CMI Entrance Exam 2026 Question Paper Solution M.Sc Data Science

FOR CMI EXAM PREPARATION

CMI 2026
Question Paper
Solution · M.Sc Data
Science
EXAM YEAR TYPE SUBJECT

CMI 2026 Question Paper Solution M.Sc Data Science

Notes · Sample Papers · Previous Year Papers · Mock Tests

m
m .co

m .co CHENNAI MATHEMATICAL INSTITUTE
s em
se g la 2026
M.Sc. Data Science Entrance Examination
Solutions
a
Part (A) - Multiple-choice questions

1. An insurance company offers a special policy covering two independent risk events in a policyholder’s life
over one year: a medical emergency and a vehicle accident. Based on historical actuarial data:
1 1
P (medical emergency) = , P (vehicle accident) =
10 20
A claim is considered valid if at least one of the two events occurs during the year. Each policyholder
m
m
pays an annual premium of Rs. 4,800, and if a valid claim is filed, the company pays out Rs. 6,000 (no
.co
.co m
payout otherwise). Which of the following statements are correct?

m s e
s e 29
(a) The probability of valid claim is 200 .
l a
g that she
l a
(b) A single policyholder
files a g
a
renews the same policy for four consecutive years. The probability
a valid claim in exactly one ofthe four years
 
equals

3 1
4 29 171
≈ 0.0095
1 200 200
(c) A family of four members each independently hold this policy for a year. The probability that
exactly one of the four members files a valid claim equals
    
4 29 1 171 3
≈ 0.3257
1 200 200
m
.co
(d) If 800 policyholders are enrolled under this policy in a given year, the expected profit of the insurance

m
e
company for the year equals Rs. 19,44,000.

Solution: (a), (c)
l as
(a) Finding P (valid claim) ag
A claim is valid if at least one event occurs. Using the complement rule:
  
1 1 9 19 171
P (no claim) = P (no medical emergency)×P (no vehicle accident) = 1 − 1− = × =
10 20 10 20 200
171 29
∴ =
P (valid claim) = 1 −
200 200
1 1 30
Simply adding 10 + 20 = 200 overcounts the case where both events occur simultaneously. The correct
approach via inclusion-exclusion gives the same result:
o m
m 29 .c
co m
1 1 1 1 20 10 1
. P (A ∪ B) = + −
10 20 10 20
× = +
200 200 200
− =
s e
200

s em (b) The correct expression for P(exactly one claim in 4 years) is: gla
g la a
a
     1 3
4 29 171
≈ 0.3257
1 200 200
instead the option is given as
    
4 29 3 171 1
≈ 0.0095
1 200 200
(c) P(exactly one of the four members files a valid claim ) is correct.
(d) Per policy the revenue is:

om .
171 29
m
E = (+4800) × + (−1200) × = Rs. 3, 930
c is 800 × 3930 = 31, 44, 000. So (d) is wrong
.policies 200 200
s e
e m
So Expected revenue for 800
l a
l as ag
ag 1
For more Question Papers, Sample Papers, Notes & Syllabus visit Page 1 of 13

2. Consider the single variable, real-valued function f (x) = x3 − 3x. Which of the following statements
about f are true?

(a) f is increasing for all x.
(b) f has exactly 2 critical points.
(c) f has exactly one inflection point.
(d) f is bounded below.

Solution: (b), (c)

3. There are 10 switches labeled 1 to 10, all initially turned OFF. You perform the following operation: at
step k, you toggle (an ON switch is turned OFF and vice versa) every switch whose label is divisible by
k. After performing 10 such steps, which switches are in ON position?

(a) 1, 4, 9
(b) 2, 4, 6, 8
(c) 1, 4, 9, 10
(d) 1, 3, 6, 9

Solution: (a)
Only switches numbered by perfect squares will be ON: 1, 4, 9. Note that a switch will be ON after an
odd number of steps. So we are looking for numbers that have an odd number of divisors including 1
and itself, and are less than 10.

