N ATIONAL U NIVERSITY OF S INGAPORE
D EPARTMENT OF S TATISTICS AND DATA S CIENCE
ST2334 P ROBABILITY AND S TATISTICS
M ID - SEMESTER T EST S AMPLE PAPER 2
(S EMESTER I, AY 2023/2024)
T IME A LLOWED : 60 MINUTES
Suggested Solutions
I NSTRUCTIONS TO S TUDENTS
1. Please write your student number only. Do not write your name.
2. This assessment contains 15 questions and comprises 7 printed pages.
3. The total marks is 25; marks are equal distributed for all questions.
4. Please answer ALL questions.
5. Calculators may be used.
6. This is an OPEN BOOK assessment. Only HARD COPIES of materials are allowed.
ST2334
1. M ULTIPLE C HOICE : C HOOSE ONE ANSWER ONLY
A ∪ (B ∩C) =?
(a) (A ∪ B) ∩ (A ∪C) (c) A ∪ B′ ∪C′
(b) (A ∪ B) ∩C (d) (A ∩ B) ∪ (A ∩C)
S OLUTION
(a)
2. F ILL IN THE BLANK
How many ways are there to choose an arbitrary number of students (including the pos-
sibility of choosing 0 student) from 6 students?
(Provide your answer in numerical form.)
S OLUTION
For each student, there are 2 possibilities: “chosen” or “not chosen”. So the total number
of possibilities is 2 × 2 × 2 × 2 × 2 × 2 = 26 = 64.
3. F ILL IN THE BLANK
Suppose
P(A′ ) = 1/2, P(B) = 3/8, and, P(B′ |A) = 3/4.
Find P(B ∩ A).
(Provide your answer in decimal form and round it to three decimal places if necessary)
S OLUTION
0.125.
3 1 3
P(B′ ∩ A) = P(B′ |A)P(A) = × = ,
4 2 8
1 3 1
P(B ∩ A) = P(A) − P(B′ ∩ A) = − = = 0.125.
2 8 8
4. F ILL IN THE BLANK
A group of 8 friends A,B,C,D,E,F,G,H go to a restaurant. Due to safe-distancing mea-
sures, the group needs to split up into two groups of 4. How many ways are there to split
the group such that A and B are together but away from C?
(Provide your answer in numerical form.)
S OLUTION
Except A, B, C, there are 5 people left. The group with A and B only has two more slots,
the groups are set if and only if we select two more people out of 5 to fillinthe slots of
5
the group with A, B, and the rest 3 are with C. So the number of ways is = 10.
2
2
ST2334
5. M ULTIPLE R ESPONSE : C HOOSE ALL ANSWERS THAT APPLY
Let A and B be two events. Which of the following statements is/are true?
(a) If A ̸= B, then P(A) ̸= P(B).
(b) If A and B are independent, then we must have P(A∪B) = 1−{1 − P(A)} {1 − P(B)}.
(c) If P(A) = 1 − P(B′ ), then P(A) = P(B).
(d) (A ∩ B′ ) ∪ (A′ ∩ B) = 0,
/ then A = B.
S OLUTION
Answer: (b), (c), (d).
6. M ULTIPLE C HOICE : C HOOSE ONE ANSWER ONLY
Consider the following statements about Peter whom you have not met before.
(A): He is not married. (C): He is married.
(B): He is not married and smokes. (D): He is married and does not smoke.
You are to assign probabilities to these statements. Which answer below is consistent
with the laws of probability?
(a) P(A) = 0.45, P(B) = 0.5, P(C) = 0.55, P(D) = 0.4
(b) P(A) = 0.45, P(B) = 0.1, P(C) = 0.6, P(D) = 0.3
(c) P(A) = 0.45, P(B) = 0.2, P(C) = 0.55, P(D) = 0.5
(d) P(A) = 0.45, P(B) = 0.4, P(C) = 0.55, P(D) = 0.6
S OLUTION
Answer: (c).
7. T RUE /FALSE
Let A and B be mutually exclusive events. If P(A) = 0.1, P(B) = 0.01, then A and B are
not independent.
• TRUE
• FALSE
S OLUTION
TRUE; If otherwise, P(A ∩ B) = P(A)P(B) > 0, which contradicts that A and B are mu-
tually exclusive.
8. T RUE /FALSE
Cumulative distribution function can not take on values greater than 1 or smaller than 0.
• TRUE
3
ST2334
• FALSE
S OLUTION
TRUE
9. F ILL IN THE BLANK
Suppose that random variable X has the cumulative distribution function given by
0, x < 0
2
x
F(x) = , 0≤x≤3
1,9 x > 3
Compute P(X = 1.5).
(Provide your answer in decimal form and round it to two decimal places if necessary.)
S OLUTION
From the c.d.f., it is a continuous random variable. Therefore P(X = 1.5) = 0.
