Probability Questions for MAH-CET | Download PDF

0
865
PROBABILITY Questions
PROBABILITY Questions

Probability Questions for MAH-CET PDF

Here you can download the important MAH – CET Probability Questions PDF by Cracku. Very Important MAH – CET 2022 and These questions will help your MAH – CET preparation. So kindly download the PDF for reference and do more practice.

Download Probability Questions for MAH-CET PDF

Enroll to MAH-CET Crash Course

Question 1: The total number of eight – digit landline telephone numbers that can be formed having at least one of their digits repeated is:

a) 98185600

b) 97428800

c) 100000000

d) None of the above

1) Answer (D)

Solution:

Total 8 digit number which can be made using all digits when repetition is allowed = $9*10^7$ = $90000000$

Total 8 digit number which can be made using all digits when repetition is not allowed = $9*9*8*7*6*5*4*3$ = $1632960$

Therefore, The total number of eight – digit landline telephone numbers that can be formed having at least one of their digits repeated =  $90000000$ – $1632960$ = $88367040$.  Hence the correct answer is None of the above (D).

Question 2: In an Engineering College in Pune, 8 males and 7 females have appeared for Student Cultural Committee selection process. 3 males and 4 females are to be selected. The total number of ways in which the committee can be formed, given that Mr. Raj is not to be included in the committee if Ms. Rani is selected, is:

a) 1960

b) 2840

c) 1540

d) None of the above

2) Answer (C)

Solution:

Mr. Raj is not to be included in the committee if Ms. Rani i.e. they both cannot be included together.
Let
i) Raj is selected and Rani is not.
Thus, the remaining 2 males can be selected $^7C_{2}$ ways and the remaining 4 females can be selected in $^6C_{4}$ ways.
Thus, the total number of ways = $^7C_{2}$*$^6C_{4}$ = $315$
ii) Raj is not selected and Rani is selected.

Thus, the remaining 3 males can be selected $^6C_{3}$ ways and the remaining 3 females can be selected in $^6C_{3}$ ways

Thus, the total number of ways = $^7C_{3}$*$^6C_{3}$ = $700$
iii) Both are not selected.

Thus, the remaining 3 males can be selected $^7C_{3}$ ways and the remaining 4 females can be selected in $^6C_{4}$ ways.
Thus, the total number of ways = $^7C_{3}$*$^6C_{4}$ = $525$
Thus, the total number of ways = $315+700+525 = 1540$
Hence, option C is the correct answer.

Question 3: The number of ways in which a mixed double tennis game can be arranged amongst 9 married couples if no husband and wife play in the same game is:

a) 1514

b) 1512

c) 3024

d) None of the above

3) Answer (B)

Solution:

There are 9 married couples. Therefore, there will be 9 men and 9 women.
First, let us select the 2 men.
2 men can be selected in 9C2 = 36 ways.

Now, the wives of these 2 men cannot be selected. Therefore, we have to select 2 women from the remaining 7 women.
2 women can be selected in 7C2 = 21 ways.

We have selected 2 men and 2 women. A team should consist of one man and one woman. Therefore, the 2 teams can be formed in 2 ways.

Therefore, the total number of ways in which the team can be selected is 36*21*2 = 1512.
Therefore, option B is the right answer.

Question 4: A medical clinic tests blood for certain disease from which approximately one person in a hundred suffers. People come to the clinic in group of 50. The operator of the clinic wonders whether he can increase the efficiency of the testing procedure by conducting pooled tests. In the pooled tests, the operator would pool the 50 blood samples and test them altogether. If the polled test was negative, he could pronounce the whole group healthy. If not, he could then test each person’s blood individually. The expected number of tests the operator will have to perform if he pools the blood samples are:

a) 47

b) 25

c) 21

d) None of the above

4) Answer (C)

Solution:

1 person in every 100 suffers from the disease.