4. Which of the following inequalities involving real valued functions are true?

(a) | sin x| ≤ |x|, ∀x ∈ R
√ √
(b) e x ≤ x, ∀x ≥ 0
(c) ln(1 + x) ≤ x, ∀x > −1
(d) tan x ≥ x, for all x where defined

Solution: (a), (c)

5. Let S be the set of all functions from {1, 2, 3, 4} → {1, 2, 3}. Which of the following are true?

(a) The number of functions in S whose image has exactly 2 elements is 3 · (24 − 2)
(b) The number of functions in S whose image has exactly 3 elements is 34 − 3 · 24 + 3
(c) The number of injective functions in S is nonzero
(d) The number of surjective functions in S with f (1) = 1 and f (2) = 2 is 5.

Solution: (a), (b), (d)
(a) is true: choose 2-element images and count the onto functions – 24 − 2. (b) is true by inclusion-
exclusion. (c) is false since domain size > codomain size. (d) is true: Since f (1), f (2) are already
determined, valid pairs for (f (3), f (4)) are (3, 3), (3, 1), (3, 2), (1, 3), (2, 3).

6. A box contains 3 red balls and 2 blue balls. Two balls are drawn one after another without replacement.
Let A be the event that both balls are red, and B the event that the first ball is red. Which of the
following are true?
3
(a) P (A) = 10
(b) P (B) = 53
(c) P (B c | Ac ) = 32 (here X c is the event that X does not take place).
(d) P (A | B) = 35

2 Notes & Syllabus visit
For more Question Papers, Sample Papers, Page 2 of 13

m
m .co

c. o Solution: (a), (b) s em
m la
se 3 2 3 3
P (A) = · = , so (a) is true. P (B) = , so (b) is true.
g
= . The event B ∩ A is a
5 4 10 5
c 3 7 c c
For (c), note that P (A ) = 1 − 10 10 the event that the first ball is blue (since
if the first ball is blue, both balls certainly cannot be red), so P (B c ∩ Ac ) = P (B c ) = 52 . Hence

P (B c ∩ Ac ) 2/5 4 2
P (B c | Ac ) = = = ̸= ,
P (Ac ) 7/10 7 3

so (c) is false.
For (d), P (A | B) = P P(A∩B) P (A) 3/10 1 3
(B) = P (B) = 3/5 = 2 ̸= 5 , so (d) is false.
m
m .co
7. Five people, A, B, C, D, E sit in a row. Two particular people, A and B, refuse to sit next to each other.

c o
Which of the following are true?
. e m
(a) The number ofm
e valid arrangements is 5! − 4! · 2
l as
l as
(b) The probability that A and B are not adjacent is 3
a g
g of arrangements (where A and B are not adjacent) in which A and D sit next to each
5

a
(c) The number
other is 36
(d) The number of arrangements (where A and B are not adjacent) in which D and E sit next to each
other is 48.

Solution: (a), (b), (c)
The total number of arrangements is 5! = 120. Treating A and B as a block gives 4! · 2 = 48 arrangements
together. So the arrangements where they are not together is 120 − 48 = 72; thus (a) is true. The
m
.co
probability of this happening is= 72/120 = 3/5, so (b) is also true. There are 48 arrangements where
A, D can sit besides each other. Now, out of these arrangements that contain DA, B or B, AD are not

em
allowed, these are 12 in number hence (c) is also correct. Option (d) is false, because we have to subtract
s
lafollowing statements are always true?
arrangements where A, B are siting next to each other.

8. Let A be an n × n real matrix. Which ofgthe
a
(a) If A is invertible, then the equation Ax = b has a unique solution for every b ∈ Rn .
(b) If Ax = 0 has only the trivial solution, then A is invertible.
(c) If A2 = 0, then A = 0.
(d) If AB = 0 for some n × n matrix B ̸= 0, then A is not invertible.

