Rd Sharma 2022 _mcqs Solutions for Class 10 Maths Chapter 16 Probability are provided here with simple step-by-step explanations. These solutions for Probability are extremely popular among Class 10 students for Maths Probability Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2022 _mcqs Book of Class 10 Maths Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2022 _mcqs Solutions. All Rd Sharma 2022 _mcqs Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 248:

Question 1:

Mark the correct alternative in each of the following:

If a digit is chosen at random from the digit 1, 2, 3, 4, 5, 6, 7, 8, 9, then the probability that it is odd, is

(a) 49

(b) 59

(c) 19

(d) 23

Answer:

GIVEN: digits are chosen from 1, 2, 3 4, 5, 6, 7, 8, 9 are placed in a box and mixed thoroughly. One digit is picked at random.

TO FIND: Probability of getting an odd digit

Total number of digits is 9

Digit that are odd number are 1,3,5,7,9

Total number of odd digits is 5

We know that PROBABILITY =

Hence probability of getting an odd digit is

Hence optionis correct

Page No 248:

Question 2:

In Q. No. 1, The probability that the digit is even is

(a) 49

(b) 59

(c) 19

(d) 23

Answer:

GIVEN: digits are chosen from 1, 2, 3, 4, 5, 6, 7, 8, 9 are placed in a box and mixed thoroughly. One digit is picked at random.

TO FIND: Probability of getting an even digit

Total number of digits is 9

Digits that are even number are 2, 4, 6, and 8

Total number of even digits is 4.

We know that PROBABILITY =

Hence probability of getting an even digit is

Hence the correct option is

 

Page No 248:

Question 3:

In Q. No. 1, the probability that the digit is a multiple of 3 is

(a) 13

(b) 23

(c) 19

(d) 29

Answer:

GIVEN: digits are chosen from 1, 2, 3, 4, 5, 6, 7, 8, 9 are placed in a box and mixed thoroughly. One digit is picked at random.

TO FIND: Probability of getting a multiple of 3

Total number of digits is 9

Digits that are multiple of 3 are 3, 6 and 9

Total digits that are multiple of 3 are 3

We know that PROBABILITY =

Hence probability of getting a multiple of 3 is

Hence the correct option is option

 

Page No 248:

Question 4:

If three coins are tossed simultaneously, then the probability of getting at least two heads, is

(a) 14

(b) 38

(c) 12

(d) 14

Answer:

GIVEN: Three coins are tossed simultaneously.

TO FIND: Probability of getting at least two head.

When three coins are tossed then the outcome will be

TTT, THT, TTH, THH. HTT, HHT, HTH, HHH

Hence total number of outcome is 8.

At least two heads means that, THH, HHT, HTH and HHH are favorable events

Hence total number of favorable outcome is 4

We know that PROBABILITY =

Hence probability of getting at least two head when three coins are tossed simultaneously is equal to

Hence the correct option is

 

Page No 248:

Question 5:

In a single throw of a die, the probability of getting a multiple of 3 is

(a) 12

(b) 13

(c) 16

(d) 23

Answer:

GIVEN: A dice is thrown once

TO FIND: Probability of getting a multiple of 3.

Total number on a dice is 6.

Numbers which are on multiple of 3 are 3 and 6

Total number of numbers which are multiple of 3 is 2

We know that PROBABILITY =

Hence probability of getting a number which are multiple of 3 is equal to

Hence the correct option is

 

Page No 248:

Question 6:

The probability of guessing the correct answer to a certain test questions is x12. If the probability of not guessing the correct answer to this question is 23, then x =

(a) 2

(b) 3

(c) 4

(d) 6

Answer:

GIVEN: Probability of guessing a correct answer to a certain question is

Probability of not guessing a correct answer to a same question

TO FIND: The value of x

CALCULATION: We know that sum of probability of occurrence of an event and probability of non occurrence of an event is 1.

