RD Sharma 2018 Solutions for Class 10 Math Chapter 16 Probability are provided here with simple step-by-step explanations. These solutions for Probability are extremely popular among class 10 students for Math Probability Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2018 Book of class 10 Math Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s RD Sharma 2018 Solutions. All RD Sharma 2018 Solutions for class 10 Math are prepared by experts and are 100% accurate.

#### Page No 16.20:

#### Question 1:

The probability that it will rain tomorrow is 0.85. What is the probability that it will not rain tomorrow?

#### Answer:

Given: Probability that it will rain

TO FIND: Probability that it will not rain

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

Hence the probability that it will not rain is

#### Page No 16.20:

#### Question 2:

A die is thrown, Find the probability of getting:

(a) a prime number

(b) 2 or 4

(c) a multiple of 2 or 3

(d) an even prime number

(e) a number greater than 5

(f) a number lying between 2 and 6

#### Answer:

GIVEN: A dice is thrown once

TO FIND

(i) Probability of getting a prime number

(ii) Probability of getting 2 or 4

(iii) Probability of getting a multiple of 2 or 3.

(iv) Probability of getting an even number

(v) Probability of getting a number greater than five.

(vi) Probability of lying between 2 and 6

Total number on a dice is 6.

(i) Prime number on a dice are 2,3,5

Total number of prime numbers on dice is 3

We know that PROBABILITY =

Hence probability of getting a prime number =

(ii) for getting 2 and 4 favorable outcome are 2

We know that PROBABILITY =

Hence probability of getting 2 or 4 =

(iii) Multiple of 2 are 3 are 2, 3, 4 and 6

Hence favorable outcome is 4

We know that PROBABILITY =

Hence probability of getting an multiple of 2 or 3 =

(iv) an even prime number is 2

Hence favorable outcome is 1

We know that PROBABILITY =

Hence probability of getting an even prime number

(v) A number greater than 5 is 6

Hence favorable outcome is 1

We know that PROBABILITY =

Hence probability of getting a number greater than 5

(vi) Total number on a dice is 6.

Number lying between 2 and 6 are 3, 4 and 5

Total number of number lying between 2 and 6 is 3

We know that PROBABILITY =

Hence probability of getting a number lying between 2 and 6 =

#### Page No 16.20:

#### Question 3:

Three coins are tossed together. Find the probability of getting:

(a) exactly two heads

(b) at most two heads

(c) at least one head and one tail.

(d) no tails

#### Answer:

GIVEN: Three coins are tossed simultaneously.

TO FIND: We have to find the following probability

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.

(i) For exactly two head we get favorable outcome as THH, HHT ,HTH

Hence total number of favorable outcome i.e. exactly two head 3

We know that PROBABILITY =

Hence probability of getting exactly two head is

(ii) In case, at least two head we have favorable outcome as TTT, THT, TTH, THH. HTT, HHT, HTH

Hence total number of favorable outcome i.e. at most two head is 7

We know that PROBABILITY =

Hence probability of getting at most two head when three coins are tossed simultaneously is equal to $\frac{7}{8}$

(iiii) At least one head and one tail we get in case THT, TTH, THH. HTT, HHT, HTH,

Hence total number of favorable outcome i.e. at least one tail and one head is 6

We know that PROBABILITY =

Hence probability of getting at least one head and one tail is equal to

(iv) No tail i.e. HHH

Hence total number of favorable outcome is 1

We know that PROBABILITY =

Hence probability of getting no tail is

#### Page No 16.20:

#### Question 4:

*A* and *B* throw a pair of dice. If *A* throws 9, find *B*'s chance of throwing a higher number.

#### Answer:

GIVEN: A pair of dice is thrown

TO FIND: Probability that the total of numbers on the dice is greater than 9

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 total of numbers on the dice greater than 9 are

(5,5), (5,6), (6,4), (4,6), (6,5) and (6,6),

Hence total number of favorable events i.e. getting the total of numbers on the dice greater than 9 is 6

We know that PROBABILITY =

Hence probability of getting the total of numbers on the dice greater than 9 is

#### Page No 16.20:

#### Question 5:

Two unbiased dice are thrown. Find the probability that the total of the numbers on the dice is greater than 10.

