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CMI M.Sc. and Ph.D. Computer Science Chapter-wise PYQs

Practice CMI M.Sc. and Ph.D. Computer Science previous year questions organized by chapter. 230+ PYQs from 10 years with detailed solutions. First chapter FREE!

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Basic Probability

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Probability Theory • 6 PYQs

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Conditional Probability and Independence

Probability Theory • 6 PYQs

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Probabilistic Bounds

Probability Theory • 1 PYQs

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Expectation and Variance

Probability Theory • 3 PYQs

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Foundations of Propositional Logic

Logic and Boolean Algebra • 11 PYQs

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Boolean Algebra

Logic and Boolean Algebra • 2 PYQs

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Special Graph Structures

Graph Theory • 3 PYQs

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Graph Definitions and Types

Graph Theory • 23 PYQs

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Minimum Spanning Trees

Graph Theory • 1 PYQs

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Matchings and Coloring

Graph Theory • 6 PYQs

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Graph Traversal

Graph Theory • 4 PYQs

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Shortest Path Algorithms

Graph Theory • 5 PYQs

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Fundamentals of Vector Spaces

Linear Algebra • 1 PYQs

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Advanced Counting Techniques

Discrete Mathematics • 6 PYQs

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Permutations and Combinations

Discrete Mathematics • 9 PYQs

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Fundamental Counting Principles

Discrete Mathematics • 8 PYQs

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Set Theory

Discrete Mathematics • 4 PYQs

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Relations

Discrete Mathematics • 7 PYQs

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Functions

Discrete Mathematics • 5 PYQs

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Mathematical Induction

Discrete Mathematics • 1 PYQs

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Basic Probability - PYQs

Free sample questions from Probability Theory

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1 Multiple Select

2022

The Telvio mobile service provider allows each customer to choose a part of their 10-digit mobile number when they get a new connection. The first two digits of the number are fixed by the company based on the customer region. The customer can choose the last four digits that they wish. The company chooses each of the remaining four digits uniformly at random and without replacement from the digits

\{0,1,2,\ldots,9\}

. Note that this means that the digits in positions 3, 4, 5 and 6 in a Telvio number are all different. What is the probability that, in the mobile number assigned to a new customer by Telvio, the digits in positions 3, 4, 5 and 6 appear in increasing order when read from left to right?

\frac{1}{4}

\frac{1}{16}

\frac{1}{24}

\frac{1}{32}

View Solution

The company chooses four distinct digits for positions

3,4,5,6

Since the digits are chosen without replacement, all four chosen digits are different. Now fix any particular set of four distinct digits. There are exactly

4!

possible orders in which these four digits can appear in the four positions. Among these

4!

orders, exactly one order is strictly increasing from left to right. Therefore, conditioned on any chosen set of four distinct digits, the probability that they appear in increasing order is

\frac{1}{4!}

That is,

\frac{1}{24}

Since this value does not depend on which four digits were chosen, the overall probability is also

\frac{1}{24}

Therefore, the correct answer is

\frac{1}{24}

2 Multiple Select

2021

In the chamber containing the Philosopher's stone, Harry sees a deck of

5

cards, each with a distinct number from

1

5

Harry removes two cards from the deck, one at a time. What is the probability that the two cards selected are such that the first card's number is exactly one more than the number on the second card?

1/5

4/25

1/4

2/5

View Solution

Let the ordered pair

(i,j)

mean that the first card drawn is

i

and the second card drawn is

j

Since two different cards are drawn from

\{1,2,3,4,5\}

the sample space is

\{(i,j)\mid i,j \in \{1,2,3,4,5\},\ i \ne j\}

The total number of outcomes is

5 \times 4 = 20

We want the first number to be exactly one more than the second. So the favorable ordered pairs are

(2,1),\ (3,2),\ (4,3),\ (5,4)

There are

4

favorable outcomes. Hence the probability is

\frac{4}{20} = \frac{1}{5}

So the correct answer is

1/5

3 Multiple Select

2020

A fair coin is repeatedly tossed. Each time a head appears,

1

rupee is added to the first bag. Each time a tail appears,

2

rupees are put in the second bag. What is the probability that both the bags have the same amount of money after

6

coin tosses?

\frac{1}{2^6}

\frac{6!}{2!\,4!\,2^6}

\frac{2^2}{2^6}

\frac{6!}{2^6}

View Solution

Let the number of heads be

h

and the number of tails be

t

Then after

6

tosses we have

h+t=6

The first bag gets

h

rupees, because each head contributes

1

rupee. The second bag gets

2t

rupees, because each tail contributes

2

rupees. For the two amounts to be equal, we need

h=2t

Together with

h+t=6

this gives

2t+t=6

3t=6

and therefore

t=2,\quad h=4

So the desired event is exactly:

\text{in 6 tosses, there are 2 tails and 4 heads}

The number of such sequences is

\binom{6}{2} = \frac{6!}{2!\,4!}

Since all

2^6

toss sequences are equally likely, the probability is

\frac{\binom{6}{2}}{2^6} = \frac{6!}{2!\,4!\,2^6}

Therefore, the correct answer is

\frac{6!}{2!\,4!\,2^6}

4 Multiple Select

2018

Let

C_n

be the number of strings

w

consisting of

n

X's and

n

Y's such that no initial segment of

w

has more Y's than X's. Now consider the following problem. A person stands in the middle of a swimming pool holding a bag of

n

red and

n

blue balls. He draws a ball out one at a time and discards it. If he draws a blue ball, he takes one step back, if he draws a red ball, he moves one step forward. What is the probability that the person remains dry?