If E is an event of occurrence and is its complementary then

According to the question we have

Hence the correct option is

Page No 248:

Question 7:

A bag contains three green marbles, four blue marbles, and two orange marbles, If a marble is picked at random, then the probability that it is not an orange marble is

(a) 14

(b) 13

(c) 49

(d) 79

Answer:

GIVEN: A bag contains 3green, 4blue and 2orange marbles

TO FIND: Probability of not getting an orange marble

Total number of balls 3+4+2=9

Total number of non orange marbles that is green and blue balls is

We know that PROBABILITY =

Hence probability of getting a non orange ball is

Hence the correct option is

 

Page No 248:

Question 8:

A number is selected at random from the numbers 3, 5, 5, 7, 7, 7, 9, 9, 9, 9 The probability that the selected number is their average is

(a) 110

(b) 310

(c) 710
 
(d) 910

Answer:

GIVEN: A number is selected from the numbers 3,5,5,7,7,7,9,9,9,9

TO FIND: Probability that the selected number is the average of the numbers

Total numbers are 10

Average of numbers

Total numbers of numbers which are average of these numbers are 3

We know that PROBABILITY =

Hence Probability that the selected number is the average of the numbers is

Hence the correct option is

 

Page No 248:

Question 9:

The probability of throwing a number greater than 2 with a fair dice is

(a) 35

(b) 25

(c) 23

(d) 13

Answer:

GIVEN: A dice is thrown once

TO FIND: Probability of getting a number greater than 2.

Total number on a dice is 6.

Number greater than 2 is 3, 4, 5 and 6

Total number greater than 2 is 4

We know that PROBABILITY =

Hence probability of getting a number greater than 2 is equal to

Hence the correct option is

 

Page No 248:

Question 10:

A card is accidently dropped from a pack of 52 playing cards. The probability that it is an ace is

(a) 14

(b) 113

(c) 152

(d) 1213

Answer:

GIVEN: One card is drawn from a well shuffled deck of 52 playing cards

TO FIND: Probability of getting an Ace

Total number of cards is 52

Cards which are Ace are 1 from each suit

Total number of Ace cards is

We know that PROBABILITY =

Hence probability of getting an Ace is equal to

Hence the correct option is

 

Page No 248:

Question 11:

A number is selected from numbers 1 to 25. The probability that it is prime is

(a) 23

(b) 16

(c) 13

(d) 56

Answer:

GIVEN: A number is selected from 1 to 25

TO FIND: Probability that, the number is a prime

Total number is 25

Numbers from 1 to 25 that are primes are 2, 3, 5, 7, 11, 13, 17, 19 and 23

Total numbers that are primes from 1 to 25 is 9

We know that PROBABILITY =

Hence probability of getting a prime number from 1 to 25 is equal to

Hence no option is correct

 



Page No 249:

Question 12:

Which of the following cannot be the probability of an event?

(a) 23

(b) -1.5

(c) 15%

(d) 0.7

Answer:

GIVEN: 4 options of probability of some events

TO FIND: Which of the given options cannot be the probability of an event?

We know that

.

As the probability of an event cannot be negative

In option (b) P=-1.5

Hence the correct answer is option

 

Page No 249:

Question 13:

If P(E) = 0.05, then P(not E) =

(a) −0.05

(b) 0.5

(c) 0.9

(d) 0.95

Answer:

Given:

TO FIND:

CALCULATION: We know that sum of probability of occurrence of an event and probability of non occurrence of an event is 1. Therefore

Hence the correct answer is option

 

Page No 249:

Question 14:

Which of the following cannot be the probability of occurence of an event?

(a) 0.2

(b) 0.4

(c) 0.8

(d) 1.6

Answer:

GIVEN: 4 options of probability of some events

TO FIND: Which of the given options cannot be the probability of an event?

We know that

As the probability of an event cannot be more than 1

Hence the correct answer is option

 

Page No 249:

Question 15:

The probability of a certain event is

(a) 0

(b) 1

(c) 1/2

(d) no existent

Answer:

GIVEN: 4 options of probability of some events

TO FIND: Which of the given options is the probability of sure event?

We know that, probability of a certain event is 1.

Hence the correct answer is option.

 

Page No 249:

Question 16:

The probability of an impossible event is

(a) 0

(b) 1

(c) 1/2

(d) non-existent

Answer:

GIVEN: 4 options of probability of some events

TO FIND: Which of the given options is the probability of impossible event?

We know that, probability of an impossible event is 0.

Hence the correct answer is option

 

Page No 249:

Question 17:

Aarushi sold 100 lottery tickets in which 5 tickets carry prizes. If Priya purchased a ticket, what is the probability of Priya winning a prize?