#### Answer:

GIVEN: A pair of dice is thrown

TO FIND: Probability that the total of numbers on the dice is greater than 10

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 total of numbers on the dice greater than 10

Is (5, 6), (6, 5) and (6, 6)

Hence total number of favorable events i.e. getting the total of numbers on the dice greater than 10 is 3

We know that PROBABILITY =

Hence probability of getting the total of numbers on the dice greater than 10 is

#### Page No 16.20:

#### Question 6:

A card is drawn at random from a pack of 52 cards. Find the probability that the card drawn is

(i) a black king

(ii) either a black card or a king

(iii) black and a king

(iv) a jack, queen or a king

(v) neither a heart nor a king

(vi) spade or an ace

(vii) neither an ace nor a king

(viii) neither a red card nor a queen.

(ix) other than an ace

(x) a ten

(xi) a spade

(xii) a black card

(xiii) the seven of clubs

(xiv) jack

(xv) the ace of spades

(xvi) a queen

(xvii) a heart

(xviii) a red card

(xix) neither a king nor a queen

#### Answer:

Given: A card is drawn at random from a pack of 52 cards

TO FIND: Probability of the following

Total number of cards = 52

(i) Cards which are black king is 2

We know that PROBABILITY =

Hence probability of getting a black king is equal to

(ii) Total number of black cards is 26

Total numbers of kings are 4 in which 2 black kings are also included

Hence total number of black card or king will be

We know that PROBABILITY =

Hence probability of getting a black cards or a king =

(iii) Total number of black and a king cards is 2

We know that PROBABILITY =

Hence probability of getting a black cards and a king is

(iv) A jack, queen or a king are 3 from each 4 suits

Total number of a jack, queen and king are 12

We know that PROBABILITY =

Hence probability of getting a jack, queen or a king is =

(v) Total number of heart cards are 13 and king are 4 in which king of heart is also included.

Total number of cards that are a heart and a kingie equal to

Hence Total number of cards that are neither a heart nor a king=

We know that PROBABILITY =

Hence probability of getting cards neither a heart nor a king =

(vi) Total number of spade cards is 13

Total number of aces are 4 in which ace of spade is included in the spade cards.

Hence total number of card which are spade or ace =

We know that PROBABILITY =

Hence probability of getting cards that is spade or an ace = $\frac{16}{52}=\overline{)\frac{4}{13}}$

(vii) Total number of ace card are 4 and king are 4

Total number of cards that are a ace and a king is equal to

Hence Total number of cards that are neither an ace nor a kin is

We know that PROBABILITY =

Hence probability of getting cards neither an ace nor a king =

(viii) Total number of red cards is 26

Total numbers of queens are 4 in which 2 red queens are also included

Hence total number of red card or queen will be

Hence Total number of cards that are neither a red nor a queen=

We know that PROBABILITY =

Hence probability of getting neither a red card nor a queen is equal to

(ix) Total number of card other than ace is

We know that PROBABILITY =

Hence probability of getting other than ace is

(x) Total number of ten is 4

We know that PROBABILITY =

Hence probability of getting a ten is

(xi) Total number of spade is 13

We know that PROBABILITY =

Hence probability of getting a spade =

(xii) Total number of black cards is 26

We know that PROBABILITY =

Hence probability of getting black cards is

(xiii) Total number of 7 of club is 1

We know that PROBABILITY =

Hence probability of getting a 7 of club is equal to

(xiv) Total number of jack are 4

We know that PROBABILITY =

Hence probability of getting jack

(xv) Total number of ace of spade is 1

We know that PROBABILITY =

Hence probability of getting a ace of spade

(xvi) Total number of queen is 4

We know that PROBABILITY =

Hence probability of getting a queen

(xvii) Total number of heart cards is 13

We know that PROBABILITY =

Hence probability of getting a heart cards =

(xviii) Total number of red cards is 26

We know that PROBABILITY =

Hence probability of getting a red cards =

#### Page No 16.21:

#### Question 7:

In a lottery of 50 tickets numbered 1 to 50, one ticket is drawn. Find the probability thatn the drawn ticket bears a prime number.

#### Answer:

GIVEN: Tickets are marked with one of the numbers 1 to 50. One ticket is drawn at random.

TO FIND: Probability of getting a prime number on the drawn ticket

Total number of tickets is 5.

Tickets marked prime number are 1,3,5,7,11,13,17,19,23,29,31,37,43,47,49

Total number of tickets marked prime is 15

We know that PROBABILITY =

Hence probability of getting a prime number on the ticket is

#### Page No 16.21:

#### Question 8:

An urn contains 10 red and 8 white balls. One ball is drawn at random. Find the probability that the ball drawn is white.

