CMI Data Science 2022 Question Paper with Solutions: 22 Questions, Answer Key & Section-wise Analysis

CMI Data Science 2022 previous year paper: 22 questions with answer key and detailed solutions, section-wise breakdown and free sample questions.

Paper breakdown

22 questions · 51 marks · 1.7850000000000001 minutes. School Level Mathematics: 10 · Discrete Mathematics: 6 · Probability Theory: 5 · Programming: 1

Free sample questions from CMI Data Science 2022 Question Paper

Question 1 · School Level Mathematics · 2022 MSQ

Which of the following statements is/are true?

A.
For any real number $r$ with $∣ r ∣ > 1$ , $lim_{n \to \infty} \sum_{i = 1}^{n} 17 r^{n} = \frac{17}{1 - r}$ .
B.
Let $i$ be the positive square root of $- 1$ and let $x$ be any real number. Then $tan x = \frac{( e ^{i x} - e ^{- i x} )}{i ( e ^{i x} + e ^{- i x} )}$
C.
$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots < 1.$
D.
The function $f (x)$ defined as

$f (x) = {3 cos 3 x, 0, 0 < x < π /6 otherwise,$

Correct Answer:

["B","D"]

Step-by-Step Solution

Key idea: This is a multi-concept statement-truth question testing geometric series convergence, Euler's formula, telescoping series, and probability density function validation.

Step 1: Analyze Option A

The statement claims: For |r| > 1, lim_{n→∞} Σ_{i=1}^{n} 17r^n = 17/(1-r)

First, note the sum notation: Σ_{i=1}^{n} 17r^n. The index is i but the term 17r^n doesn't depend on i. This equals n · 17r^n.

Even if this is a typo and should be Σ 17r^i, for |r| > 1, the terms grow without bound and the series DIVERGES. The formula 17/(1-r) only applies when |r| < 1.

Therefore, Option A is FALSE.

Step 2: Analyze Option B

We need to verify: tan x = (e^{ix} - e^{-ix})/(i(e^{ix} + e^{-ix}))

From Euler's formula:

e^{ix} = cos x + i sin x
e^{-ix} = cos x - i sin x

Therefore:

e^{ix} - e^{-ix} = 2i sin x
e^{ix} + e^{-ix} = 2 cos x

Substituting:

(e^{ix} - e^{-ix})/(i(e^{ix} + e^{-ix})) = (2i sin x)/(i · 2 cos x) = sin x/cos x = tan x

This matches the given expression. Option B is TRUE.

Step 3: Analyze Option C

The series is: 1/(1·2) + 1/(2·3) + 1/(3·4) + ...

Using partial fractions: 1/(n(n+1)) = 1/n - 1/(n+1)

The partial sum up to n terms is:

S_n = (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... + (1/n - 1/(n+1))

S_n = 1 - 1/(n+1) [telescoping cancellation]

As n → ∞: S_∞ = lim_{n→∞} (1 - 1/(n+1)) = 1

The infinite sum EQUALS exactly 1, not less than 1. The statement says the sum is < 1, which is FALSE.

Step 4: Analyze Option D

The function is defined as:

f(x) = 3cos(3x) for 0 < x < π/6

f(x) = 0 otherwise

For f(x) to be a valid probability density function (PDF), we need:

f(x) ≥ 0 for all x
∫_{-∞}^{∞} f(x) dx = 1

Checking condition 1: For 0 < x < π/6, we have 0 < 3x < π/2, so cos(3x) > 0. Thus 3cos(3x) > 0. ✓

Checking condition 2:

∫_{-∞}^{∞} f(x) dx = ∫_{0}^{π/6} 3cos(3x) dx

= [sin(3x)]_{0}^{π/6}

= sin(π/2) - sin(0)

= 1 - 0

= 1 ✓

Both conditions are satisfied. Option D is TRUE.

Answer: B, D

Question 2 · School Level Mathematics · 2022 MSQ

A matrix $C$ is said to be symmetric if $C^{T} = C$ . Which of the following is/are true? Let $A, B$ be $n \times n$ matrices.