(a) 1920

(b) 125

(c) 120

(d) 1720

Answer:

GIVEN: 100 lottery tickets were sold in which 5 tickets carry prize

TO FIND: Probability of Priya winning a prize

Total number of tickets is100

Total number of prize carrying tickets is 5

We know that PROBABILITY =

Hence probability of Priya winning a prize is equal to

The correct answer is option

Page No 249:

Question 18:

A number is selected from first 50 natural numbers. What is the probability that it is a multiple of 3 or 5?

(a) 1325

(b) 2150

(c) 1225

(d) 2350

Answer:

GIVEN: A number is selected from 50 natural numbers

TO FIND: Probability that the number selected is a multiple of 3 or 5

Total number is 50

Total numbers which are multiple of 3 or 5 up to 50 natural numbers are

3,6,5,9,10,12,15,18,20,21,24,25,27,30,33,35,36,39,40,42,45,48,50

Total number which are multiple of 3 or 5 up to 50 natural numbers are23

We know that PROBABILITY =

Hence probability that the number selected is a multiple of 3 or 5 is equal to

The correct answer is option

 

Page No 249:

Question 19:

A month is selected at random in a year. The probability that it is March or October, is

(a) 112

(b) 16

(c) 34

(d) None of these

Answer:

GIVEN: A month is selected at random in a year.

TO FIND: Probability that it is March or October

Total months in a year is 12

Hence total number of favorable outcome is 2 i.e. March or October

We know that PROBABILITY =

Hence probability that the month selected is March or October is equal to

Hence the correct option is

 

Page No 249:

Question 20:

From the letters of the word ''MOBILE",  a letter is selected. The probability that the letter is a vowel, is

(a) 13

(b) 37

(c) 16

(d) 12

Answer:

GIVEN: A letter is selected from the word “MOBILE”

TO FIND: Probability that the letter chosen is a vowel

Total letter in the word “MOBILE” is 6

Vowels in the word “MOBILE” are ‘O’.’I’,’E’

Hence total number of favorable outcome is 3 i.e. ‘O’.’I’,’E’

We know that PROBABILITY =

Hence probability that the letter chosen is a vowel

Hence the correct option is

Page No 249:

Question 21:

Mark the correct alternative in each of the following:

A die is thrown once. The probability of getting a prime number is

(a) 23                         (b) 13                         (c) 12                         (d) 16                           [CBSE 2013]

Answer:

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.

∴ Total number of outcomes = 6

The favourable outcomes are 2, 3 and 5.

So, the number of favourable outcomes are 3.

∴ P(getting a prime number) = Favourable number of outcomesTotal number of outcomes=36=12

Hence, the correct answer is option C.

Page No 249:

Question 22:

Mark the correct alternative in each of the following:

The probability of getting an even number, when a die is thrown once is

(a) 12                      (b) 13                      (c) 16                      (d) 56                  [CBSE 2013]

Answer:

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.

∴ Total number of outcomes = 6

The favourable outcomes are 2, 4 and 6.

So, the number of favourable outcomes are 3.

∴ P(getting an even number) = Favourable number of outcomesTotal number of outcomes=36=12

Hence, the correct answer is option A.

Page No 249:

Question 23:

Mark the correct alternative in each of the following:

A box contains 90 discs, numbered from 1 to 90. If one disc is drawn at random from the box, the probability that it bears a prime number less than 23, is

(a) 790                       (b) 1090                       (c) 445                       (d) 989                        [CBSE 2013]

Answer:

There are 90 discs numbered from 1 to 90.

∴ Total number of outcomes = 90

The prime numbers less than 23 are 2, 3, 5, 7, 11, 13, 17 and 19.

So, the favourable number of outcomes are 8.

∴ P(disc drawn bears a prime number less than 23) = Favourable number of outcomesTotal number of outcomes=890=445

Hence, the correct answer is option C.

Page No 249:

Question 24:

Mark the correct alternative in each of the following:

The probability that a number selected at random from the numbers 1, 2, 3, ..., 15 is a multiple of 4, is

(a) 415                          (b) 215                          (c) 15                          (d) 13                        [CBSE 2014]                        

Answer:

The total number of given numbers is 15.

∴ Total number of outcomes = 15

Among the given numbers, the multiples of 4 are 4, 8 and 12.

So, the favourable number of outcomes are 3.

∴ P(number selected is a multiple of 4) = Favourable number of outcomesTotal number of outcomes=315=15

Hence, the correct answer is option C.



