
value, and 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 engine size and fuel consumption, and (ii)
the median, spread and outliers of fuel consumption within one vehicle class. (2 marks)
3(d)(i). Fifteen independent observations are assumed normal with unknown mean and variance. The
sample mean is 74.2 and sample variance is 6.25. Compute a 95% confidence interval for the mean. You
may use t14,0.975 = 2.145. (4 marks)
3(d)(ii). Test H0 : µ = 75 against H1 : µ ̸= 75 and calculate the two-sided p-value. State the conclusion
at level 0.05. (5 marks)
3(e). Define a p-value, explain how it is interpreted, and state when it is uniformly distributed. (3
marks)
T otal: 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 = E[X] and the minimum is Var(X). Any median of X minimises
absolute loss.
3 marks; A:2, B:1
3(b). Write bθ − θ = ( bθ − Ebθ) + ( Ebθ − θ). Squaring and taking expectations, the cross term vanishes.
Hence
E(bθ − θ)2 = Var(bθ) + {Ebθ − θ}2.
3 marks; A:3
3(c). (i) A scatterplot. (ii) A boxplot.
2 marks; A:2
3(d)(i). Using ( ¯X − µ)/(S/√n) ∼ t14, with s = 2.5,
74.2 ± 2.145 2.5√
15 = 74.2 ± 1.384,
so the interval is approximately (72.82, 75.58).
4 marks; A:4
3(d)(ii).
t = 74.2 − 75
2.5/
√
15 = −1.239,
with 14 degrees of freedom. The two-sided p-value is approximately 0.236. Since p > 0.05, fail to reject
H0; the data do not provide significant evidence that µ ̸= 75.
5 marks; A:3, C:1, D:1
3(e). It is the probability, assuming H0, of obtaining a test statistic at least as extreme as the observed
one in the direction(s) specified by H1. Small values are evidence against H0, not the probability that
H0 is true. For a continuous test statistic and an exact p-value, the p-value is U (0, 1) under H0.
3 marks; B:2, C:1
Question total: 20 marks
4
Correlation, regression diagnostics, bootstrap and MLE
MATH40005 Probability and Statistics Resit Preparation
Question 4
4(a). Suppose X ∼ N (5, 9) and Y ∼ N (4, 4) with Corr(X, Y ) = −1/3. Let Z = X − 2Y . Compute
Corr(X, Z). (4 marks)
4(b). A simple linear regression has R2 = 0.94 and the following residual plot. Does it fit well? Suggest
an improvement.
0 2 4 6 8 10 12 14 16 18 20
−1
−0.5
0
0.5
xi
Residual ˆεi
(3 marks)
4(c). Give a real-world example in which linear regression can investigate a relationship between two
quantities. (2 marks)
4(d). Two unrelated annual time series have sample correlation 0.95. Can causation be concluded? (2
marks)
4(e)(i). The data are {12, 18, 21, 27, 30}. Is {18, 18, 30, 12, 27} a bootstrap sample? Justify. (2 marks)
4(e)(ii). Is {12, 18, 21, 27} a bootstrap sample of the same data? (2 marks)
4(f). Independent observations come from U [θ, θ + 2]. Find all maximum likelihood estimates of θ. (5
marks)
T otal: 20 marks
5
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 4: solution and marking guide
4(a). Cov(X, Y ) = ( −1/3)(3)(2) = −2. Thus
Cov(X, Z) = 9 − 2(−2) = 13 ,
while
Var(Z) = 9 + 4(4) − 4 Cov(X, Y ) = 9 + 16 + 8 = 33 .
Therefore Corr(X, Z) = 13 /(3
√
33).
4 marks; A:2, B:1, C:1
4(b). No. The pronounced U-shape is systematic, so the errors are not behaving as independent mean-
zero noise despite the high R2. A nonlinear transformation or a polynomial term (for example x2) should
be considered.
3 marks; C:1, D:2
4(c). For example, house price versus floor area. The fitted slope quantifies the expected price change
per unit area and the fitted line can be used for prediction, subject to model assumptions.
2 marks; D:2
4(d). No. A high sample correlation may arise from a common time trend, confounding or coincidence.
Correlation alone does not establish a causal mechanism.
2 marks; C:2
4(e)(i). Yes. It has the same size as the original and every value is sampled from the original, with
repetition allowed.
2 marks; B:2
4(e)(ii). No. A nonparametric bootstrap sample must have the same sample size, here five.
2 marks; B:2
4(f). The likelihood equals 2−n when θ ≤ xi ≤ θ + 2 for every i, and 0 otherwise. The constraints are
max
i
xi − 2 ≤ θ ≤ min
i
xi.
If this interval is nonempty, every point in it is an MLE; otherwise the observed sample is incompatible
with the model.
5 marks; B:2, D:3
Question total: 20 marks
5