Question 1
MATH40005 Probability and Statistics Resit Preparation
Question 1
1(a). A fair four-sided die is rolled and then a fair coin is tossed. Describe a probability space (Ω, F, P) for the experiment.
(3 marks)
1(b). Assume birthdays are independent and uniform over 365 days. Find the probability that among 12 people exactly one
pair shares a birthday and all other birthdays are distinct. (5 marks)
1(c). A disease has prevalence 0.08. A test has sensitivity 0.85 and false-positive rate 0.10. Find P(D | +). (4 marks)
1(d). A nine-character code contains exactly four letters, three digits and two distinct symbols selected from six symbols.
Letters and digits may repeat. Count the codes. (4 marks)
1(e). T wenty-two identical units are distributed among six labelled teams, each receiving at least two. Count the allocations.
(4 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, 4} × {H, T }, F = P(Ω).
All eight outcomes are equally likely, so for A ⊆ Ω define P(A) = |A|/8.
3 marks; A:3
1(b). Choose the two people in the pair, choose their shared day, and assign distinct different days to the other ten people:
P =
12
2
365 (364)10
36512 =
12
2
365 · 364 · 363 · · · 355
36512 .
5 marks; A:2, C:1, D:2
1(c). The positive probability is
P(+) = 0 .85(0.08) + 0.10(0.92) = 0 .16.
Therefore
P(D | +) = 0.85(0.08)
0.16 = 0.425.
4 marks; A:1, B:1, D:2
1(d). Choose the positions and then the characters. The two distinct symbols can be assigned to their ordered positions in
6 · 5 ways. Hence the count is
9!
4!3!2! 264 103 (6 · 5).
4 marks; A:2, B:2
1(e). After giving two units to each team, distribute the remaining ten without restriction. The number is
10 + 6 − 1
6 − 1
=
15
5
.
4 marks; A:2, B:1, C:1
Question total: 20 marks
2
Question 2
MATH40005 Probability and Statistics Resit Preparation
Question 2
2(a). For x, y ≥ 0, let
FX,Y (x, y) = (1 − e−x)(1 − e−3y),
and let the joint CDF be zero whenever x < 0 or y < 0. Find the marginal CDFs and densities, identify the distributions, and
decide independence. (6 marks)
2(b). If U ∼ U(0, 1) and λ > 0, identify the distribution of T = − log U /λ. (4 marks)
2(c). Let N ∼ Poi(λ) and let Y1, Y2, . . . be i.i.d., independent of N , with mean m and variance v. For S = PN
i=1 Yi (with
S = 0 if N = 0), derive ES and Var(S). (6 marks)
2(d). Let G ∼ Geom(p) on {1, 2, . . .}. Find its probability generating function and state its domain of convergence.
(4 marks)
Total: 20 marks
3
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 2: solution and marking guide
2(a). Letting the other argument tend to infinity gives
FX (x) = 1 − e−x, f X (x) = e−x, F Y (y) = 1 − e−3y, f Y (y) = 3 e−3y
for nonnegative arguments. Thus X ∼ Exp(1) and Y ∼ Exp(3). Since FX,Y = FX FY , the variables are independent.
6 marks; A:3, B:2, C:1
2(b). For t ≥ 0,
FT (t) = P(− log U ≤ λt) = P(U ≥ e−λt) = 1 − e−λt.
Thus T ∼ Exp(λ).
4 marks; A:2, B:1, D:1
2(c). Conditionally on N ,
E(S | N ) = N m, Var(S | N ) = N v.
Therefore
ES = E(N m) = λm,
and by the law of total variance,
Var(S) = E(N v) + Var(N m) = λv + λm2 = λE(Y 2
1 ).
6 marks; A:1, B:1, C:2, D:2
2(d). For |s(1 − p)| < 1,
GG(s) =
∞X
k=1
skp(1 − p)k−1 = ps
1 − (1 − p)s .
Thus the power series converges for |s| < 1/(1 − p) (and in particular for |s| ≤ 1).
4 marks; A:2, B:1, D:1
Question total: 20 marks
3
Question 3
MATH40005 Probability and Statistics Resit Preparation
Question 3
3(a). For a normal sample of size 16, the realised sample variance is s2 = 10 . Compute a 90% confidence interval for σ2.