Page No 250:

Question 25:

Mark the correct alternative in each of the following:

Two different coins are tossed simultaneously. The probability of getting at least one head is

(a) 14                            (b) 18                            (c) 34                            (d) 78                            [CBSE 2014]

Answer:

When two coins are tossed simultaneously, all possible outcomes are HH, HT, TH, TT.

∴ Total number of outcomes = 4

The favourable outcomes are HH, HT, TH.

So, the favourable number of outcomes are 3.

∴ P(getting at least one head) = Favourable number of outcomesTotal number of outcomes=34

Hence, the correct answer is option C.

Page No 250:

Question 26:

Mark the correct alternative in each of the following:

If two different dice are rolled together, the probability of getting an even number on both dice is

(a) 136                         (b) 12                         (c) 16                         (d) 14                     [CBSE 2014]               
   

Answer:

When two dice are rolled together, all possible outcomes are

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

∴ Total number of outcomes = 36

The favourable outcomes are (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4) and (6, 6).

So, the number of favourable outcomes are 9.

∴ P(getting even number on both dice) = Favourable number of outcomesTotal number of outcomes=936=14

Hence, the correct answer is option D.

Page No 250:

Question 27:

Mark the correct alternative in each of the following:

A number is selected at random from the numbers 1 to 30. The probability that it is a prime number is

(a) 23                            (b) 16                            (c) 13                            (d) 1130                            [CBSE 2014]

Answer:

Total number of outcomes = 30

The prime numbers from 1 to 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.

So, the favourable number of outcomes are 10.

∴ P(selected number is a prime number) = Favourable number of outcomesTotal number of outcomes=1030=13

Hence, the correct answer is option C.

Page No 250:

Question 28:

Mark the correct alternative in each of the following:

A card is drawn at random from a pack of 52 cards. The probability that the drawn card is not an ace is

(a) 113                            (b) 913                            (c) 413                            (d) 1213                            [CBSE 2014]

Answer:

Total number of possible outcomes = 52

There are 4 ace cards in a pack of cards.

∴ Number of non ace cards in the pack of cards = 52 − 4 = 48

So, the favourable number of outcomes are 48.

∴ P(drawn card is not an ace card) = Favourable number of outcomesTotal number of outcomes=4852=1213

Hence, the correct answer is option D.

Page No 250:

Question 29:

Mark the correct alternative in each of the following:

Two dice are thrown together. The probability of getting the same number on both dice is

(a) 12                      (b) 13                      (c) 16                      (d) 112                         [CBSE 2012]

Answer:

When two dice are thrown together, all possible outcomes are

(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

∴ Total number of outcomes = 36

The favourable outcomes are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6).

So, the number of favourable outcomes are 6.

∴ P(getting the same number on both dice) = Favourable number of outcomesTotal number of outcomes=636=16

Hence, the correct answer is option C.

Page No 250:

Question 30:

In a family of 3 children, the probability of having at least one boy is

(a) 78

(b) 18

(c) 58

(d) 34

Answer:

Considering G for ‘girl’ and B for ‘boy’.

In a family of 3 children, there are 8 possible outcomes as listed below - 
GGG, GGB, GBG, BGG, BBG, BGB, GBB, BBB

Let E be the event ‘getting at least one boy'.

The outcomes favourable to event E, 'at least one boy’ are (GGB), (GBG), (BGG), (BBG), (BGB), (GBB), (BBB).
So, the number of outcomes favourable to E is 7.
PE=Number of favourable outcomesTotal number of outcomes=78

Thus, the probability of having at least one boy is 78.
Hence, the correct answer is option (a).
 

Page No 250:

Question 31:

Mark the correct alternative in each of the following:

A bag contains cards numbered from 1 to 25. A card is drawn at random from the bag. The probability that the number on this card is divisible by both 2 and 3 is

(a) 15                            (b) 325                            (c) 425                            (d) 225                            [CBSE 2014]

Answer:

The total number of cards in the bag is 25.

∴ Total number of outcomes = 25

The numbers from 1 to 25 which are divisible by both 2 and 3 are 6, 12, 18 and 24.

So, the favourable number of outcomes are 4.

∴ P(number on the drawn card is divisible by both 2 and 3) = Favourable number of outcomesTotal number of outcomes=425

Hence, the correct answer is option C.