Probability of a person being healthy = $\ \frac{\ 99}{100}$

In a group of 50 people if the test is positive, then he could then test each person’s blood individually otherwise he will consider that the entire group is healthy

The number of tests =50+1 = 51

The probability that all the people in the group are healthy = $^{50}C_{50}\ \times\ \left(\ \frac{\ 99}{100}\right)^{50}$

= approx 0.605

So the probability that atleast one person suffers in a group of 50 = 1-0.605= 0.395

Expected number of tests = 51*0.395 + 0.605*1

= 20.145+0.605

=20.75 = 21 tests.

C is the correct answer.

Question 5: The game of “chuck-a-luck” is played at carnivals in some parts of Europe. Its rules are as follows: if you pick a number from 1 to 6 and the operator rolls three dice. If the number you picked comes up on all three dice, the operator pays you €3; if it comes up on two dice, you are paid €2; and if it comes up on just one die, you are paid €1. Only if the number you picked does not come up at all, you pay the operator €1. The probability that you will win money playing in this game is:

a) 0.52

b) 0.753

c) 0.42

d) None of these

5) Answer (C)

Solution:

There are 3 ways to win money in the game.
The number you picked can come up in one dice, 2 dice or 3 dice.

The probability of the number you picked coming in all three dice = (1/6)*(1/6)*(1/6) = 1/216
The probability of the number picked coming on 2 dice = 3C2*(5/6)(1/6)(1/6) = 15/216
The probability of the number picked coming on 1 dice = 3C1*(5/6)(5/6)(1/6) = 75/216

Probability of winning = 1/216 + 15/214 + 75/216  = 91/216 = 0.421.
Therefore, option C is the right answer.

Take Free MAH-CET mock tests here

Instructions

Answer the questions based on the following information.

Rajat is sales manager of Dubin Computers Ltd. and looks after Delhi market. The company sells laptops in India. He is currently trying to select a distributor for coming five years. The distributor ensures that the products are accessible to the customers in the market. Market share of a company depends on the coverage by the distributor. The total profit potential of the entire laptop market in Delhi is Rs. 5 crores in the current year and present value of next four years’ cumulative profit potential is Rs. 15 crores. The first choice for Rajat is to enter into long-term contract with a distributor M/s Jagan with whom Dubin has done business in the past, and whose distribution system reaches 55 percent of all potential customers. At the last moment, however, a colleague suggests Rajat to consider signing a one-year contract with other distributors. Distributors M/s Bola and M/s James are willing to be partner with Dubin. Although a year ago M/s Bola’s and M/s James’s coverage reached only 40 and 25 percent of customers respectively, they claim to have invested heavily in distribution resources and now expect to be able to reach 60 percent and 75 percent of customers respectively. The probability of M/s Bola’s claim and M/s James’s claim to be true is 0.60 and 0.20 respectively. The knowledge about distributors’ coverage will evolve over time. The assumption is that the true level of coverage offered by the new distributors could be discovered, with certainty, through a one-year trail, and this trail will reveal exactly one of the two levels of coverage: for example in case of M/s Bola – 40 percent (as it was last year) or 60 percent (as claimed). In addition, it is also assumed that whatever the coverage is for both distributors, it will not change over time. Rajat narrows down on three choices, which are as follows:

Choice 1. Give a five year contract to the familiar distributor M/s Jagan.
Choice 2. Give a one year contract to the new distributor M/s Bola, and base next year’s decision to renew contract with M/s Bola on observed coverage for next four years or enter into a four years’ contract with M/s Jagan.
Choice 3. Give a one-year contract to the new distributor M/s James, and base next year’s decision to renew contract with M/s James on observed coverage for next four years or enter into a four years contract with M/s Jagan..