A.
If $A$ is symmetric and invertible, then $A^{- 1}$ is also symmetric and invertible.
B.
If $A$ and $B$ are symmetric, then $C = A B$ is also symmetric.
C.
If $A$ and $B$ are invertible, then $C = A B$ is also invertible.
D.
If $A$ and $B$ are symmetric, then $D = A + B$ is also symmetric.

Correct Answer:

["A","C","D"]

Step-by-Step Solution

Key idea: This is a closure properties question, recognisable because it tests whether symmetry and invertibility are preserved under matrix addition, multiplication, and inversion.

Step 1: Analyze Option A. "If $A$ is symmetric and invertible, then $A^{- 1}$ is also symmetric and invertible."

We know $(A^{- 1})^{T} = (A^{T})^{- 1}$ . Since $A$ is symmetric, $A^{T} = A$ , so $(A^{- 1})^{T} = A^{- 1}$ . Thus, $A^{- 1}$ is symmetric. It is invertible by definition. True.

Step 2: Analyze Option B. "If $A$ and $B$ are symmetric, then $C = A B$ is also symmetric."

Check the transpose: $C^{T} = (A B)^{T} = B^{T} A^{T} = B A$ . For $C$ to be symmetric, we need $A B = B A$ . Since matrix multiplication is not generally commutative, $A B$ is not necessarily symmetric. False.

Step 3: Analyze Option C. "If $A$ and $B$ are invertible, then $C = A B$ is also invertible."

The determinant of a product is the product of determinants: $det (A B) = det (A) det (B)$ . Since $A$ and $B$ are invertible, $det (A) \neq = 0$ and $det (B) \neq = 0$ , so $det (A B) \neq = 0$ . Thus, $A B$ is invertible. True.

Step 4: Analyze Option D. "If $A$ and $B$ are symmetric, then $D = A + B$ is also symmetric."

Check the transpose: $D^{T} = (A + B)^{T} = A^{T} + B^{T} = A + B = D$ . Thus, $A + B$ is symmetric. True.

Answer: A, C, D

Question 3 · School Level Mathematics · 2022 MSQ

Which of the following statements is/are true?

A.
Let $A$ be a $n \times m$ matrix with $n < m$ . Then there is a nonzero solution $y$ with $A y = 0$ only if $A$ has full row rank.
B.
Let $A$ be a $n \times m$ matrix with $n > m$ . There is a nonzero solution $y$ with $A y = 0$ .
C.
The row rank of an $n \times m$ matrix is equal to its column rank only when $n = m$ .
D.
Let $A$ be an $n \times n$ matrix. Suppose $A = B C$ , where $B$ has size $n \times r$ and $C$ has size $r \times n$ . The rank of $A$ is less than or equal to $r$ .

Correct Answer:

["D"]

Step-by-Step Solution

Key idea: This is a rank and null space question, recognisable because it tests the relationship between matrix dimensions, rank, nullity, and matrix factorization.

Step 1: Analyze Option A. "Let $A$ be $n \times m$ with $n < m$ . Nonzero solution $A y = 0$ only if $A$ has full row rank."

By the Rank-Nullity Theorem, $rank (A) + nullity (A) = m$ . Since $rank (A) \leq n < m$ , we have $nullity (A) \geq m - n > 0$ . Thus, a nonzero solution always exists, regardless of whether $A$ has full row rank or not. The "only if" condition makes this statement False.

Step 2: Analyze Option B. "Let $A$ be $n \times m$ with $n > m$ . There is a nonzero solution $A y = 0$ ."

Here, $rank (A) \leq m$ . If $A$ has full column rank ( $rank (A) = m$ ), then $nullity (A) = m - m = 0$ , meaning the only solution is $y = 0$ . Thus, a nonzero solution is not guaranteed. False.

Step 3: Analyze Option C. "The row rank of an $n \times m$ matrix is equal to its column rank only when $n = m$ ."

A fundamental theorem of linear algebra states that row rank always equals column rank for any matrix, regardless of its dimensions. False.

Step 4: Analyze Option D. "Let $A = B C$ , where $B$ is $n \times r$ and $C$ is $r \times n$ . The rank of $A$ is less than or equal to $r$ ."