Page No 250:

Question 32:

A number x is chosen at random from the numbers −3, −2, −1, 0, 1, 2, 3 the probability that | x | < 2 is

(a) 57

(b) 27
 
(c) 37

(d) 17

Answer:

GIVEN: A number x is chosen from the numbers −3, −2, −1, 0, 1, 2 and 3

TO FIND: Probability of getting

Total numbers are 7

Number x such that are −1, 0, 1

Total numbers x such are 3

We know that PROBABILITY =

Hence the probability of getting a number x such thatis equal to

Hence the correct option is

 

Page No 250:

Question 33:

If a number x is chosen from the numbers 1, 2, 3, and a number y is selected from the numbers 1, 4, 9. Then, P(xy < 9)

(a) 79

(b) 59

(c) 23

(d) 19

Answer:

GIVEN: x is chosen from the numbers 1, 2, 3 and y is chosen from the numbers 1, 4, 9

TO FIND: Probability of getting

We will make multiplication table for x and y such that

So the numbers such that is 5

We know that PROBABILITY =

Hence

Hence the correct option is

Page No 250:

Question 34:

The probability that a non-leap year has 53 sundays, is

(a) 27

(b) 57

(c) 67

(d) 17

Answer:

GIVEN: A non leap year

TO FIND: Probability that a non leap year has 53 Sundays.

Total number of days in non leap year is 365days

Hence number of weeks in a non leap year is

In a non leap year we have 52 complete weeks and 1 day which can be any day of the week e.g. Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday

To make 53 Sundays the additional day should be Sunday

Hence total number of days is 7

Favorable day i.e. Sunday is 1

We know that PROBABILITY =

Hence probability that a non leap year has 53 Sundays is

Hence the correct option is

Page No 250:

Question 35:

In a single throw of a pair of dice, the probability of getting the sum a perfect square is

(a) 118

(b) 736

(c) 16

(d) 29

Answer:

GIVEN: A pair of dice is thrown

TO FIND: Probability of getting the sum a perfect square

Let us first write the all possible events that can occur

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6),

Hence total number of events is

Favorable events i.e. getting the sum as a perfect square are

(1,3), (2,2), (3,1), (3,6), (4,5), (5,4), (6,3)

Hence total number of favorable events is 7

We know that PROBABILITY =

Hence probability of getting the sum a perfect square is

Hence the correct option is option

Page No 250:

Question 36:

What is the probability that a non-leap year has 53 Sundays?


(a) 67

(b) 17

(c) 57

(d) None of these

Answer:

GIVEN: A non leap year

TO FIND: Probability that a non leap year has 53 Sundays.

Total number of days in a non leap year is 365days

Hence number of weeks in a non leap year is

In a non leap year we have 52 complete weeks and 1 day which can be any day of the week i.e. Sunday, Monday, Tuesday, Wednesday, Thursday, Friday and Saturday

To make 53 Sundays the additional day should be Sunday

Hence total number of days which can be any day is7

Favorable day i.e. Sunday is 1

We know that PROBABILITY =

Hence probability that a non leap year has 53 Sundays is

Hence the correct option is

Page No 250:

Question 37:

Two numbers 'a' and 'b' are selected successively without replacement in that order from the integers 1 to 10. The probability that ab is an integer, is

(a) 1745

(b) 15

(c) 1790

(d) 845

Answer:

We have a set of natural numbers from 1 to 10 where andare two variables which can take values from 1 to 10.

So, total number of possible combination of and so that is a fraction without replacement are:

Similarly we have 9 such sets of 10 elements each. So total number of possible combination,

Now the possible combination which makes an integer without replacement are-

Therefore the probability thatis an integer,

The correct answer is option



Page No 251:

Question 38:

Two dice are rolled simultaneously. The probability that they show different faces is

(a) 23

(b) 16

(c) 13

(d) 56

Answer:

GIVEN: A pair of dice is thrown

TO FIND: Probability of getting different faces

Let us first write the all possible events that can occur

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6),

Hence total number of events

Favorable events i.e. getting different faces of both dice are

(1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5),

Hence total number of favorable events i.e. getting different faces of both dice is 30

We know that PROBABILITY =

Hence probability of getting different faces of both dice is

Hence the correct option is

Page No 251:

Question 39:

What is the probability that a leap year has 52 Mondays?

(a) 27

(b) 47

(c) 57

(d) 67
 

Answer:

GIVEN: A leap year

TO FIND: Probability that a leap year has 52 Mondays.