Question 6: If the distributor M/s James claims a coverage of 55% instead of 75% and probability of this claim to be true is 0.70 instead of 0.20 then which of the following statement is true?

a) Choice 1 is more profitable than Choice 2

b) Choice 2 is more profitable than Choice 3

c) Choice 3 is more profitable than Choice 1

d) None of the above

6) Answer (B)

Solution:

We are left with 3 choices.

Choice 1:

The first choice is to give the contract to M/S Jagan. In this case, we know that Jagan’s market reach is 55%. It has been given that the total profit potential is 5 crores in the present year and 15 crores in the next 4 years.

Therefore, the expected value of profit earned for choice 1 is 0.55*(5+15) = Rs.11 crore.

Choice 2:

Give the contract to M/s Bola for one year and based on the performance, renew the contract with him for the next 4 years or give M/S Jagan the contract for the next 4 years.

Let us assume that M/S Bola retains the contract for all 5 years. Rajat will renew the contract only if M/S Bola’s claim that their market reach is 60% is true. The probability of the claim being true is 0.6.

Therefore, the EV of return if M/S Bola bags the contract for all 5 years = 0.6*0.6*(5+15) = Rs. 7.2 crores.

Let us assume that M/S Bola’s claim is false. The probability of the claim being false is 1-0.6 = 0.4.
Now, if the claim is false, Rajat will terminate the contract by the end of the year and will partner with M/S Jagan for the next 4 years. Also, we have historic data that M/S Bola reaches 40% of the customers. Even if the claim is false, the laptops will reach 40% of the customers in the first year and 55% of the customers from the second year (Since M/S Jagan will bag the contract).

Therefore, the EV of profit in this case is 0.4*0.4*5+0.4*0.55*15 = 0.8 + 3.3 = Rs.4.1 crores.

Therefore, the total EV if M/S Bola bags the contract the first year is 7.2+4.1 = Rs.11.3 crores.

It has been given in this question that M/S James claims a coverage of 55% and the probability of this being true is 0.7.

Choice 3:

Give the contract to M/s James for one year and based on the performance, renew the contract with him for the next 4 years or give M/S Jagan the contract for the next 4 years.

Let us assume that M/S James retains the contract for all 5 years. Rajat will renew the contract only if M/S Jame’s claim that their market reach is 55% is true. The probability of the claim being true is 0.7.

Therefore, the EV of return if M/S James bags the contract for all 5 years = 0.7*0.55*(5+15) = Rs. 7.7 crores.

Let us assume that M/S James’s claim is false. The probability of the claim being false is 1-0.7 = 0.3.
Now, if the claim is false, Rajat will terminate the contract by the end of the year and will partner with M/S Jagan for the next 4 years. Also, we have historic data that M/S James reaches 25% of the customers. Even if the claim is false, the laptops will reach 25% of the customers in the first year and 55% of the customers from the second year (Since M/S Jagan will bag the contract).

Therefore, the EV of profit in this case is 0.3*0.25*5+0.3*0.55*15 = 0.375 + 2.475 = Rs.2.85 crores.

Therefore, the total EV if M/S Bola bags the contract the first year is 7.7+2.85 = Rs.10.55 crores.

EV of choice 1 = Rs. 11 crores
EV of choice 2 = Rs. 11.3 crores
EV of choice 3 = Rs. 10.55 crores

Arranging the choices by EV, we get, Choice 2 > Choice 1 > Choice 3.
Choice 2 is more profitable than choice 3. Therefore, option B is true and hence, option B is the right answer.

Question 7: McDonald’s ran a campaign in which it gave game cards to its customers. These game cards made it possible for customers to win hamburgers, French fries, soft drinks, and other fast-food items, as well as cash prizes. Each card had 10 covered spots that could be uncovered by rubbing them with a coin. Beneath three of these spots were “No Prize” signs. Beneath the other seven spots were names of the prizes, two of which were identical. For, example, one card might have two pictures of a hamburger, one picture of a coke, one of French fries, one of a milk shake, one of a $5, one of $1000, and three “No Prize” signs. For this card the customer could win a hamburger. To win on any card, the customer had to uncover the two matching spots (which showed the potential prize for that card)before uncovering a “No Prize”; any card with a “No Prize” uncovered was automatically void. Assuming that the two matches and the three “No Prize” signs were arranged randomly on the cards, what is the probability of a customer winning?

a) 0.10

b) 0.15

c) 0.12

d) None of the above

7) Answer (A)

Solution:

Case 2: When we win by uncovering just 3 spots.