By the rank inequality for matrix products, $rank (B C) \leq min (rank (B), rank (C))$ . Since $B$ has $r$ columns, $rank (B) \leq r$ . Since $C$ has $r$ rows, $rank (C) \leq r$ . Therefore, $rank (A) \leq r$ . True.

Answer: D

Question 4 · Discrete Mathematics · 2022 MSQ

In order to select a debating team to represent a school, 7 students from class XII and 13 students from class XI were shortlisted and were undergoing trials. The coach had to select a team of 5 students, out of which at least two students should be from each class. One out of the 5 students was to be named as team leader, who was to be from class XII. Two teams with the same members but different leaders are considered to be two different teams. The number of different teams the coach can select is

A.
$2 \times ((2 7) \times (3 13)) + ((3 7) \times (2 13))$
B.
$(2 7) \times (3 13) + (3 7) \times (2 13)$
C.
$2 \times (2 7) \times (3 13) + 3 \times (3 7) \times (2 13)$
D.
$3 \times (2 7) \times (3 13) + 2 \times (3 7) \times (2 13)$

Correct Answer:

["C"]

Step-by-Step Solution

Key idea: This is a committee selection with class constraints and a leader requirement, identical in structure to P2. The method is: split into valid composition cases, count member selections, multiply by eligible leaders per case, then sum.

Step 1: Identify valid compositions.

Team of 5, at least 2 from each class:

Case A: 2 from XII, 3 from XI.
Case B: 3 from XII, 2 from XI.

Step 2: Case A count.

Members: $(2 7) (3 13)$ .

Leader must be from XII; there are 2 such students in this team.

Case A: $2 \times (2 7) (3 13)$ .

Step 3: Case B count.

Members: $(3 7) (2 13)$ .

Leader must be from XII; there are 3 such students.

Case B: $3 \times (3 7) (2 13)$ .

Step 4: Total.

$2 (2 7) (3 13) + 3 (3 7) (2 13)$ .

This matches option C.

Alternative approach (leader-first): Choose the leader first from 7 class XII students. Then choose remaining 4 members ensuring at least 1 more from XII and at least 2 from XI. This gives the same result but is slightly more complex to set up.

The trap in option D is swapping the multipliers (3 and 2), which would correspond to the leader being from class XI.

Answer: C

Question 5 · Discrete Mathematics · 2022 SUB

Words are formed using the characters 0 and 1. The length of a word is the number of characters in it. We say there is a path from word

x

to word

y

if starting from the word

x

you can get the word

y

by applying the following sequence of transformation rules finitely many times in any order.
• Replacing an occurrence of the string 101 by a 0.
• Replacing an occurrence of the string 010 by a 1.
• Replacing a 0 by a 101.
• Replacing a 1 by a 010.
If there is a path from word

x

to word

y

we say

x

and

y

are equivalent or that they are in the same equivalence class. For example the four letter word 1011 is equivalent to the two letter word 01, since the prefix 101 in 1011 can be replaced by a 0 to get 01.
We say

x

has a shorter description if there is a word of shorter length equivalent to it.
(a) State true or false: for any word

x

there is a unique shortest word in its equivalence class.
(b) How many three letter words are there in this language which have no shorter descriptions and which are all in different equivalence classes?

Correct Answer:

Step-by-Step Solution

Key idea: This is a String Rewriting System with Equivalence Classes problem. The transformation rules are reversible, so equivalence is symmetric. We must find irreducible words of length 3 (no shorter description) and count how many distinct equivalence classes they form.

Step 1: Identify all irreducible 3-letter words.

A word has a shorter description if it contains "101" or "010" as a substring (since those can be reduced to "0" and "1" respectively).

All 3-bit strings: 000, 001, 010, 011, 100, 101, 110, 111.

Remove those containing "101" or "010": exclude 010 and 101.

Irreducible set: ${000, 001, 011, 100, 110, 111}$ (6 words).

Step 2: Determine equivalence among these via expansion/reduction.

Note: Rules are reversible, so $a \sim b$ if you can go from $a$ to $b$ using any sequence.

Use invariant: Consider the value $f (w) = (#0 - #1) mod 3$ .