#### Answer:

GIVEN: A bag contains10 red, and 8 white balls

TO FIND: Probability that one ball is drawn at random and getting a white ball

Total number of balls

Total number of white balls is 8

We know that PROBABILITY =

Hence probability of getting a white ball is

#### Page No 16.21:

#### Question 9:

A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is :

(a) white?

(b) red?

(c) black?

(d) not red?

#### Answer:

GIVEN: A bag contains 3 red, 5 black and 4 white balls

TO FIND: Probability of getting a

(i) White ball

(ii) Red ball

(iii) Black ball

(iv) Not red ball

Total number of balls

(i) Total number white balls is 4

We know that PROBABILITY =

Hence probability of getting white ball =

(ii) Total number red balls are 3

We know that PROBABILITY =

Hence probability of getting red ball is

(iii) Total number of black balls is 5

We know that PROBABILITY =

Hence probability of getting black ball

(iv) Total number of non red balls are 4 white balls and 5 black balls i.e.

We know that PROBABILITY =

Hence probability of getting non red ball

#### Page No 16.21:

#### Question 10:

What is the probability that a number selected from the numbers 1, 2, 3, ...,15 is a multiple of 4?

#### Answer:

GIVEN: Numbers are from 1 to 15. One number is selected

TO FIND: Probability that the selected number is multiple of 4

Total number is 15

Numbers that are multiple of 4 are 4,8,12,

Total number which is multiple of 4 is 3

We know that PROBABILITY =

Hence probability of selecting a multiple of 4 is

#### Page No 16.21:

#### Question 11:

A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the probability that ball drawn is not black?

#### Answer:

GIVEN: A bag contains 6 red, 8 black and 4 white balls and a ball is drawn at random

TO FIND: Probability that the ball drawn is not black

Total number of balls

Total number of black balls is 8

We know that PROBABILITY =

Probability of getting a black ball

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

Hence

#### Page No 16.21:

#### Question 12:

A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the probability that ball drawn is not black?

#### Answer:

GIVEN: A bag contains 7 red, and 5 white balls and a ball is drawn at random

TO FIND: Probability that the ball drawn is white

Total number of balls

Total number of white balls is 5

We know that PROBABILITY =

Probability of getting a white ball

#### Page No 16.21:

#### Question 13:

Tickets numbers from 1 to 20 are mixed up and a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 7?

#### Answer:

GIVEN: Tickets are marked from 1 to 20 are mixed up. One ticket is picked at random.

TO FIND: Probability that the ticket bears a multiple of 3 or 7

Total number of cards is 20

Cards marked multiple of 3 or 7 are 3, 6, 7, 9, 12, 14, 15 and18

Total number of cards marked multiple of 3 or 7 are 8

We know that PROBABILITY =

Hence probability of getting a, multiple of 3 or 7 is

#### Page No 16.21:

#### Question 14:

In a lottery there are 10 prizes and 25 blanks. What is the probability of getting a prize?

#### Answer:

GIVEN: In a lottery there are 10 prizes and 25 blanks.

TO FIND: Probability of winning a prize

Total number of tickets is

Total number of prize carrying tickets is 10

We know that PROBABILITY =

Hence probability of winning a prize is

#### Page No 16.21:

#### Question 15:

If the probability of winning a game is 0.3, what is the probability of loosing it?

#### Answer:

Given: Probability of winning a game

TO FIND: Probability of losing the game

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

Hence the probability of losing the game is

#### Page No 16.21:

#### Question 16:

A bag contains 5 black, 7red and 3 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is :

(i) red

(ii) black or white

(iii) not black

#### Answer:

GIVEN: A bag contains 7 red, 5 black and 3 white balls and a ball is drawn at random

TO FIND: Probability of getting a

(i) Red ball

(ii) Black or white ball

(iii) Not black ball

Total number of balls

(i) Total number red balls are 7

We know that PROBABILITY =

Hence probability of getting a red ball is equal to

(ii) Total number of black or white balls is

We know that PROBABILITY =

Hence probability of getting white or black ball

(iii) Total number of black balls is 5

We know that PROBABILITY =

Hence probability of getting black ball

We know that sum of probability of occurrence of an event and probability of non

occurrence of an event is 1

Hence the probability of getting non black ball

#### Page No 16.21:

#### Question 17:

A bag contains 4 red, 5 black and 6 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is :

