ACTL1101: 00001 3
1. Problem
You must create a team of four people, formed of: one General, one Captain, and two
Admin Staff (whose roles are exactly identical). There are 7 individuals available, and they
can all do every role. How many different teams are possible?
Solution
You need to pick 4 people out of 7. After doing that, you need to assign the roles (but two
roles are identical), so in total you have
7!
3!4!
× 4!
2!
= 420.
2. Problem
Josephine is making drinks (again). She has at her disposal 5 alcoholic beverages and 8
non-alcoholic beverages. If a drink must be made of exactly one alcohol and at most two
non-alcohols (including the possibility of 0 non-alcohols), how many different drinks can she
make?
Solution
There are 5 possibilities for the alcohol and 1+8+8·7/2 = 37 possibilities for the non-alcohol
(the ‘1’ is the possibility of 0 non-alcohol). So, in total:
5× 37 = 185.
3. Problem
Find Pr[X ≤ 0.83] for X a continuous random variable (0 ≤ X ≤ 1) with density: f(x) =
2.3333333x1.3333333, for 0 ≤ x ≤ 1. [Express your answer as a decimal number from 0 to 1,
e.g. 0.5, NOT 50%, and NOT 1/2]
Solution
We have that F (x) = x2.3333333, hence, F (0.83) = 0.64741.
4. Problem
Let X ∼ Binomial(n, p). You know that the expectation of X is 15.6 and its standard deviation
is 2.498. What is the numerical value of p? [Express your answer as a decimal number from
0 to 1, e.g. 0.5, NOT 50%, and NOT 1/2]
Solution
We have
Var[X]
E[X]
=
np(1− p)
np
= 1− p,
so p = 1− σ2X /E[X] = 0.6.
5. Problem
In the course ACTN1001, 18% of students got above 85 in their Quiz and 25% of students
got a HD as their final grade. Also, 15% of students got above 85 in their Quiz and a HD.
Given that a student has not obtained above 85 in their Quiz, what is the probability this
student got a HD? [Express your answer as a decimal number from 0 to 1, e.g. 0.5, NOT
50%, and NOT 1/2]
Solution
We have
Pr[HD|QZC ] = Pr[HD ∩QZ
C ]
Pr[QZC ]
=
0.25− 0.15
1− 0.18 = 0.1219512.
6. Problem
An insurer created four categories of drivers and their percentages in the population are as
follows: good (50%), mediocre (30%), bad (15%), dangerous(5%). On average per year,
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those types of driver claim to the insurer the following amounts: 0 (good), 600 (mediocre),
1900 (bad), 7000 (dangerous). What is the expected per driver yearly insurance claim in
this population?
Solution
We use the Law of Total Expectation:
E[C] = E[C|good]P[good] + E[C|med]P[med] + E[C|bad]P[bad] + E[C|dan]P[dan] = 815
7. Problem
A sample {X1, X2, X3, X4, X5, X6} stemming from an Poisson(λ) distribution is used to
estimate the parameter λ. If you observe
x1 = 9, x2 = 11, x3 = 7, x4 = 13, x5 = 10, x6 = 9
what is your best estimate of the value of λ?
(a) 59.00
(b) 9.83
(c) 33.48
(d) 0.10
(e) 69.40
Solution
We have that λ̂ = X¯, and here x¯ = 9.8333333, so that is our estimate.
(a) False
(b) True
(c) False
(d) False
(e) False
8. Problem
The rate of interest effective per year is 28%. What is the equivalent rate of interest effective
per month? [The choices are expressed in decimal values, i.e. 0.0500 ≡ 5.00%.]
(a) 0.0267
(b) 0.0153
(c) 0.0130
(d) 0.0208
(e) 0.0233
Solution
We simply must find i such that (1 + i)12 = 1 + 0.28, i.e.
(1.28)1/12 − 1 = 0.02078.