Solution: (a), (b), (d)
For option (c) there are straightforward counterexamples.
o m
9. Howm c
. integers?
c. (a)o 20 m
many 3-element subsets of {1, 2, . . . , 8} contain exactly one pair of consecutive

m s e
s e (b) 36 g l a
g la a
a (c) 30
(d) 56

Solution: (c)
There are exactly 7 pairs of the form (i, i + 1). We split the counting in 2 cases. First we count the
subsets that contain either {1, 2} or {7, 8}. In order to extend these subsets to a 3-element subset there
are exactly 5 choices each (since for the first pair 3 is not allowed for the second 6 is not allowed). Then
we count subsets containing one of the remaining 5 pairs. In this case, the third element can neither be

m
i − 1 nor i + 2. So we are left with 4 choices. The total count is (2 × 5) + (5 × 4) = 30.
.
.co s e m
em
10. Let a and b be the outcomes of two independent rolls of a fair die. Define X = max{a, b}. Which of the
following are true?
s l a
g la ag
a 3 Notes & Syllabus visit
For more Question Papers, Sample Papers, Page 3 of 13

(a) P (X = 6) = 11
36
1
(b) P (X = 1) = 36
(c) E[X] > 4
8
(d) P (X ≤ 3) = 36

Solution: (a), (b), (c)
11 1
P (X = 6) = 36 and P (X = 1) = 36 are standard counts, so (a), (b) are true. One can compute
E[X] ≈ 4.47 > 4, so (c) is true. For X ≤ 3, both rolls must be ≤ 3, giving 9/36, so (d) is false.

11. Starting with the number n = 1, we generate a sequence of numbers. In the second step we replace 1 by
either 2 = 2 × 1 or 3 = 2 × 1 + 1. In general, replace n by either 2n or 2n + 1 to get a new value for n .
Which of the following numbers can be obtained as values of n in this fashion?

(a) 7
(b) 10
(c) 15
(d) 2026

Solution: (a), (b), (c), (d)
Every positive integer can be obtained by reversing the process: if even divide by 2, if odd subtract 1,
eventually reaching 1. Hence all listed numbers are reachable.

12. Arrange the numbers 1, 2, 3, 4, 5 in a row. We say a number i, 1 ≤ i ≤ 5 is well-placed in the given
arrangement if it appears in position i, starting from the left. For example, in the arrangement 4, 1, 3, 2, 5,
the numbers 3 and 5 are well-placed. We are interested in arrangements where no number is well-placed.
How many such arrangements exist?

(a) 20
(b) 44
(c) 60
(d) 45

Solution: (b)
Use inclusion–exclusion:
Total number of arrangements is 5! = 120.
Let Ai be the set of arrangements where i is well-placed. Then

|Ai | = 4!, |Ai ∩ Aj | = 3!, |Ai ∩ Aj ∩ Ak | = 2!, etc.

Hence the number is          
5 5 5 5 5
5! − 4! + 3! − 2! + 1! − 0!
1 2 3 4 5
= 120 − 5 · 24 + 10 · 6 − 10 · 2 + 5 · 1 − 1
= 120 − 120 + 60 − 20 + 5 − 1 = 44.

13. Three statements are made:

• S1: “Exactly one of S2 and S3 is true.”
• S2: “S1 is false.”
• S3: “S2 is true.”

Which of the following are possible?

(a) Only S1 is true.

4 Notes & Syllabus visit
For more Question Papers, Sample Papers, Page 4 of 13

m
m .co

m .co (b) Only S2 is true.
s e m
se (c) Only S3 is true.
g l a
(d) Exactly two statements are true. a
Solution: (d)
If S1 is true, consistency fails. If S1 is false, then S2 and S3 must both be true, which is consistent. Thus
exactly two statements are true.