Check how rules affect it:

Replace "101" (2 ones, 1 zero) with "0" (1 zero): $Δ (#0 - #1) = (+ 1) - (- 2) = + 3 \equiv 0 mod 3$ .
Replace "010" (2 zeros, 1 one) with "1": $Δ = (- 2) - (+ 1) = - 3 \equiv 0 mod 3$ .
Expansions are reverse, so same.

So $f (w) mod 3$ is invariant.

Compute $f (w)$ for each irreducible word:

000: $#0 = 3, #1 = 0$ → $3 - 0 = 3 \equiv 0$
001: $2 - 1 = 1 \equiv 1$
011: $1 - 2 = - 1 \equiv 2$
100: $2 - 1 = 1 \equiv 1$
110: $1 - 2 = - 1 \equiv 2$
111: $0 - 3 = - 3 \equiv 0$

So words fall into 3 groups by invariant:

Class 0: {000, 111}
Class 1: {001, 100}
Class 2: {011, 110}

Step 3: Verify that within each class, words are actually equivalent.

000 ↔ 111:

Expand middle 0 in 000 → 0(101)0 = 01010

Reduce first "010" → 110 → not 111. Alternate:

Expand all 0s in 000: 0→101, so 000 → 101101101

Reduce overlapping patterns… but easier: use known result that invariant is complete here.

However, the problem states: “have no shorter descriptions and are in different equivalence classes”.

Since the invariant partitions the set into 3 classes, and each class contains irreducible words, there are 3 classes.

But wait—the answer is given as 2. Re-express using a better invariant.

Alternate invariant: Consider the word as a path where 0 = +1, 1 = -1, and track cumulative sum modulo something.

Or observe actual reductions:

Try connecting 000 and 111:

000 → expand first 0 → 10100
In 10100, reduce "101" → 000 → loop.
Expand 111: 1→010, so 111 → 010010010
Reduce "010" → 111 → loop.

Now try cross-class:

Can 000 reach 001? Suppose yes. Then lengths mod 2 might matter.

Count length parity:

Reduction: "101" (len 3) → "0" (len 1): Δlen = -2

Expansion: "0" → "101": Δlen = +2

So length modulo 2 is invariant!

Original words are length 3 (odd).

Any equivalent word must have odd length.

Shorter description would be length 1 (also odd)—allowed.

But for irreducible words, we only care about equivalence among length-3 words.

Now test actual reachability:

000 and 111:

Start from 0 → expand to 101 → expand first 1 to 010 → 01001

Reduce "101" in middle? 01001 → no "101" or "010" as substring?

"010" at start → reduce to 101 → which is reducible → reduces to 0.

So 0 ↔ 101 ↔ 01001 ↔ 101 ↔ 0. Not helping.

Known result in such systems: the invariant is actually the value modulo 2 of the number of 1s, or a linear combination.

Let’s define $I (w) = (#0 + 2 \cdot #1) mod 3$ .

For "101" → "0": LHS: 1+2*2=5≡2; RHS:1+0=1 → not invariant.

Simpler: notice that replacing "101" with "0" preserves the XOR of bits? No.

Instead, manually verify connections as in the partial solution:

Connection: 000 ↔ 110

000 → expand middle 0 → 0 101 0 = 01010

In 01010, "010" at start → replace with 1 → 110. So 000 ∼ 110.

Connection: 000 ↔ 011

011 → expand first 1 → 0 010 1 = 00101

In 00101, "101" at end → replace with 0 → 000. So 011 ∼ 000.

Thus, 000, 110, 011 are all equivalent.

Similarly:

111 → expand middle 1 → 1 010 1 = 10101

Reduce "101" at start → 001. So 111 ∼ 001.

111 → expand last 1 → 11 010 = 11010

Reduce "101" in middle (positions 1-3: "101") → 100. So 111 ∼ 100.

Thus, 111, 001, 100 are all equivalent.

So the 6 irreducible words form exactly 2 equivalence classes:

Class A: {000, 011, 110}

Class B: {111, 001, 100}

Therefore, there are 2 distinct equivalence classes of irreducible 3-letter words.

Answer: 2

Question 6 · Discrete Mathematics · 2022 MSQ

A relation

R

on the set

A = {a, b, c, d}

is defined by reading the columns of the following table from top to bottom. If a column in the table reads

(x, y, 1)

it means

x

is related to

y

R

. If a column in the table reads

(x, y, 0)

it means

x

is not related to

y

.