Total number of days in leap year is 366days

Hence number of weeks in a leap year is

In a leap year we have 52 complete weeks and 2 day which can be any pair of the day of the week i.e.

(Sunday, Monday)

(Monday, Tuesday)

(Tuesday, Wednesday)

(Wednesday, Thursday)

(Thursday, Friday)

(Friday, Saturday)

(Saturday, Sunday)

To make 52 Mondays the additional days should not include Monday

Hence total number of pairs of days is 7

Favorable day i.e. in which Mondays is not there is 5

We know that PROBABILITY =

Hence probability that a leap year has 52 Mondays is equal to

Hence the correct option is

Page No 251:

Question 40:

If a two digit number is chosen at random, then the probability that the number chosen is a multiple of 3, is

(a) 310

(b) 29100

(c) 13

(d) 725

Answer:

GIVEN: A two digit number is chosen at random

TO FIND: Probability that the number chosen is a multiple of 3

Total two digit numbers is 90

Two digit Numbers multiple of 3 are 12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99

Hence total two digit numbers multiple of 3 are 30

We know that PROBABILITY =

Hence probability that the number chosen is a multiple of 3 is equal to

Hence the correct option is

 

Page No 251:

Question 41:

​​​​​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.

(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): The probability that a leap year has 53 Sundays is 27.
Statement-2 (Reason):  The probability that a non-leap year has 53 Sundays is 17.

Answer:

Statement-2 (Reason):  The probability that a non-leap year has 53 Sundays is 17.
The number of days in a non-leap year (or a common year) = 365 days.
So, there are 52 weeks and 1 day in 365 days.

There are 52 Sundays in a non-leap year. But 1 day apart from those 52 weeks can be either a Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, or Sunday.

Let E be the event of getting 53 Sundays in a non-leap year.

Here, the number of favourable outcomes = 1
Total number of outcomes = 7 
PE=Number of favourable outcomesTotal number of outcomes=17

Thus, Statement-2 is true.

Statement-1 (Assertion): The probability that a leap year has 53 Sundays is 27.
The number of days in a leap year (or a common year) = 366 days.
So, there are 52 weeks and 2 days in 366 days.

There are 52 Sundays in a leap year. But 2 days apart from those 52 weeks can be either a Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, or Sunday.

Let E be the event of getting 53 Sundays in a leap year.

Here, the number of favourable outcomes = 2
Total number of outcomes = 7 
PE=Number of favourable outcomesTotal number of outcomes=27

Thus, Statement-1 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).

Page No 251:

Question 42:

​​​​​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.

(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): When a die is rolled, the probability of getting a number which is a multiple of 3 and 5 both is zero.
Statement-2 (Reason):  The probability of an impossible event is zero.

Answer:

Statement-2 (Reason):  The probability of an impossible event is zero.
Let the possible number of outcomes be n.
Number of favourable outcomes for an impossible event is 0.
PE=Number of favourable outcomesTotal number of outcomes=0n=0

Thus, Statement-2 is true.

Statement-1 (Assertion): When a die is rolled, the probability of getting a number which is a multiple of 3 and 5 both is zero.

When a die is rolled, the possible outcomes are 1, 2, 3, 4, 5, 6.
∴ Number of  possible outcomes = 6
First multiple of both 3 and 5 is 15.
∴ Number of favourable outcomes = common multiples of 3 and 5 = 0       
⇒ The event of 'getting a number which is a multiple of 3 and 5 both' is an impossible event.

∴ P(getting a number which is a multiple of 3 and 5 both ) = 0     [From statement (1)]

Thus, Statement-1 is true.
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Page No 251:

Question 43:

​​​​​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.

(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): A cubical die is rolled. The probability of getting a composite number is 13.
Statement-2 (Reason):  In a throw of a cubical die, the probability of getting a prime number is 23.

Answer:

Statement-2 (Reason):  In a throw of a cubical die, the probability of getting a prime number is 23.
When a die is rolled, the possible outcomes are 1, 2, 3, 4, 5, 6.
∴ Number of  possible outcomes = 6

Let E be the event of 'getting a prime number'.
Thus, the favourable outcomes are 2, 3 and 5.
∴ Number of favourable outcomes = 3 

   PE=Number of favourable outcomesTotal number of outcomes=36=12 

Thus, Statement-2 is false.

