Probability topic an important part of the CAT. You can expect close to 2-3 questions in the latest 22 question format of the CAT Quant section. In this article, we will look into some Probability Questions for the CAT Exam. If you want to practice these important probability questions, you can download the PDF, which is completely Free.

**Probability**is often one of the most feared topics among the candidates. It is not a very difficult topic if you understand the basics of Probability well.- Probability-based questions appear in the CAT test almost every year. A lot of aspirants avoid this topic but remember that one can definitely solve the easy questions on Probability if one is thorough with the basics. Therefore, practising questions with Probability should not be avoided.
- The chances of occurring or not occurring an event should be determined based on the number of favourable and not favourable conditions.
- Here we are giving some very important probability questions, which also include questions from the
**CAT previous papers**. The candidates are advised to try each question on their own and later go through the solutions given below.

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**Question 1:Â **How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?

a)Â 499

b)Â 500

c)Â 375

d)Â 376

**Question 2:Â **What is the number of distinct terms in the expansion of $(a + b + c)^{20}$?

a)Â 231

b)Â 253

c)Â 242

d)Â 210

e)Â 228

**Question 3:Â **There are 6 boxes numbered 1,2,â€¦ 6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is

a)Â 5

b)Â 21

c)Â 33

d)Â 60

**Question 4:Â **A graph may be defined as a set of points connected by lines called edges. Every edge connects a pair of points. Thus, a triangle is a graph with 3 edges and 3 points. The degree of a point is the number of edges connected to it. For example, a triangle is a graph with three points of degree 2 each. Consider a graph with 12 points. It is possible to reach any point from any point through a sequence of edges. The number of edges, e, in the graph must satisfy the condition

a)Â $11 \leq e \leq 66$

b)Â $10 \leq e \leq 66$

c)Â $11 \leq e \leq 65$

d)Â $0 \leq e \leq 11$

**Question 5:Â **N persons stand on the circumference of a circle at distinct points. Each possible pair of persons, not standing next to each other, sings a two-minute song one pair after the other. If the total time taken for singing is 28 minutes, what is N?

[CAT 2004]

a)Â 5

b)Â 7

c)Â 9

d)Â None of the above

**Question 6:Â **In the adjoining figure, the lines represent one-way roads allowing travel only northwards or only westwards. Along how many distinct routes can a car reach point B from point A?

a)Â 15

b)Â 35

c)Â 120

d)Â 336

**Question 7:Â **A new flag is to be designed with six vertical stripes using some or all of the colours yellow, green, blue and red. Then, the number of ways this can be done such that no two adjacent stripes have the same colour is

a)Â 12 Ã— 81

b)Â 16 Ã— 192

c)Â 20 Ã— 125

d)Â 24 Ã— 216

**Question 8:Â **Consider the set S = { 1, 2, 3, .., 1000 }. How many arithmetic progressions can be formed from the elements of S that start with l and end with 1000 and have at least 3 elements?

a)Â 3

b)Â 4

c)Â 6

d)Â 7

e)Â 8

**Question 9:Â **There are 6 tasks and 6 persons. Task 1 cannot be assigned either to person 1 or to person 2; task 2 must be assigned to either person 3 or person 4. Every person is to be assigned one task. In how many ways can the assignment be done?

[CAT 2006]

a)Â 144

b)Â 180

c)Â 192

d)Â 360

e)Â 716

**Question 10:Â **In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls, and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is

a)Â 200

b)Â 216

c)Â 235

d)Â 256

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**Question 11:Â **Let S be the set of five-digit numbers formed by the digits 1, 2, 3, 4 and 5, using each digit exactly once such that exactly two odd positions are occupied by odd digits. What is the sum of the digits in the rightmost position of the numbers in S?

a)Â 228

b)Â 216

c)Â 294

d)Â 192

**Question 12:Â **Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over person-to-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English, and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose?

a)Â 5

b)Â 10

c)Â 9

d)Â 15

**Question 13:Â **The figure below shows the network connecting cities A, B, C, D, E and F. The arrows indicate permissible direction of travel. What is the number of distinct paths from A to F?

a)Â 9

b)Â 10

c)Â 11

d)Â None of these

**Question 14:Â **One red flag, three white flags and two blue flags are arranged in a line such that,

A. no two adjacent flags are of the same colour

B. the flags at the two ends of the line are of different colours.

In how many different ways can the flags be arranged?

a)Â 6

b)Â 4

c)Â 10

d)Â 2

**Question 15:Â **Sam has forgotten his friendâ€™s seven-digit telephone number. He remembers the following: the first three digits are either 635 or 674, the number is odd, and the number nine appears once. If Sam were to use a trial and error process to reach his friend, what is the minimum number of trials he has to make before he can be certain to succeed?