\frac{C_n}{2^{2n}}

\frac{C_n}{\binom{2n}{n}}

\frac{n \cdot C_n}{(2n)!}

\frac{n \cdot C_n}{\binom{2n}{n}}

View Solution

Each sequence of draws is determined only by the order of the

n

red balls and

n

blue balls. So the total number of possible color sequences is

\binom{2n}{n}

because we only need to choose which

n

of the

2n

positions contain the red balls. Now the person remains dry exactly when, at every stage, he has never moved too far backward. This corresponds to the condition that in every initial segment, the number of red moves is at least the number of blue moves. If we encode

X=\text{red}

and

Y=\text{blue}

then the number of such favorable sequences is precisely

C_n

by the definition given in the question. Therefore the required probability is

\frac{C_n}{\binom{2n}{n}}

Hence the correct answer is

\frac{C_n}{\binom{2n}{n}}

5 Multiple Select

2017

An FM radio channel has a repository of

10

songs. Each day, the channel plays

3

distinct songs that are chosen randomly from the repository. Mary decides to tune in to the radio channel on the weekend after her exams. What is the probability that no song gets repeated during these

2

days?

\binom{10}{3}^2 \cdot \binom{10}{6}^{-1}

\binom{10}{6}\cdot \binom{10}{3}^{-2}

\binom{10}{3}\cdot \binom{7}{3}\cdot \binom{10}{3}^{-2}

\binom{10}{3}\cdot \binom{7}{3}\cdot \binom{10}{6}^{-1}

View Solution

On day

1

, the channel chooses any

3

distinct songs out of

10

. So the number of possibilities for day

1

\binom{10}{3}

If no song is to be repeated on day

2

, then after choosing the

3

songs of day

1

, only

10-3=7

songs remain available for day

2

. So the number of valid choices for day

2

\binom{7}{3}

Hence the number of favourable outcomes is

\binom{10}{3}\binom{7}{3}

Now count the total number of possible choices over the two days without any restriction. Each day independently has

\binom{10}{3}

choices, so the total number of outcomes is

\binom{10}{3}\binom{10}{3}=\binom{10}{3}^2

Therefore the required probability is

\frac{\binom{10}{3}\binom{7}{3}}{\binom{10}{3}^2}

which is exactly

\binom{10}{3}\cdot \binom{7}{3}\cdot \binom{10}{3}^{-2}

Hence the correct answer is option (c).

6 Multiple Select

2016

Varsha lives alone and dislikes cooking, so she goes out for dinner every evening. She has two favourite restaurants, Dosa Paradise and Kababs Unlimited, which she travels by local train. The train to Dosa Paradise runs every

10

minutes, at

0

10

20

30

40

and

50

minutes past the hour. The train to Kababs Unlimited runs every

20

minutes, at

8

28

and

48

minutes past the hour. She reaches the station at a random time between

7{:}15

pm and

8{:}15

pm and chooses between the two restaurants based on the next available train. What is the probability that she ends up in Kababs Unlimited?

\frac{1}{5}

\frac{1}{3}

\frac{2}{5}

\frac{1}{2}

View Solution

Varsha chooses the restaurant whose train comes next. So we must find all arrival times for which the next train to Kababs Unlimited comes before the next train to Dosa Paradise. Between

7{:}15

and

8{:}15

, the relevant Dosa Paradise trains are at

7{:}20,\ 7{:}30,\ 7{:}40,\ 7{:}50,\ 8{:}00,\ 8{:}10,\ 8{:}20

and the relevant Kababs Unlimited trains are at

7{:}28,\ 7{:}48,\ 8{:}08,\ 8{:}28

Now compare the next available train in each interval. Kababs Unlimited is chosen exactly when the arrival time lies in these intervals:

7{:}20 \text{ to } 7{:}28

7{:}40 \text{ to } 7{:}48

8{:}00 \text{ to } 8{:}08

Each of these intervals has length

8 \text{ minutes}

So the total favourable time is

8 + 8 + 8 = 24 \text{ minutes}

The total time window is from

7{:}15

8{:}15

, which has length

60 \text{ minutes}

Therefore the required probability is

\frac{24}{60} = \frac{2}{5}

Hence the correct answer is

\frac{2}{5}

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PYQs are organized chapter-wise across 38 topics, making it easy to practice specific areas. Each question shows its year and slot.

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We have 10 years of CMI M.Sc. and Ph.D. Computer Science PYQs from 2016 to 2025, totaling 230+ questions.

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