_ _ P _ _ _ _ _ _ _

From the first two uncovered spots 1 will show up P. Out of remaining 7 spots, 3 spots will be filled by No prize. Total number of ways = 2C1*7C3*5!

There are a total of 10 spots out of which 3 are of one type (No Prize), 2 are of one time (The one which will give us prize) and 5 are different. Therefore, total number of combination in which we can uncover these spots = $\dfrac{10!}{2!*3!}$

We win if we win the two same card before any of the No prize spot. We can win by uncovering just 2 spots and a maximum of 7 spots. Let ‘P’ denotes the occurrence of winner card.

Case 1: When we win by uncovering just 2 spots.

P P _ _ _ _ _ _ _ _

Out of remaining 8 spots, 3 spots will be filled by No prize and 5 with different signs. Total number of ways = 8C3*5!

Case 2: When we win by uncovering just 3 spots.

_ _ P _ _ _ _ _ _ _

From the first two uncovered spots 1 will show up P. Out of remaining 7 spots, 3 spots will be filled by No prize. Total number of ways = 2C1*7C3*5!

 

Case 3: When we win by uncovering just 4 spots.

_ _  _ P _ _ _ _ _ _

From the first three uncovered spots 1 will show up P. Out of remaining 6 spots, 3 spots will be filled by No prize. Total number of ways = 3C1*6C3*5!

Case 4: When we win by uncovering just 5 spots.

_ _ _ _ P _ _ _ _ _

From the first four uncovered spots 1 will show up P. Out of remaining 5 spots, 3 spots will be filled by No prize. Total number of ways = 4C1*5C3*5!

Case 5: When we win by uncovering just 6 spots.

_ _ _ _ _ P _ _ _ _

From the first five uncovered spots 1 will show up P. Out of remaining 4 spots, 3 spots will be filled by No prize. Total number of ways = 5C1*4C3*5!

Case 6: When we win by uncovering just 7 spots.

_ _ _ _ _ _ P _ _ _

From the first six uncovered spots 1 will show up P. Out of remaining 3 spots, 3 spots will be filled by No prize. Total number of ways = 6C1*3C3*5!

Hence, the probability that a customer will win = $\dfrac{8C3*5!+2C1*7C3*5!+3C1*6C3*5!+4C1*5C3*5!+5C1*4C3*5!+6C1*3C3*5!}{\dfrac{10!}{2!*3!}}$

$\Rightarrow$ $\dfrac{3!*2!*5!(56+70+60+40+20+6)}{10!}$

$\Rightarrow$ $\dfrac{1}{10}$. Therefore, option A is the correct answer.

Question 8: While packing for a business trip Mr. Debashis has packed 3 pairs of shoes, 4 pants, 3 half-pants, 6 shirts, 3 sweater and 2 jackets. The outfit is defined as consisting of a pair of shoes, a choice of “lower wear” (either a pant or a half-pant), a choice of “upper wear” (it could be a shirt or a sweater or both) and finally he may or may not choose to wear a jacket. How many different outfits are possible?

a) 567

b) 1821

c) 743

d) None of the above

8) Answer (D)

Solution:

Let us find out the number of ways in which an outfit can be selected.
An outfit is defined as a pair of shoes, an upper wear, and a lower wear.
A pair of shoes can be selected in 3 ways.
There are 4 pants and 3 half-pants.
A pant or a half-pant can be selected in 7 ways.
A shirt can be selected in 6 ways.
A sweater can be selected in 3 ways.
A shirt and a sweater can be selected in 18 ways.
Therefore, an upper wear can be selected in 6+3+18 = 27 ways.
A jacket can be worn in 1 (no jacket is selected) + 2 (one of the 2 jackets is selected) = 3 ways.