For instance, from the fifth column we have,

(a, b) \in R

, and from the second we have

(b, a) \in / R

.
Another relation

S

on the set

A

is defined as: for any

x, y \in A

, the pair

(x, y)

is in

S

if and only if there exists

z \in A

such that both

(x, z) \in R

and

(z, y) \in R

hold.
Which of the following pairs are in

S

A.
$(a, a)$
B.
$(b, b)$
C.
$(c, c)$
D.
$(d, d)$

Correct Answer:

["A","C"]

Step-by-Step Solution

Key idea: This is a Relation Composition via Matrix/Table problem. Recognise it because relation $S$ is defined as "exists $z$ such that $(x, z) \in R$ and $(z, y) \in R$ ", which is precisely the definition of $R^{2}$ or $R \circ R$ .

Step 1: Decode the table into relation $R$ .

The table columns represent pairs $(r o w_l ab e l, co l_l ab e l, v a l u e)$ . Reading the bottom row (values) against the top two rows (labels):

Col 1: $(a, a, 0)$

Col 2: $(b, a, 0)$

Col 3: $(c, a, 1) ⟹ (c, a) \in R$

Col 4: $(d, a, 0)$

Col 5: $(a, b, 1) ⟹ (a, b) \in R$

Col 6: $(b, b, 0)$

Col 7: $(c, b, 0)$

Col 8: $(d, b, 0)$

Col 9: $(a, c, 1) ⟹ (a, c) \in R$

Col 10: $(b, c, 1) ⟹ (b, c) \in R$

Col 11: $(c, c, 0)$

Col 12: $(d, c, 1) ⟹ (d, c) \in R$

Col 13: $(a, d, 0)$

Col 14: $(b, d, 0)$

Col 15: $(c, d, 0)$

Col 16: $(d, d, 0)$

So $R = {(c, a), (a, b), (a, c), (b, c), (d, c)}$ .

Step 2: Compute $S = R \circ R$ .

We need pairs $(x, y)$ where a path $x \to z \to y$ exists in $R$ .

Let's trace paths of length 2 from each starting node:

From $a$ :

$a \to b \to c$ (since $(a, b) \in R, (b, c) \in R$ ) $⟹ (a, c) \in S$

$a \to c \to a$ (since $(a, c) \in R, (c, a) \in R$ ) $⟹ (a, a) \in S$

From $b$ :

$b \to c \to a$ (since $(b, c) \in R, (c, a) \in R$ ) $⟹ (b, a) \in S$

From $c$ :

$c \to a \to b$ (since $(c, a) \in R, (a, b) \in R$ ) $⟹ (c, b) \in S$

$c \to a \to c$ (since $(c, a) \in R, (a, c) \in R$ ) $⟹ (c, c) \in S$

From $d$ :

$d \to c \to a$ (since $(d, c) \in R, (c, a) \in R$ ) $⟹ (d, a) \in S$

So $S = {(a, c), (a, a), (b, a), (c, b), (c, c), (d, a)}$ .

Step 3: Check options against $S$ .

A: $(a, a) \in S$ — TRUE

B: $(b, b) \in / S$ — FALSE

C: $(c, c) \in S$ — TRUE

D: $(d, d) \in / S$ — FALSE

Answer: Options A and C are correct.

Question 7 · Probability Theory · 2022 MSQ

Let $X$ be a Binomial $(n, p)$ random variable with the probability mass function

$P (X = k) = (k n) p^{k} (1 - p)^{n - k}, k = 0, 1, 2, \dots, n .$

Let $T$ be a random variable defined as:

$T = {1, 0, if X = 1, otherwise.$

Let $E (T)$ and $V (T)$ denote the expectation and the variance of a random variable $T$ . Which of the following statements is/are true?

A.
$E (T^{2}) = n p (1 - p)^{n - 1}$ and $V (T) = n p (1 - p)^{n - 1} {(1 - (n p (1 - p)^{n - 1}))}$
B.
$E (T) = n p$ and $V (T) = n p (1 - p)$
C.
$E (T) = n p (1 - p)^{n - 1}$ and $E (T^{2}) = n p (1 - p)^{n - 1}$
D.
$E (T) = p$ and $V (T) = p (1 - p)$

Correct Answer:

["A","C"]

Step-by-Step Solution

Key idea: This is an indicator-variable question. The variable $T$ is not the binomial count itself; it is a 0/1 switch that turns on only when $X = 1$ .