(a) False
(b) False
(c) False
ACTL1101: 00001 5
(d) True
(e) False
9. Problem
Your net salary is 5000 per year (paid at the end of each year). Because you live in your
parents’ basement, you have no expenses and invest all of you salary in Dogecoin, which
yields a constant rate of 18% per year (effective). What is the total value of your investments
after exactly 5 years (immediately after you receive the 5th year of salary and invest it)?
Solution
For i = 0.18 the effective yearly rate, we solve this by finding the present value of all the
salary money, then accruing it 5 years into the future:
5000 · (1− v5)/i · (1 + i)5 = 3.5771049 · 104.
10. Problem
A loan of amount 5000 is to be repaid with one single payment of 9500, exactly 9 years
from now. What is the continuously compounded interest rate (δ) in place?
(a) 0.0739
(b) 0.1000
(c) 0.0766
(d) 0.0807
(e) 0.0713
Solution
We have:
5000 · e9δ = 9500
δ =
1
9
· log(9500/5000) = 0.07132.
(a) False
(b) False
(c) False
(d) False
(e) True
11. Problem
You just bought a car of price L through a loan which you repay with monthly repayments (in
arrears) of 55.242, paid over a five year period (60 payments in total). The nominal interest
rate is 10% (annual), compounded monthly. What is L?
Solution
Let r = 0.1/12 be the effective rate per month, we solve:
L = 55.242 · 1− v
60
r
= 2600.
12. Problem
The effective rate of interest on a 6 years loan is 76% (effective over the whole period, not
per year). What is the annual continuously compounded rate (δ) in place?
(a) 0.09422
(b) 0.09880
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(c) 0.07129
(d) 0.09039
(e) 0.07382
Solution
We have:
e6δ = 1 + 0.76
δ = log(1.76)/6
δ = 0.09422.
(a) True
(b) False
(c) False
(d) False
(e) False
13. Problem
A researcher wants to establish if a new vaccine is effective at preventing a given disease.
She frames her hypothesis test as:
H0 : the vaccine is not effective versus H1 : the vaccine is effective.
In this context, which statement(s) is(are) TRUE?
(a) A Type II error means: failing to recognise the vaccine as effective, when in fact it is
(b) A Type I error means: falsely concluding the vaccine is effective
(c) If H0 is rejected, then H1 is accepted
(d) A Type I error means: concluding the vaccine is not effective, while in fact it is
(e) If H0 is not rejected, then H0 is accepted as true
Solution
(a) True
(b) True
(c) True
(d) False
(e) False
14. Problem
You have two packs of Skittles. The first pack contains a number n1 of the delicious candies,
and the second pack contains n2 of them, where n2 > n1. Every Skittle can be red, or
not, so that the total number of red Skittles in any given pack of size n is modelled by
a Binomial(n,p) random variable, where p > 0 is an unknown parameter. Denote by X
the (random) number of red Skittles in the first pack, and Y the (random) number of red
Skittles in the second pack. Out of the following list of estimators for p, which one(s) is(are)
unbiased?
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(a) Yn1
(b) Y−Xn1+n2
(c) X+Yn1
(d) Xn2
(e) Yn2
Solution
(a) False
(b) False
(c) False
(d) False
(e) True
15. Problem
A sequence of independent random variables X1, . . . , Xn stems from the Uniform(0, θ) dis-
tribution, where θ is an unknown parameter. Consider the following two estimators for θ:
θ̂n = 2X¯, θ̂
∗
n = maximum(X1, . . . , Xn)
Assume that the sample size n ≥ 4. Which statement(s) is(are) TRUE (select all which
apply)?
(a) MSE[θ̂n] = Var[θ̂n]
(b) The variance of θ̂∗n is smaller than that of θ̂n.
(c) MSE[θ̂∗n] = Var[θ̂∗n]
(d) Both estimators are asymptotically unbiased.
(e) θ̂∗n is biased.
Solution
(a) True
(b) True
(c) False
(d) True
(e) True
16. Problem
Right after Valentine’s day, Josephine decides to take a pregnancy test. We can picture this
situation as a hypothesis test where:
H0 : she is not pregnant versus H1 : she is pregnant.