14. A subset (possibly empty) of {1, 2, 3, 4, 5, 6} is chosen uniformly at random. What is the probability that
the chosen subset contains at least one even number and at least one odd number?

m
.co
1
m
(a)

.co
2
3
e m
(b)
4
em l as
(c)
7
l as ag
(d)
8
49 ag
64
Solution: (d)
Total subsets = 26 = 64. Subsets with only even elements: 23 = 8. Only odd: 23 = 8. Empty counted
twice, subtract once: 8 + 8 − 1 = 15. Good subsets = 64 − 15 = 49. Probability = 49/64.

15. A student writes the following code to compute the average of numbers A[1], . . . , A[n]:
sum = 0
m
.co
for i = 1 to n :

em
sum = sum + A [ i ]
avg = sum / n
return avg
las
g
a this piece of code are correct?
Which of the following statements about

(a) The code always returns the correct average.
(b) The output depends on the order of the input values
(c) In general, the code doesn’t calculate the correct average.
(d) The variable avg is unnecessarily updated.

m
Solution: (a), (d)

m .co
The code correctly computes the average. The line that updates avg can safely be moved out of the loop.

.co
16. In the following code, the operator % denotes the remainder after integer division.
m function mystery ( n ) { s e m
s e g la
g la result = 0; a
a for i from 1 to n {
if ( i % 2 == 0) {
result = result + i ;
} else {
result = result - i ;
}
}
return ( result );
}
m .
What does mystery(10) return?
.co s e m
s em l a
ag
(a) −5

g la
a 5 Notes & Syllabus visit
For more Question Papers, Sample Papers, Page 5 of 13

(b) 5
(c) 10
(d) 0

Solution: (b)
The code computes 0 − 1 + 2 − 3 + 4 − 5 + 6 − 7 + 8 − 9 + 10 = 5.

Questions 17 to 19 are based on the following code.
The arguments to the function enigma(A, n) in the code below are: (i) an integer array A indexed from 0,
and (ii) the number n of elements in A.
function enigma (A , n ) {
for i from 0 to n - 2 {
for j from 0 to n - 2 - i {
if ( A [ j ] > A [ j +1]) {
temp = A [ j ];
A [ j ] = A [ j +1];
A [ j +1] = temp ;
}
}
}
return ( A [0]);
}
Answer the next three questions about this function.

17. What does enigma([3, 1, 4, 2, 5], 5) return?
(a) 1
(b) 3
(c) 4
(d) 5
Solution: (a)
The function sorts the array in ascending order and returns the first (smallest) element.
18. Which of the following statements about the function enigma(A,n) are true?

(a) After execution, the array A is sorted in non-decreasing order.
(b) The function returns the smallest element of the array.
(c) The function modifies the input array A.
(d) The function breaks down if the array contains repeated elements.

Solution: (a), (b), (c)
19. Which of the following statements about the number of operations performed in enigma(A,n) are true?
Note that assigning a value to a variable is not considered as an operation.

(a) The number of comparisons performed is proportional to n2 .
(b) The total number of operations depends on the initial order of the array.
(c) The function always performs the same number of comparisons for a given n.
(d) The function may terminate early if the array is already sorted.

Solution: (a), (c)
Options (a) and (c) are true, since this function is similar to the bubble sort and the loops are fixed.
Option (b) is false because the number of comparisons is fixed. Similarly, (d) is also false, all the
comparisons will be performed.

6 Notes & Syllabus visit
For more Question Papers, Sample Papers, Page 6 of 13

m
m .co

c. o20. The argument to the function steps(n) is a positive integer n. Thesoperator
e m % denotes the remainder
m after integer division. la
se function steps ( n ) { a g
count = 0;
while ( n != 1) {
if ( n % 2 == 0) {
n = n / 2;
} else {
n = 3 * n + 1;
}
count = count + 1;
m
}
m .co
.co e m
return ( count );
}
e m l as
la s
What does steps(6) return?
ag
(a) 6 a g
(b) 8
(c) 3
(d) 4

Solution: (b)
Trace: 6 → 3 → 10 → 5 → 16 → 8 → 4 → 2 → 1, which is 8 steps.

m
m .co
s e
l a
ag

m
m .co
m .co s e m
s e g la
g la a
a

m .
.co s e m
s em l a
g la ag
a 7 Notes & Syllabus visit
For more Question Papers, Sample Papers, Page 7 of 13

Part (B) - Short-answer questions

For questions in part (B), you have to write your answer with a short explanation in the space
provided for the question in your answer sheet. If you need more space, you may continue on the
pages provided for rough work. Any such overflows must be clearly labeled.