Therefore, the total number of ways in which an outfit can be worn = 3*7*27*3 = 1701.
As the answer is not among the given choices, option D is the right answer.

Question 9: The internal evaluation for Economics course in an Engineering programme is based on the score of four quizzes. Rahul has secured 70, 90 and 80 in the first three quizzes. The fourth quiz has ten True-False type questions, each carrying 10 marks. What is the probability that Rahul’s average internal marks for the Economics course is more than 80, given that he decides to guess randomly on the final quiz?

a) 12/1024

b) 11/1024

c) 11/256

d) 12/256

9) Answer (B)

Solution:

Rahul has to score either 90 or 100 marks in the fourth quiz in order to have average more than 80.

So,there will be two cases:

Case 1: If Rahul scores 90 marks

Then 9 out of 10 will be correct and those 9 correct answers can be in any order. So, total ways of arranging them is $\frac{10!}{9!}$

And the probability of choosing either a right answer or wrong answer is $\frac{1}{2}$

Hence, the probability of getting 9 answers correct is: $\frac{10!}{9!}$  x $(\frac{1}{2})^{10}$

Case 2: If Rahul scores 100 marks

Then 10 out of 10 will be correct. So, total ways of arranging them is $\frac{10!}{10!}$ = 1

And the probability of choosing either a right answer or wrong answer is $\frac{1}{2}$

Hence, the probability of getting all 10 answers correct is: $(\frac{1}{2})^{10}$

So, the final probability is $\frac{10!}{9!}$  x $\frac{1}{2^{10}}$ + $\frac{1}{2^{10}}$ = $\frac{11}{1024}$

Question 10: During the essay writing stage of MBA admission process in a reputed B-School, each group consists of 10 students. In one such group, two students are batchmates from the same IIT department. Assuming that the students are sitting in a row, the number of ways in which the students can sit so that the two batchmates are not sitting next to each other, is:

a) 3540340

b) 2874590

c) 2903040

d) None of the above

10) Answer (C)

Solution:

Consider the case where batchmates are sitting together and then subtract those cases from total no. of cases.

When 10 students are arranged in a line, total arrangements possible are 10!

Now considering batchmates sitting together,

B1 B2  _ _ _ _ _ _ _ _

Then total arrangements possible are 9!*2!

Total ways = 10! – 9!*2!

9!*8 =  2903040

Question 11: In the board meeting of a FMCG Company, everybody present in the meeting shakes hand with everybody else. If the total number of handshakes is 78, the number of members who attended the board meeting is:

a) 7

b) 9

c) 11

d) 13

11) Answer (D)

Solution:

Total handshakes are given by: N$C_2$

N$C_2$ = 78

$\frac{N*(N-1)}{2}$ = 78

N=13

Hence, option D is the correct answer.

Question 12: In a reputed engineering college in Delhi, students are evaluated based on trimesters. The probability that an Engineering student fails in the first trimester is 0.08. If he does not fail in the first trimester, the probability that he is promoted to the second year is 0.87. The probability that the student will complete the first year in the Engineering College is approximately:

a) 0.8

b) 0.6

c) 0.4

d) 0.7

12) Answer (A)

Solution:

The probability that the student passes in the first trimester is 0.92

Now given that if the student passes in the first trimester then probability of moving into second second year is 0.87

Hence, the probability of completing first year is 0.92 x 0.87 = 0.80

Question 13: A playschool contains 4 boys and y girls. On every Wednesday during winter, five students, of which at least three are boys, go to Zoological Garden, a different group being sent every week. At the Zoological Garden, each boy in the group is given a ball. If the total number of balls distributed is 368, then the value of y is

a) 5

b) 6

c) 7

d) 8

13) Answer (D)