a)Â 10000

b)Â 2430

c)Â 3402

d)Â 3006

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**Question 16:Â **There are three cities A, B and C. Each of these cities is connected with the other two cities by at least one direct road. If a traveller wants to go from one city (origin) to another city (destination), she can do so either by traversing a road connecting the two cities directly, or by traversing two roads, the, first connecting the origin to the third city and the second connecting the third city to the destination. In all there are 33 routes from A to B (including those via C). Similarly there are 23 routes from B to C (including those via A). How many roads are there from A to C directly?

a)Â 6

b)Â 3

c)Â 5

d)Â 10

**Question 17:Â **If there are 10 positive real numbers $n_1 < n_2 < n_3 … < n_{10}$ , how many triplets of these numbers $(n_1, n_2, n_3 ), ( n_2, n_3, n_4 )$ can be generated such that in each triplet the first number is always less than the second number, and the second number is always less than the third number?

a)Â 45

b)Â 90

c)Â 120

d)Â 180

**Question 18:Â **Ten straight lines, no two of which are parallel and no three of which pass through any common point, are drawn on a plane. The total number of regions (including finite and infinite regions) into which the plane would be divided by the lines is

a)Â 56

b)Â 255

c)Â 1024

d)Â not unique

**Question 19:Â **In how many ways is it possible to choose a white square and a black square on a chessboard so that the squares must not lie in the same row or column?

a)Â 56

b)Â 896

c)Â 60

d)Â 768

**Question 20:Â **There are six boxes numbered 1, 2, 3, 4, 5, 6. Each box is to be filled up either with a white ball or a black ball in such a manner that at least one box contains a black ball and all the boxes containing black balls are consecutively numbered. The total number of ways in which this can be done equals.

a)Â 15

b)Â 21

c)Â 63

d)Â 64

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**Probability Questions – Answers & Solutions:**

**1)Â AnswerÂ (D)**

We have to essentially look at numbers between 1000 and 4000 (including both).

The first digit can be either 1 or 2 or 3.

The second digit can be any of the five numbers.

The third digit can be any of the five numbers.

The fourth digit can also be any of the five numbers.

So, total is 3*5*5*5 = 375.

However, we have ignored the number 4000 in this calculation and hence the total is 375+1=376

**2)Â AnswerÂ (A)**

The power is 20.

20 has to be divided among a, b and c. This can be done in $^{20+3-1}C_{3-1}$ = $^{22}C_2$ = 231

Option a) is the correct answer.

**3)Â AnswerÂ (B)**

If there is only 1 green ball, it can be done in 6 ways

If there are 2 green balls, it can be done in 5 ways.

.

.

.

If there are 6 green balls, it can be done in 1 way.

So, the total number of possibilities is 6*7/2 = 21

**4)Â AnswerÂ (A)**

Take any 12 points.

The maximum number of edges which can be drawn through these 12 points are $^{12}C_2$ = 66

The minimum number of edges which can be drawn through these 12 points are 12-1 = 11 as the resulting figure need not be closed. It might be open.

**5)Â AnswerÂ (B)**

Total number of pairs is $^{N}C_2$. Number of pairs standing next to each other = N. Therefore, number of pairs in question = $^{N}C_2$ – N

= 28/2 = 14.

If N = 7,

$^{7}C_2$ – 7 =Â 21 – 7 = 14.

N = 7.

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**6)Â AnswerÂ (B)**

The person has to take 3 steps north and 4 steps west, in whatever way he travels.

Total steps = 7, 3 north and 4 west.

Number of ways = 7!/(4!3!) = 35

**7)Â AnswerÂ (A)**

The number of ways of selecting a colour for the first stripe is 4. The number of ways of selecting a colour for the second stripe is 3. Similarly, the number of ways of selecting colours for the third, fourth, fifth and sixth stripes are 3, 3, 3 and 3 respectively.

The total number of ways of selecting the colours is, therefore, 4*3*3*3*3*3 = 12*81.

**8)Â AnswerÂ (D)**

The nth term is a + (n-1)d

1000 = 1 + (n-1)d

So, (n-1)d = 999

999 = 3^3 * 37

So, the number of factors is 4*2 = 8

Since there should be at least 3 terms in the series, d cannot be 999.

So, the number of possibilities is 7

**9)Â AnswerÂ (A)**

If the first task is assigned to either person 3 or person 4, the second task can be assigned in only 1 way. If the first task is assigned to either person 5 or person 6, the second task can be assigned in 2 ways. Therefore, the number of ways in which the first two tasks can be assigned is 2*1 + 2*2 = 6.

The other 4 tasks can be assigned to 4 people in 4! ways.

The total number of ways of assigning the 6 tasks is, therefore, 6*4! = 144.