1. Let A be a 2 × 2 real matrix such that A2 = I. Determine all possible values of trace(A).
Solution: The matrix satisfies the characteristic polynomial.

A2 − tA + dI = 0,

where t is the trace and d is the determinant. Substitute A2 = I and d = ±1 in the above equation to
get the three possible values of t.
Alternatively, the given matrix equation implies that each eigen value satisfies λ2 = 1. So the eigenvalues
come from the set {+1, −1}. Hence the trace could be −2, 0 or +2.

2. How many binary strings of length 8 contain exactly three 1s such that no two 1s are adjacent.
Solution: In particular, the string should

contain 5 zeros, which create 6 gaps (before, between and
6
after. Choose 3 positions from 6 gaps: 3 = 20. We have to choose 3 gaps to insert a single 1 each.

Questions (3) and (4) are based on the following information
Let f : R → R be a continuous function that satisfies

f (x + y) = f (x) · f (y), ∀x, y ∈ R,

and that f (0) = 1. The following two questions are based on this information.

3. Show that for any rational number r, we have f (r) = (f (1))r . (Hint: Consider the reduced expression
r = p/q, where p, q are integers).
Solution: First, assume q = 1. Then f (p) = f (1 + · · · + 1) = f (1) × · · · × f (1) = (f (1))p . Let q be any
integer. Then f (qr) = f (p) = (f (1))p . Now, qr is also r added q times, hence f (qr) = (f (r))q . But,
(f (r))q = (f (1))p implies the answer.

4. Assuming that f (x) = (f (1))x holds for all real x, determine all values of f (1) for which f is strictly
increasing.
Solution: Let f (1) = a. Assumption implies that

f (x) = ax x ∈ R.

We want the function to be strictly increasing, hence x < y should imply ax < ay , this is possible only
when a > 1.

5. Determine all natural numbers n ≥ 1 such that
n
X n
X
k divides k2 .
k=1 k=1

You may express your answer as a sequence of integers, giving the first few integers n for which the
statement is true.
Solution: Let S1 be the first sum and S2 be the second sum. We want S2 /S1 to be an integer:

S2 n(n + 1)(2n + 1)/6 2n + 1
= = .
S1 n(n + 1)/2 3

Now, 2n + 1 is a multiple of 3 if and only if n is of the form 3k + 1. First few integers are 1, 4, 7, 10 etc.

Question (6) and (7) are based on the following information

8 Notes & Syllabus visit
For more Question Papers, Sample Papers, Page 8 of 13

m
m .co

c. o A list (a , . . . , a ) is an arrangement of natural numbers 1, 2, . . . , n insaerandom
m order, with all orderings
m equally likely. We say that a position i is a new maximum if a >laa for all j < i. For example, in the
se
1 n

ag
i j
list
(3, 1, 4, 2, 5)
positions 1, 3, 5 are new maxima. Note that position 1 is always a new maximum. Now answer the two
questions below based on this information.

6. For a given i, determine the probability that position i is a new maximum.
Solution: Exactly one of the first i entries must be maximum, so the probability is 1i .

m
7. What is the expected number of new maxima in a list?

m
Let Xi be a random variable which takes the valueP
1 when the ith position is a new maximum, and the
.co
.co
value 0 otherwise. Then the random variable X = ni=1 Xi denotes the total number of new maxima in
m s e m
e
a list. The expected number of new maxima is thus calculated as follows:
s l a
l a ag
X
E[X] = E[Xi ]

ag =
X
P (Xi = 1)
X1
= .
i

Questions (8) and (9) are based on the following description.
In the pseudocode below, the function Steps(n) takes a positive integer n as input. For positive integers
a and b, the expression a % b evaluates to the remainder obtained when a is divided by b.