Solution:

Total 5 students out of 4 boys and y girls can be chosen in two ways :

(1)  3 Boys and 2 Girls = $4c_{3}*$y$c_{2}$

(2)  4 Boys and 1 Girls = $4c_{4}*$y$c_{1}$

So total number of balls given to boys :

3*$4c_{3}*$y$c_{2}$+4*$4c_{4}*$y$c_{1}$ = 368

$6(y)(y-1)+4y$ = 368

6$y^{2}-2y-368$ = 0

3$y^{2}-y-184$ = 0

$(y-8)(3y+23)$ = 0

$y=8$,  y $\neq$ $ \frac{-23}{3}$

Question 14: Which of the following statements regarding arrangement of the word ‘RIYADH’ is/are true:
i. Two vowels can be arranged together in 120 ways
ii. Vowels do not occur together in 240 ways
Which of the above statements are true?

a) Statement (i) only

b) Statement (ii) only

c) Both statements (i) and (ii)

d) None of the above

14) Answer (D)

Solution:

Let’s check both the statements one by one:

(i) Two vowels together –  (2!)*(5!) = 2*120 = 240

                                                

So we can say that statement (i) is wrong

(ii) Two vowel not together =  Total ways – Vowels together

= (6!) – (2!)*(5!) = 720 – 240 = 480

hence statement (ii) is also incorrect.

Question 15: In a True/False quiz, 4 marks are awarded for each correct answer and 1 mark is deducted for each wrong answer. Amit, Benn and Chitra answered the same 10 questions, and their answers are given below in the same sequential order.
AMIT     T T F F T T F T T F
BENN    T T T F F T F T T F
CHITRA T T T T F F T F T T
If Amit and Benn both score 35 marks each then Chitra’s score will be:

a) 10

b) 15

c) 20

d) 25

e) None of the above

15) Answer (A)

Solution:

Both Amit and Ben scored 35 marks. 4 marks are awarded for a correct answer and 1 mark is deducted for an incorrect answer.
Let the number of questions that Amit got right be ‘x’.
=> 4x -(10-x) = 35
5x = 45
x = 9.
Therefore, Amit must have made only 1 mistake. The same must have been the case with Ben too.
The responses given by the 3 persons are as follows:

AMIT     T T F F T T F T T F
BENN    T T T F F T F T T F
CHITRA T T T T F F T F T T

Amit and Ben have given different responses for question 3 and question 5. Therefore, one of them must be wrong in each of these questions. Also, Amit must have given the correct answer for one of these 2 questions and Ben must have answered the other one correct, since both of them got 9 questions correct.

Let us assume Amit has given an incorrect response for question 3 and Ben has given an incorrect response for question 5.
In this case, Chitra’s would have given 4 correct responses (questions 1,2,3, and 9). Chitra’s score would have been 4*4 – 6 = 10.

Let us assume Amit has given an incorrect response for question 5 and Ben has given an incorrect response for question 3.
In this case, Chitra’s would have given 4 correct responses (questions 1,2,5, and 9). Chitra’s score would have been 4*4 – 6 = 10.

Therefore, Chitra’s score should have been 10 and hence, option A is the right answer.

Question 16: A dice is rolled twice. What is the probability that the number in the second roll will be higher than that in the first?

a) 5/36

b) 8/36

c) 15/36

d) 21/36

e) None of the above

16) Answer (C)

Solution:

A die is rolled twice.
The number of combinations that can occur = 6*6 = 36.
We have to find the probability of the second roll being higher than the first.
If we select 2 numbers out of the 6 and arrange them in ascending order, then we will obtain the scenario in which the number obtained in the second roll will be greater than the number obtained in the first roll.