**10)Â AnswerÂ (A)**

Number of games in which both the players are girls = $^GC_2$ where G is the number of girls

$^GC_2 = 45$

$^{10}C_2 = 45$

So, G = 10

Similarly, number of games in which both the players are boys = $^BC_2$, where B is the number of boys

$^BC_2 = 190$

$^{20}C_2 = 190$

So, B = 20

So, number of games in which one player is a boy and the other player is a girl is 20*10 = 200

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**11)Â AnswerÂ (B)**

When the odd numbers occupy places 1 and 3, only 2 or 4 can be in the 5th place. Odd numbers can occupy places 1 and 3 in 3C2*2! = 6 ways. When 2 is at the 5th place, the other odd number and 4 can be arranged in the remaining places in 2 ways. So, 2 occurs at the end 6*2 = 12 times. Similarly, 4 occurs 12 times.

If odd numbers occupy places 1 and 5, then 2 or 4 should come in the 3rd place. The other two numbers can then be arranged in 2 ways in the remaining blanks. So, if 1 is in the first place and 5 is in the 5th place, the other numbers can be arranged in 2*2 = 4 ways. Similar for 1 and 3; 5 and 1; 3 and 1; 5 and 3; 3 and 5. So, 5 occurs 8 times, 1 8 times and 3 8 times. Similar is the case when odd numbers are placed in 3rd and 5th places.

On the whole, 4 occurs 12 times, 2 occurs 12 times, 5, 3 and 1 each occur 16 times. The total is, therefore, 48+24+80+48+16 = 216

**12)Â AnswerÂ (C)**

Consider there are 6 people numbered 1-3 englishmen and 3-6 frenchmen, let 3 know both english and french.

First call would be between 1-3 then 2-3 such that 3 know secret of all 3 englishmen.

Let 3 call 4 .

Similarly there would be call between 4-5 then 4-6 such that 4 know secret of all 3 frenchmen.

Now 3 would call 4 . Such that 3 and 4 would know secret of all 6 members.

Now to let this know to 1,2,5,6 more 4 calls would be required.

Hence, minimum calls required would be 9.

**13)Â AnswerÂ (B)**

The distinct paths are:

A -> D -> C -> F

A -> D -> E -> F

A -> D -> C -> E -> F

A -> B -> D -> C -> F

A -> B -> D -> E -> F

A -> B -> D -> C -> E -> F

A -> B -> C -> F

A -> B -> E -> F

A -> B -> F

A -> B -> C -> E -> F

So, the total number of distinctÂ paths is 10

**14)Â AnswerÂ (A)**

The three white flags can be arranged in the following two ways:

__ W __ W __ W or W __ W __ W __

In the blanks, the 2 blue and one red flag can be arranged in 3 ways.

So, the total number of arrangements is 2*3 = 6

**15)Â AnswerÂ (C)**

Consider cases : 1) Last digit is 9: No. of ways in which the first 3 digits can be guessed is 2. No. of ways in which next 3 digits can be guessed is 9*9*9. So in total the number of ways of guessing = 2*9*9*9 = 1458.

2) Last digit is not 9: the number 9 can occupy any of the given position 4, 5, or 6, and there shall be an odd number at position 7.

So in total, the number of guesses = 2*3*(9*9*4) = 1944+1458 = 3402

**16)Â AnswerÂ (A)**

The possible roads are:

A -> B Let this be x

A -> C Let this be y

B -> C Let this be z

From the information given, x + yz = 33 -> (1)

z + xy = 23 -> (2)

From the options, if y = 10, x =2 and z = 3 from (2), but it doesn’t satisfy (1)

If y = 5, x = 4 and z = 3 from (2) but they don’t satisfy (1)

A possible set of numbers for (x,y,z) are (3,6,5)

Number of roads from A -> C = 6

**17)Â AnswerÂ (C)**

For any selection of three numbers, there is only one way in which they can be arranged in ascending order.

So, the answer is $^{10}C_3 = 120$

**18)Â AnswerÂ (A)**

If there are ‘m’ non-parallel lines, then theÂ maximumÂ numberÂ of regions into which the plane is divided isÂ given by

[m(m+1)/2]+1

In this case, ‘m’ = 10

So, the number of regions into which the plane is divided is (10*11/2) + 1 = 56

**19)Â AnswerÂ (D)**

First a black square can be selected in 32 ways. Out of remaining rows and columns, 24 white squares remain. 1 white square can them be chosenÂ in 24 ways. So total no. of ways of selection is 32*24 = 768.

**20)Â AnswerÂ (B)**

Total ways when all 6 boxes have only black balls = 1

Total ways when 5 boxes have black balls = 2

Total ways when 4 boxes have black balls = 3

Total ways when 3 boxes have black balls = 4

Total ways when 2 boxes have black balls = 5

Total ways when only 1 box has black ball = 6

So total ways of putting a black ball such that all of them come consecutively = (1+2+3+4+5+6) = 21

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