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The test returns ‘not pregnant’. However, some time mid-November Josephine gives birth
to a beautiful baby (and it is known she did not engage in sexual activities after taking the
test). Which statement(s) is(are) TRUE (select all which apply)?
(a) A Type I error has been committed.
(b) A Type II error has been committed.
(c) A Type I error has not been committed.
(d) The p-value of the pregnancy test was lower than the significance level.
(e) Both a Type I and a type II errors have been committed.
Solution
(a) False
(b) True
(c) True
(d) False
(e) False
17. Problem
Let n ≥ 3 be an integer. Let y be the present value of a regular annuity (in arrears, payments
of 1) for n periods. Let z be the present value of an annuity due (payments of 1) for n periods.
Both are computed on the same effective interest rate i > 0. Given this information, which
of the following statement(s) is(are) necessarily TRUE (select all which apply)?
(a) y + 1 = z
(b) y + z < n
(c) y + 1 > z
(d) y + z < 2n
(e) y < z
Solution
(a) False
(b) False
(c) True
(d) True
(e) True
18. Problem
The time it takes (in months) for a new update of your favourite video game to be released
is modelled by a random variable X > 0 with:
fX(x) =
β
2
√
x
e−β
√
x
for x > 0, where β > 0 is an unknown parameter.
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(a) (1pt) Show that the CDF, FX(x), has the following expression:
FX(x) = 1− e−β
√
x.
(b) (2pt) Only for this sub-question, assume that β = 1.6. If an update of the game was
just released, what is the probability you need to wait more than 1 month (X > 1) but
less than 3 months (X < 3) for the next update to be released?
(c) (2pt) Consider an independent sample X1, . . . , Xn of ‘time-for-updates’. Considering
that E[X] = 2/β2, propose an estimator for the parameter β, and explain your choice.
(d) (3pt) Again consider an independent sample X1, . . . , Xn, with n = 19. You use the
sample mean X¯ as a test statistic to test:
H0 : β = 1 versus H1 : β < 1.
You observe a sample mean of x¯ = 3.11 from that sample. At level of significance α = 0.05,
conduct the hypothesis test. You must include the numerical computation of the p-value and
your final decision. You can consider that the sample mean X¯ is Normally distributed, and
also that Var[X|H0] = 20.
Solution
(a) The quickest way is to check the derivative of F (x), which is
F ′(x) = −β
2
x−1/2(−e−β
√
x) =
β
2
√
x
e−β
√
x = f(x).
This is sufficient, though we could also go the other way:
F (x) =
∫ x
0
f(t)dt
=
β
2
∫ x
0
1√
t
e−β
√
tdt
And via a change of variable u =
√
t which yields 2du = 1√
t
dt, we get
F (x) =
β
2
∫ √x
0
2e−βudu
= β
(
− 1
β
e−βu
) ∣∣∣√x
0
= −(e−β
√
x − 1) = 1− e−β
√
x.
(b) This is given by
Pr[1 < X < 3] = F (3)− F (1)
= −e−β
√
3 + e−β
√
1
= −0.0625818 + 0.2018965
= 0.1393148
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(c) We solve the equation for β, i.e.
E[X] = 2/β2 =⇒ β2 = 2/E[X] =⇒ β =
√
2/E[X].
Next, because we can expect the sample mean X¯ to be fairly close to the theoretical
expectation E[X], we obtain our estimator by substituting E[X] by X¯ in the above
equation, i.e.
β̂ =
√
2
X¯
.
(d) Under H0, the variance of X¯ is 20n . Large values of our test statistic would make
us reject H0, in favor of H1 (because a large expectation corresponds to a small β).
Hence, the p-value is:
p-value = Pr[X¯ ≥ 3.11|H0]
= Pr
[
Z ≥ 3.11− 2√
20/n
|H0
]
= Pr [Z ≥ 1.0818942]
= 0.1396498
Since the p-value is BIGGER than the significance level (0.05), then we DO NOT
REJECT H0.