10. F ILL IN THE BLANK
Let X be a random variable, whose cumulative distribution function is given by
0, x<0
0.2, 0 ≤x<2
F(x) = 0.6, 2 ≤ x < 3 .
0.7, 3 ≤ x < 5
1 x≥5
Compute E(X).
(Provide your answer in decimal form and round it to two decimal places if necessary.)
S OLUTION
X is a discrete RV, whose p.m.f. is given by f (x) = 0.2, 0.4, 0.1, 0.3, for x = 0, 2, 3, 5.
E(X) = 0 × 0.2 + 2 × 0.4 + 3 × 0.1 + 5 × 0.3 = 2.6.
11. F ILL IN THE BLANK
Let X have probability mass function given by the following table.
x 0 2 5 6
f (x) 0.3 0.5 0.1 0.1
Compute E(X).
(Provide your answer in decimal form and round it to two decimal places if necessary.)
S OLUTION
E(X) = 0(0.3) + 2(0.5) + 5(0.1) + 6(0.1) = 2.1.
4
ST2334
12. M ULTIPLE C HOICE : C HOOSE ONE ANSWER ONLY
Let X be a random variable. Which of the following statement is INCORRECT?
(a) If P(X = 1) = 0.1 and E(X) exists, then we must have E(X 2 ) > (E(X))2 .
(b) If V (X) > 0, then for any x, P(X = x) < 1.
(c) By the definition of the random variable, the range of X is a subset of R; therefore, it
is impossible that P(X = x) = 0 for any x ∈ R.
(d) There are cases under which E(X) does not exist.
S OLUTION
Answer: (c)
• (a) is correct because V (X) = E(X 2 ) − [E(X)]2 and P(X = 1) = 0.1 implies V (X) >
0 (since otherwise P(X = E(X)) = 1).
• (b) is correct with similar reason to (a).
• For any continuous RV, we must have P(X = x) = 0 for any x ∈ R.
• For example, X is a random variable with p.d.f.
1
(
, for |x| ≥ 1,
f (x) = 2x2
0, elsewhere
Then
Z 1
1 1
Z ∞
E(X) = x · 2 dx + x· dx = −∞ + ∞,
−∞ 2x 1 2x2
which implies E(X) does not exist.
13. F ILL IN THE BLANK
A service station has both self-service and full-service islands. On each island, there is a
single regular unleaded pump with two hoses. Let X denote the number of hoses being
used on the self-service island at a particular time, and let Y denote the number of hoses
on the full-service island in use at that time. The joint probability mass function of X and
Y is given in the table below.
y
x
0 1 2
0 0.10 0.04 0.02
1 0.08 0.20 0.06
2 0.06 0.14 0.30
Compute E(X|Y = 1).
(Provide your answer in decimal form and round it to two decimal places if necessary.)
S OLUTION
1.26.
5
ST2334
Focusing on the column of y = 1, the sum of these numbers leads to P(Y = 1) = 0.04 +
0.20 + 0.14 = 0.38. So, we obtain the conditional distribution of X|Y = 1:
P(X = 0|Y = 1) = 0.04/0.38;
P(X = 1|Y = 1) = 0.20/0.38;
P(X = 2|Y = 2) = 0.14/0.08.
Therefore, we obtain
E(X|Y = 1) = 0(0.04/0.38) + 1(0.20/0.38) + 2(0.14/0.38) = 0.48/0.38 = 1.26.
14. T RUE /FALSE
Let f (x, y) be the joint probability function of a discrete random vector (X,Y ). If fX (1) =
0, then f (1, y) = 0 for any y being a real number.
• TRUE
• FALSE
S OLUTION
TRUE
fX (1) = 0 implies ∑ f (1, y) = 0, but f (1, y) ≥ 0; therefore f (1, y) = 0 for all y ∈ RY ,
y∈RY
and thus f (1, y) = 0 for all real numbers y.
15. The joint probability function of (X,Y ) is given by
1
(
(x + y) 0 ≤ x ≤ 2; 0 ≤ y ≤ 2
f (x, y) = 8
0 elsewhere
Compute P(Y ≥ 1|X ≥ 1).
(Provide your answer in decimal form and round it to three decimal places if necessary.)
S OLUTION
0.6.
We need to compute
P(Y ≥ 1; X ≥ 1)
P(Y ≥ 1|X ≥ 1) = .
P(X ≥ 1)
We shall evaluate the numerator and denominator separately.
The marginal density of X is given by
Z 2
1 1 1
fX (x) = (x + y)dx = (2x + 2) = (x + 1).
0 8 8 4
Therefore
Z 2
1
P(X ≥ 1) = (x + 1)dx = 0.625.
1 4
6
ST2334
For the numerator:
Z 2Z 2
1
P(X ≥ 1;Y ≥ 1) = (x + y)dxdy = 0.375.
1 1 8
As a consequence
0.375
P(Y ≥ 1|X ≥ 1) = = 0.6.
0.625
E ND OF PAPER