m
function Steps ( n ) {

.co
count = 0;

m
while ( n > 1) {
if ( n % 2 == 0) {
s e
n = n / 2;
l a
} else {
n = n - 1; ag
}
count = count + 1;
}
return ( count );
}

8. Compute the value of Steps(13).
m
om .co
Solution:

c.This 13 → 12 → 6 → 3 → 2 → 1
e m
takes 5 steps. Hence Steps(13) = 5.
l s
em9. Let n be a positive integer with exactly k digits in its binary representation.
a What is the maximum
s
la possible value of Steps(n)? For which value(s) of n (expressed in terms g
a of k) is this maximum attained?
ag
Solution: Each step either divides by 2 (removes a binary digit) or subtracts 1 (reduces the number of
1’s in the binary representation of n). The number of steps is

(number of bits − 1) + (number of 1’s − 1).

This is maximized when all k bits are 1, i.e., n = 2k − 1. In that case,

Steps(n) = (k − 1) + (k − 1) = 2k − 2.

o m m .
c
. a , a , . . . , a is processed one by one. Two variables M and c are se
Questions (10) and (11) are based on the following information.

maintained as follows: se
m
A finite sequence of n ≥ 1 integers 1 2 n
l a
la a g
a g 9
For more Question Papers, Sample Papers, Notes & Syllabus visit Page 9 of 13

• Initially, M = a1 and c = 1.
• For each subsequent number ai , if ai > M , set M = ai and c = 1; if ai = M , set c = c + 1; otherwise,
do nothing.

10. For the input sequence 3, 1, 4, 1, 5, 9, 2, 6, 5, 9, write down the values of M and c after all elements have
been processed.
Solution: Here is a trace through the sequence step by step:

Step 1 2 3 4 5 6 7 8 9 10
ai 3 1 4 1 5 9 2 6 5 9
M 3 3 4 4 5 9 9 9 9 9
c 1 1 1 1 1 1 1 1 1 2

After all elements have been processed, M = 9 and c = 2.

11. What do the variables M and c represent at the end of processing? Justify your answer in one or two
sentences.
Solution: At the end of processing, M is the maximum value in the sequence, and c is the number of
times this maximum value appears in the sequence. This is because M is updated to ai whenever a new
maximum is encountered (resetting c to 1), and c is incremented whenever the current maximum is seen
again; all elements smaller than the current M leave both variables unchanged.

12. Find the sum of all integers from 1 to 100 which are neither multiples of 3 nor multiples of 7.
Solution: The sum of all integers from 1 to 100 is 50 ∗ 101 = 5050.
We subtract the sum 3 + 6 + . . . + 99 = 3(1 + 2 + . . . + 33) = 3 × 17 × 33 = 1683.
We also subtract 7 + 14 + . . . + 98 = 7(1 + 2 + . . . + 14) = 7 × 7 × 15.
We add back 21 + 42 + . . . + 84 = 21(1 + 2 + . . . + 4) = 21 × 10.
The answer is 5050 − (3 × 17 × 33) − (7 × 7 × 15) + 21 × 10 = 2842

Questions (13) and (14) are based on the following information.
Let p be a prime and n be a natural number such that

p divides n2 + n + 1.

13. Find all possible remainders of n3 upon division by p.
Solution: We have

n2 + n + 1 ≡ 0 mod p
(n − 1)(n2 + n + 1) ≡ 0(n − 1) mod p
n3 − 1 ≡ 0 mod p.

Thus, we have n3 ≡ 1 (mod p); the only possible remainder is 1.