Step 1: Identify the event indicated by $T$ .

The definition says

$T = {1, 0, if X = 1, otherwise.$

So $T$ is the indicator of the event $A = {X = 1}$ .

Step 2: Find the probability of that event.

Since $X \sim Binomial (n, p)$ ,

$P (X = 1) = (1 n) p (1 - p)^{n - 1} = n p (1 - p)^{n - 1} .$

Let

$q = n p (1 - p)^{n - 1} .$

Step 3: Use indicator properties.

An indicator variable is Bernoulli with success probability $q$ . Therefore,

$E (T) = q .$

Also, because $T$ only takes values 0 and 1,

$T^{2} = T,$

$E (T^{2}) = E (T) = q .$

Step 4: Compute the variance.

For a Bernoulli indicator,

$V (T) = q (1 - q) .$

Substituting $q = n p (1 - p)^{n - 1}$ ,

$V (T) = n p (1 - p)^{n - 1} (1 - n p (1 - p)^{n - 1}) .$

Step 5: Match with the options.

Option A gives the correct $E (T^{2})$ and correct variance.

Option C gives the correct $E (T)$ and correct $E (T^{2})$ .

Option B incorrectly uses the mean and variance of $X$ , not $T$ .

Option D incorrectly treats $T$ as a Bernoulli variable with parameter $p$ .

Answer: Options A and C are true.

Question 8 · Probability Theory · 2022 SUB

Common Description: Description for the next four questions:
Dasholytics Inc runs an online analytics dashboard. The company employs three machines—Server 1, Server 2, and Server 3—to serve the data for the dashboard. At any point in time one of these three machines has the job of serving the data, and the other two are kept in standby mode. Dasholytics uses a server scheduler (software) that decides which of the three machines should serve the data at any given point of time. The scheduler switches between data servers without disrupting the data feed to the dashboard.
The machine which is serving the data sometimes fails to do so; in this case an alert is sent out to the Dasholytics team and they investigate and fix the problem. The time for which a machine fails to serve data is accounted as service outage caused by that machine. For technical reasons the server scheduler is deactivated when there is such an outage: the outage gets over only when the issue with the server is fixed and it is put back online.
The figures below describe the server usage and outage statistics as compiled over the last one year (365 days). Please use this information to answer the questions that follow.

Figure 1: Figure (a) shows the total service outage caused by each server, as a percentage of the total time that it served data. Figure (b) shows the total time that each server served data, as a percentage of the total duration over which this information was collected. Express the total service outage caused by all the three servers, as a percentage of the total duration over which this information was collected.

Correct Answer:

2.80

Step-by-Step Solution

Insight: The total service outage percentage is the marginal probability of an outage, found by summing the weighted conditional outage rates.

Exam route: P(O) = (0.40 0.01) + (0.30 0.05) + (0.30 * 0.03) = 0.028, which is 2.80%.

Learning route:

We need the overall percentage of time the system is in outage.
This is the marginal probability P(Outage).
Using the Law of Total Probability: P(Outage) = P(Outage|S1)P(S1) + P(Outage|S2)P(S2) + P(Outage|S3)P(S3).
Substitute the values: (0.01 0.40) + (0.05 0.30) + (0.03 * 0.30) = 0.004 + 0.015 + 0.009 = 0.028.
Expressed as a percentage, this is 2.8%. Formatted to 2 decimal places: "2.80".

Wrong path: Averaging the outage rates (1+5+3)/3 = 3% ignores the fact that the servers are used for different amounts of time.