2 numbers out of 6 numbers can be selected in 6C2 = 15 ways. The numbers can be arranged in ascending order in only one way.
Therefore, the required probability is 15/36.

Therefore, option C is the right answer.

Question 17: An institute has 5 departments and each department has 50 students. If students are picked up randomly from all 5 departments to form a committee, what should be the minimum number of students in the committee so that at least one department should have representation of minimum 5 students?

a) 11

b) 15

c) 21

d) 41

e) None of the above.

17) Answer (C)

Solution:

We have to employ the pigeon hole principle to solve this problem.
The maximum number of students who can be picked from each department such that 5 students are not selected from the same department is 4.
Therefore, after 4 students from each department are selected (i.e., 4*5 = 20 students in total), the 21st student selected will be the fifth student to be selected from one of the 5 departments. Therefore, 20+1 = 21 students should be selected in total to ensure that at least five students from one of the departments is selected. Therefore, option C is the right answer.

Question 18: Sara has just joined Facebook. She has 5 friends. Each of her five friends has twenty five friends. It is found that at least two of Sara’s friends are connected with each other. On her birthday, Sara decides to invite her friends and the friends of her friends. How many people did she invite for her birthday party?

a) $\geq105$

b) $\leq123$

c) $<125$

d) $\geq100$ and $\leq125$

e) $\geq105$ and $\leq123$

18) Answer (B)

Solution:

Sarah has 5 friends and each of her 5 friends have 25 friends.
The 5 friends of Sarah have Sarah as one of their 25 friends. Therefore, apart from Sarah the five friends will have 24 friends each.
If no 2 friends know each other, then Sarah will have to invite 5*24 + 5  = 120 + 5 = 125 persons.
Sarah knows that at least 2 of her friends are connected to each other.
Let the 5 friends be A, B, C, D, and E. Let us consider that A and B are friends.
A will be counted twice (first as one among the 5 persons and second as one of the friends of B).
B will also be counted twice (first as one among the 5 persons and second as one of the friends of A).
Subtracting these 2 friends, we can infer that the maximum number of persons that Sarah could have invited to the party is 123.

The minimum number of friends that Sarah could have invited is obtained when all the friends are connected to each other and their friends are also the same. In this case, the five friends will have 20 common friends, will be friends with each other (4) and Sarah (1).
The minimum number of persons Sarah could have invited to the party = 20+5 = 25.
Option B is more appropriate that option C.
Therefore, option B is the right answer.

Question 19: A coin of radius 3 cm is randomly dropped on a square floor full of square shaped tiles of side 10 cm each. What is the probability that the coin will land completely within a tile? In other words, the coin should not cross the edge of any tile.

a) 0.91

b) 0.5

c) 0.49

d) 0.36

e) 0.16

19) Answer (E)

Solution:

The radius of the coin is 3 cm.
So, if the coin should not cross the edge, the centre of the coin should at least be 3 cm away from the edge of the tile.

In the given diagram the center of the circle should lie in the blue region. As we can see, the area in which the centre of the coin can fall is a square of side 10-3-3 = 4 cm.
Therefore, the area in which the centre of the coin can fall is 16 square cm.
Area of the tile = 100 square cm.
Required probability = 16/100 = 0.16.
Therefore, option E is the right answer.

Question 20: All words formed by permutations of the word `WARE’ are arranged in a list according to the dictionary ordering. The position of the word ‘WEAR’ in this list is at number

a) 20

b) 21

c) 22

d) 23

20) Answer (B)

Solution:

All the words starting with the letter A =3!

All the words starting with the letter E =3!

All the words starting with the letter R =3!

All the words starting with the letter WA =2!

All the words starting with the letter WEAR =1

$\therefore$ position of the word ‘WEAR’ in this list is at number 3!*3 + 2+1

=21

Hence B is the correct answer.

Take MAH-CET Mock Tests

Enroll to CAT 2022 course

LEAVE A REPLY

Please enter your comment!
Please enter your name here