You may useχ2
15,0.05 = 7.261 and χ2
15,0.95 = 24.996. (4 marks)
3(b). Two independent normal samples are assumed to have a common unknown variance. Sample 1: n = 12 , ¯x = 72 ,
s2
x = 9. Sample 2: m = 10, ¯y = 68, s2
y = 16. Test H0 : µ1 = µ2 against a two-sided alternative, calculate the p-value, and
conclude at level 0.05. You may use t20,0.975 = 2.086. (7 marks)
3(c). Forty hypotheses are tested with family-wise level 0.05. Using Bonferroni, state the per-test threshold and determine
which of
0.0008, 0.0015, 0.00002, 0.010
are significant. (4 marks)
3(d). For the ordered data
4.1, 4.3, 4.8, 5.0, 5.2, 5.4, 5.5, 5.7, 5.9, 6.1, 9.5,
exclude the median when forming the lower and upper halves. Compute the median, quartiles and IQR, and identify Tukey
outliers. (5 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 interval is 15(10)
24.996 , 15(10)
7.261
≈ (6.00, 20.66).
4 marks; A:3, B:1
3(b). The pooled variance is
s2
p = 11(9) + 9(16)
20 = 12.15.
The standard error is s
s2p
1
12 + 1
10
≈ 1.4925,
so
t = 72 − 68
1.4925 ≈ 2.680,
with 20 degrees of freedom. The two-sided p-value is approximately 0.0144. Since this is below 0.05 (and |t| > 2.086),
reject H0.
7 marks; A:1, B:2, C:1, D:3
3(c). The Bonferroni threshold is 0.05/40 = 0 .00125. Therefore 0.0008 and 0.00002 are significant; 0.0015 and 0.010 are
not.
4 marks; A:2, B:1, C:1
3(d). The median is 5.4. The lower-half median is q0.25 = 4 .8 and the upper-half median is q0.75 = 5 .9. Thus IQR = 1 .1,
with fences
4.8 − 1.5(1.1) = 3 .15, 5.9 + 1.5(1.1) = 7 .55.
Hence 9.5 is the only outlier.
5 marks; A:2, B:2, D:1
Question total: 20 marks
4
Question 4
MATH40005 Probability and Statistics Resit Preparation
Question 4
4(a). Independent observations are modelled as U [θ1, θ2], where 0 < θ 1 < θ 2 are unknown. Assuming the observed sample
is not constant, find the MLEs of θ1 and θ2. (6 marks)
4(b). In simple linear regression with fixed xi and independent errors of variance σ2, show that
Cov(Yi, bβ1) = σ2(xi − ¯x)
Sxx
.
(5 marks)
4(c). Interpret the normal Q–Q plot below. Does it support a normal model? (3 marks)
4(d). Describe a percentile bootstrap confidence interval for a population correlation coefficient. (3 marks)
4(e). Why can two annual time series have a very high sample correlation even when neither influences the other? (3 marks)
−3 −2 −1 0 1 2 3
−4
−2
0
2
4
Theoretical quantiles
Sample quantiles
Total: 20 marks
5
Worked solution and marking guidance
MATH40005 Probability and Statistics Resit Preparation
Question 4: solution and marking guide
4(a). The likelihood is
L(θ1, θ2) = ( θ2 − θ1)−n1{θ1 ≤ x(1), x (n) ≤ θ2}.
Within the feasible region, it is maximised by making the interval as short as possible, so
bθ1 = x(1) = min
i
xi, bθ2 = x(n) = max
i
xi.
6 marks; A:2, C:2, D:2
4(b). Since
bβ1 = 1
Sxx
nX
j=1
(xj − ¯x)Yj,
we obtain
Cov(Yi, bβ1) = 1
Sxx
X
j
(xj − ¯x) Cov(Yi, Yj).
Independence makes all terms with j ̸= i zero, while Var(Yi) = σ2. Hence the result follows.
5 marks; A:1, B:1, C:2, D:1
4(c). The middle observations are close to the reference line, but both tails bend away strongly: the lower tail is too low and
the upper tail too high. This is evidence of heavier tails than a normal distribution, so the plot does not support a normal
model.
3 marks; B:2, D:1
4(d). Resample the observed pairs (xi, yi) with replacement, always keeping each pair intact. Compute the sample correlation
for each bootstrap sample. The empirical 2.5% and 97.5% quantiles of these bootstrap correlations form a percentile 95%
interval.
3 marks; A:1, B:1, C:1
4(e). Both may share a time trend, respond to a third factor, or simply align over a short interval. Correlation does not identify
a causal mechanism and is especially vulnerable to spurious association in trending time series.
3 marks; A:2, B:1
Question total: 20 marks
5