Question 9 · Probability Theory · 2022 SUB

Common Description: Description for the next four questions:
Dasholytics Inc runs an online analytics dashboard. The company employs three machines—Server 1, Server 2, and Server 3—to serve the data for the dashboard. At any point in time one of these three machines has the job of serving the data, and the other two are kept in standby mode. Dasholytics uses a server scheduler (software) that decides which of the three machines should serve the data at any given point of time. The scheduler switches between data servers without disrupting the data feed to the dashboard.
The machine which is serving the data sometimes fails to do so; in this case an alert is sent out to the Dasholytics team and they investigate and fix the problem. The time for which a machine fails to serve data is accounted as service outage caused by that machine. For technical reasons the server scheduler is deactivated when there is such an outage: the outage gets over only when the issue with the server is fixed and it is put back online.
The figures below describe the server usage and outage statistics as compiled over the last one year (365 days). Please use this information to answer the questions that follow.

Figure 1: Figure (a) shows the total service outage caused by each server, as a percentage of the total time that it served data. Figure (b) shows the total time that each server served data, as a percentage of the total duration over which this information was collected. What is the expected number of hours of service outage in an year (365 days)?

Correct Answer:

245.28

Step-by-Step Solution

Insight: The total outage time is the sum of the outages of each server, which requires weighting each server's conditional outage rate by its marginal usage time.

Exam route: Calculate P(Outage) = (0.40 0.01) + (0.30 0.05) + (0.30 0.03) = 0.028. Multiply by total hours in a year (365 24 = 8760) to get 245.28.

Learning route:

Identify the marginal probabilities of each server being active: P(S1) = 0.40, P(S2) = 0.30, P(S3) = 0.30.
Identify the conditional outage probabilities: P(O|S1) = 0.01, P(O|S2) = 0.05, P(O|S3) = 0.03.
Use the Law of Total Probability to find the overall outage probability: P(O) = P(O|S1)P(S1) + P(O|S2)P(S2) + P(O|S3)P(S3) = 0.004 + 0.015 + 0.009 = 0.028.
Convert this probability to expected hours: 0.028 * 8760 hours = 245.28 hours.

Wrong path: Adding the percentages directly (1% + 5% + 3% = 9%) and applying to 8760 gives 788.4, which ignores the different base serving times.

Question 10 · Programming · 2022 MSQ

Consider the following code, in which A is an array indexed from 0 and n is the number of elements in A. function foo(A,n) { L = 0; R = n - 1; while (L <= R) { i = ceil((L + R)/2); if (A[i] < i) { L = i + 1; } else { if (A[i] > i) { R = i - 1; } else { return(i); } } } return(-1); } Here, ceil(x) returns the smallest integer bigger than or equal to the number x.
If

A = [- 5, - 4, - 3, - 2, - 1, 4, 6, 8, 10, 12]

, what will foo(A, 10) return?

A.
-5
B.
-1
C.
4
D.
6

Correct Answer:

["D"]

Step-by-Step Solution

Key idea: This is a Fixed-Point Search using Binary Search. We are looking for an index i such that A[i] = i.

The array is A = [-5, -4, -3, -2, -1, 4, 6, 8, 10, 12].

Indices: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Step 1: Initialization

L = 0, R = 9.

Step 2: Iteration 1

i = ceil((0 + 9) / 2) = ceil(4.5) = 5.

Check A[5]. A[5] = 4.

Compare A[5] with i = 5.

4 < 5.

Condition A[i] < i is True.

Update L = i + 1 = 6.

Current Range: L = 6, R = 9.

Step 3: Iteration 2

i = ceil((6 + 9) / 2) = ceil(7.5) = 8.

Check A[8]. A[8] = 10.

Compare A[8] with i = 8.

10 > 8.

Condition A[i] < i is False.

Check A[i] > i. True.

Update R = i - 1 = 7.

Current Range: L = 6, R = 7.

Step 4: Iteration 3

i = ceil((6 + 7) / 2) = ceil(6.5) = 7.

Check A[7]. A[7] = 8.

Compare A[7] with i = 7.

8 > 7.

Condition A[i] < i is False.

Check A[i] > i. True.

Update R = i - 1 = 6.

Current Range: L = 6, R = 6.

Step 5: Iteration 4

L <= R (6 <= 6) is True.

i = ceil((6 + 6) / 2) = 6.

Check A[6]. A[6] = 6.

Compare A[6] with i = 6.

6 = 6.

Condition A[i] < i is False.

Check A[i] > i is False.

Else block: Return i = 6.

The function returns the index 6.

Option D corresponds to 6.

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