Statement-1 (Assertion): A cubical die is rolled. The probability of getting a composite number is 13.

When a die is rolled, the possible outcomes are 1, 2, 3, 4, 5, 6.
∴ Number of  possible outcomes = 6

Let E be the event of 'getting a composite number'.
Thus, the favourable outcomes are 4 and 6.
∴ Number of favourable outcomes = 2 

PE=Number of favourable outcomesTotal number of outcomes=26=13 

Thus, Statement-1 is true.

So, Statement-1 is true, and Statement-2 is false.

Hence, the correct answer is option (c).

Page No 251:

Question 44:

​​​​​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.

(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): A number is selected from the numbers 1, 2, 3, ..., 10. The probability that it is a root of the equation x2-2x+1=0 is 15.
Statement-2 (Reason):  The equation x2 – 2x + 1 = 0 has two roots.

Answer:


Statement-2 (Reason):  The equation x2 – 2x + 1 = 0 has two roots.

 x2  2x + 1 =0  x2  x -x+ 1 =0 xx-1 -1x-1=0x-1 x-1=0x-1=0 or x-1=0x=1 or x=1
So, the  equation x2 – 2x + 1 = 0 has two equal roots, i.e., 1.
Thus, Statement-2 is true.

Statement-1 (Assertion): A number is selected from the numbers 1, 2, 3, ..., 10. The probability that it is a root of the equation x2-2x+1=0 is 15.

Possible outcomes are 1, 2, 3, ..., 10.
∴ Number of possible outcomes = 10

Let E be the event of 'getting a root of the equation x2-2x+1=0'.
From statement-2, equation x2-2x+1=0 has only one root, i.e. 1.
Favourable outcomes = 1

∴ Number of favourable outcomes = 1
PE=Number of favourable outcomesTotal number of outcomes=110

Thus, Statement-1 is false.

Hence, the correct answer is option (d).

Page No 251:

Question 45:

​​​​​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.

(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): A four digit number is formed using the digits 1, 2, 5, 6 and 8 without repetition. The probability that it is an even number is 35.
Statement-2 (Reason): The units digit of even number is also an even number.

Answer:


Statement-2 (Reason): The units digit of even number is also an even number.

The unit digits which are divisible by 2 are even numbers.
So, 0, 2, 4, 6, 8 are even numbers and any number having them at units digit is also an even number.

Thus, Statement-2 is true.

Statement-1 (Assertion): A four digit number is formed using the digits 1, 2, 5, 6 and 8 without repetition. The probability that it is an even number is 35.

Possible units digit that the numbers formed using the digits 1, 2, 5, 6 and 8 without repetition will have =  1, 2, 5, 6 and 8
∴  Number of possible outcomes = 5 

Let E be the event of 'getting an even number'.
Favourable outcomes = The numbers having even digits at units = 2, 6 and 8
∴  Number of favourable outcomes = 3

PE=Number of favourable outcomesTotal number of outcomes=35
Thus, Statement-2 is true.
But Statement-2 is not a correct explanation for Statement-1.

So, Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).

 

 

Page No 251:

Question 46:

​​​​​Each of the following questions contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has following four choices (a), (b), (c) and (d), only one of which is the correct answer. Mark the correct choice.

(a) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is false, Statement-2 is true.
Statement-1 (Assertion): Avni and Manvi were born in the year 2000. The probability that they have the same birthday is 1366.
Statement-2 (Reason):  Leap year has 366 days.

Answer:

Statement-2 (Reason):  Leap year has 366 days.

Number of days in a non-leap year = 365
A leap year has 1 day more than a non-leap year.
Number of days in a leap year = 365 + 1 = 366

Thus, Statement-2 is true.

Statement-1 (Assertion): Avni and Manvi were born in the year 2000. The probability that they have the same birthday is 1366.
Since 2000 is divisible by 4, 100 and 400, it is a leap year and number of days in a leap year = 366

∴ Number of possible outcomes = 366

Let E be the event of 'getting the same birthday for Avni and Manvi'.

Since there can be only 1 day in the whole year when they will share their birthday's on the same day.
∴ Number of favourable outcomes = 1
PE=Number of favourable outcomesTotal number of outcomes=1366

Thus, Statement-1 is true.
Also, Statement-2 is a correct explanation for Statement-1.

So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (b).



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