14. Suppose p > 3. Show that the numbers 1, n, n2 leave distinct remainders upon division by p.
Solution: We know that n3 leaves residue 1 after dividing by p. Hence, we first need to show that neither
n nor n2 leave remainder 1. This is straightforward since p > 3. For example, If n leaves remainder
1 then n2 + n + 1 leaves remainder 3, which is divisible by p. If n2 leaves remainder 1 then p divides
n2 − 1 = (n − 1)(n + 1) which further implies p divides n − 1 or it divides n + 1. The first case is already
covered and the second contradicts the given information. Now n and n2 can’t leave the same remainder,
otherwise p will divide n2 − n, equivalently p divides n or it divides n2 − 1, neither of which is possible.

15. Find all two-digit positive integers that are equal to three times the sum of their digits.
Solution: Let the number be 10a + b. Then 10a + b = 3(a + b). Which leads to b = 7a 2 . Hence a has to
be even. Trying all the even digits we get that 27 is the only number satisfying the given condition.

10 Notes & Syllabus visit
For more Question Papers, Sample Papers, Page 10 of 13

m
m .co

c. o16. At a gathering of 5 people (named A, B, C, D, E), each person recordsshow e mmany other people they shook
m hands with. Each number must be between 0 and 4. Find one possible la set of 5 recorded numbers, with
se at least 2 repeated values, but not all values equal, and describe g
a a possible configuration of handshakes.
Solution: There are several such possibilities. For example, (2, 2, 2, 1, 1) is realized by A ↔ B, B ↔
C, C ↔ A, D ↔ E. Or, (3, 3, 2, 2, 2) is realized by A ↔ B, B ↔ C, C ↔ A, A ↔ E, B ↔ D. Any valid
sequence, realized by a configuration is correct.

17. An LLM has generated 3 deep learning models M1 , M2 , M3 . Exactly one of these models is faulty. After
running internal diagnostic tests each model produced the following output:

(a) M1 : Model M2 is faulty.
m
.co
(b) M2 : Model M3 is faulty.
m
.co e m
(c) M3 : Model M1 is not faulty.

You know that onlymfaulty models produce false statements and only non-faulty models produce
e model is faulty? Explain. l astrue
l a s
statements. Which
a g
Solution. g
a Try out the possibility that each model is the faulty one, and apply the stated rules.
Case 1: M is faulty. Then M and M are non-faulty. Since M is non-faulty, its statement (M is
1 2 3 2 3
faulty) must be true — but M3 is non-faulty. Contradiction.
Case 2: M2 is faulty. Then M1 and M3 are non-faulty. Since M1 is non-faulty, its statement (M2 is
faulty) must be true. Since M2 is faulty, its statement (M3 is faulty) must be false, i.e. M3 is not faulty.
Since M3 is non-faulty, its statement (M1 is not faulty) must be true. All three statements are consistent.
Case 3: M3 is faulty. Then M1 and M2 are non-faulty. Since M1 is non-faulty, its statement (M2 is
faulty) must be true — but M2 is non-faulty. Contradiction.
m
.co
Therefore M2 is the faulty model.

18. Let A be an n × n real matrix such that
s eAm= A.
la
2

g
a then Ax = x.
(a) Show that if x is in the image of A,
(b) Show that every vector v ∈ Rn can be uniquely written as

v = u + w,

where Au = u and Aw = 0.

Solution: Suppose x = Ay for some y. Then Ax = A(Ay) = A2 y = Ay = x. Given v, define u := Av
and w := v − Av. Then straightforward application shows Au = u and Aw = 0. Uniqueness is also
m
m
straightforward, just assume that there 2 different expressions v = u1 + w1 = u2 + w2 for the identity
.co
.co m
satisfied by A and then compare.

m bility
Let X , X , X be independent and identically distributed Uniform(0, θ) a s e
s e19. 1 2 3
g l random variables with proba-

l a density function (pdf),
a
ag
1
f (x) = , 0 < x < θ,
θ
where unknown θ > 0. Let T = max(X1 , X2 , X3 ).

(a) Derive the cdf and pdf of T
(b) Find E(T ) and Var(T ).

Solution:

m .
.co
Part (a):
e m
em l as
las ag
ag 11 Notes & Syllabus visit
For more Question Papers, Sample Papers, Page 11 of 13

x
Since Xi ∼ Uniform(0, θ), the CDF of each Xi is F (x) = for 0 ≤ x ≤ θ. The CDF of T =
θ
max(X1 , X2 , X3 ) is:

FT (t) = P (T ≤ t) = P (max(X1 , X2 , X3 ) ≤ t)
= P (X1 ≤ t, X2 ≤ t, X3 ≤ t)
= P (X1 ≤ t) P (X2 ≤ t) P (X3 ≤ t) (by independence)
 3
t
= , 0≤t≤θ
θ

Differentiating, the CDF of T is:
3t2
fT (t) = , 0 ≤ x ≤ θ,
θ3
we have the pdf of T .
Part (b):

The expected value is:
Z θ Z θ
3t2 3 3 θ4 3θ
E(T ) = t· dt = 3 t3 dt = · =
0 θ3 θ 0 θ3 4 4

We compute E(T 2 ):
Z θ Z θ
2 3t2 2 3 3 θ5 3θ2
E(T ) = t · 3 dt = 3 t4 dt = · =
0 θ θ 0 θ3 5 5
Therefore:
 2  
2 3θ22 3θ 3θ2 9θ2 48 − 45 3θ2
Var(T ) = E(T ) − [E(T )] = − = − = θ2 =
5 4 5 16 80 80

20. Chebyshev’s Inequality states: For any random variable X with mean µ and variance σ 2 , and for
any k > 0:
1 1
P (|X − µ| ≥ kσ) ≤ 2 equivalently, P (|X − µ| < kσ) ≥ 1 − 2
k k
1 1
Let X ∼ Exponential(λ) with mean and variance 2 . A student claims:
λ λ
 
1 2 3
P X− < ≥
λ λ 4

Is this claim true? Verify by computing the exact probability and comment on what your answer reveals
about Chebyshev’s inequality.
Solution:

Step 1: Verify the claim using Chebyshev’s inequality.
1 1 1 2
For X ∼ Exponential(λ), we have µ = and σ = . The event X − < is of the form |X − µ| < kσ
λ λ λ λ
with k = 2. By Chebyshev’s inequality:
 
1 2 1 3
P X− < ≥1− 2 =
λ λ 2 4

So the student’s claim is consistent with Chebyshev’s inequality. But this only guarantees the bound;
3
it does not confirm whether is close to or far from the true probability.
4

Step 2: Compute the exact probability.

12 Notes & Syllabus visit
For more Question Papers, Sample Papers, Page 12 of 13

m
m .co

c. o The event X − 1 < 2 is equivalent to: s e m
m a
se λ λ
−
1
<X<
3
ag l
λ λ
3
Since X ≥ 0, this reduces to 0 ≤ X <
. Therefore:
λ
  Z 3/λ h i3/λ
1 2 −λx −λx
P X− < = λe dx = − e = 1 − e−3 ≈ 0.9502
λ λ 0 0

Step 3: Compare and comment.
m
m .co
.co
3
m
−3
= 0.75 ≤ 1−e ≈ 0.9502 ✓

m
|4 {z }
| {z }
s e
a
Exact probability

s e Chebyshev lower bound
l
g lower
3 la a
The claim is true; the exact probability 1 − e −3
≈ 0.95 comfortably exceeds the Chebyshev
bound of g
a4
. The gap between 0.75 and 0.95 reveals two things: (i) Chebyshev’s inequality is valid but
conservative, since it must hold for any distribution with finite mean and variance, and (ii) when the
specific distribution is known, the exact probability can be substantially tighter than the bound suggests.

m
m .co
s e
l a
ag

m
m .co
m .co s e m
s e g la
g la a
a

m .
.co s e m
s em l a
g la ag
a 13 Notes & Syllabus visit
For more Question Papers, Sample Papers, Page 13 of 13