(a) white

(b) red

(c) not black

(d) red or white

#### Answer:

GIVEN: A bag contains 4 red, 5 black and 6white balls and a ball is drawn at random

TO FIND: Probability of getting a

(i) white ball

(ii) red ball

(iii) not black ball

(iv) red or white

Total number of balls

(i) Total number white balls are 6

We know that PROBABILITY =

Hence probability of getting white a ball is

(ii) Total number of red are 4

We know that PROBABILITY =

Hence probability of getting red a ball is equal to

(iii) Total number of black balls are 5

We know that PROBABILITY =

Hence probability of getting black ball

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

Hence the probability of getting non black ball is

(iv) Total number of red or white balls is

We know that PROBABILITY =

Hence probability of getting white or red ball

#### Page No 16.21:

#### Question 18:

One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting:

(i) a king of red suit

(ii) a face card

(iii) a red face card

(iv) a queen of black suit

(v) a jack of hearts

(vi) a spade

#### Answer:

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

TO FIND: Probability of following

Total number of cards is 52

(i) Cards which are king of red suit are 2

Total number of Cards which are king of red suit is 2

Number of favorable event i.e. Total number of Cards which are king of red suit is 2

We know that PROBABILITY =

Hence probability of getting cards which are king of red suit is

(ii) Total number of face cards are 12

Number of favorable event i.e. total number of face cards is 12

We know that PROBABILITY =

Hence probability of getting face cards is

(iii) Total number of red face cards are 6

Number of favorable events i.e. total number of red face cards is 6

We know that PROBABILITY =

Hence probability of getting red face cards is

(iv) Total number of queen of black suit cards is 2

Total Number of favorable event i.e. total number of queen of black suit cards is 2

We know that PROBABILITY =

Hence probability of getting cards which are queen of black suit cards is

(v) Total number of jack of hearts is 1

We know that PROBABILITY =

Hence probability of getting cards which are jack of heart is equal to

(vi) Total number of spade cards are 13

Total Number of favorable event i.e. total number of queen of black suit cards are 13

We know that PROBABILITY =

Hence probability of getting spade cards is

#### Page No 16.21:

#### Question 19:

Five cards−ten, jack, queen, king, and an ace of diamonds are shuffled face downwards. One card is picked at random.

(i) What is the probability that the card is a queen?

(ii) If a king is drawn first and put aside, what is the probability that the second card picked up is the ace?

#### Answer:

GIVEN: Five cards-ten, jack, queen, king and Ace of diamond are shuffled face downwards

TO FIND: Probability of following

Total number of cards are5

(i) Cards which is a queen

Total number of Cards which are queen is 1

Number of favorable event i.e. Total number of Cards which are queen is 1

We know that PROBABILITY =

Hence probability of getting cards which are queen

(ii) If a king is drawn first and put aside then

Total number of cards is 4

Number of favorable event i.e. Total number of ace card is 1

We know that PROBABILITY =

Hence probability of getting ace cards

#### Page No 16.21:

#### Question 20:

A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is :

(i) red

(ii) black

#### Answer:

GIVEN: A bag contains 3 red, and 5 black balls. A ball is drawn at random

TO FIND: Probability of getting a

(i) red ball

(ii) white ball

Total number of balls

(i) Total number red balls are 3

We know that PROBABILITY =

Hence probability of getting red ball is equal to

(ii) Total number of black ball are 5

We know that PROBABILITY =

Hence probability of getting black ball

#### Page No 16.21:

#### Question 21:

A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number, 1, 2, 3, ..., 12 as shown in following figure. What is the probability that it will point to:

(i) 10?

(ii) an odd number?

(iii) a number which is multiple of 3?

(iv) an even number?

#### Answer:

GIVEN: A game of chance consists of spinning an arrow which is equally likely to come to rest pointing number 1,2,3…..12

TO FIND: Probability of following

Total number on the spin is 12

(i) Favorable event i.e. to get 10 is 1

Total number of Favorable event i.e. to get 10 is 1

We know that PROBABILITY =

Hence probability of getting a10

(ii) Favorable event i.e. to get an odd number are 1,3,5,7,9,11,

Total number of Favorable event i.e. to get a prime number is 6

We know that PROBABILITY =

Hence probability of getting a prime number is

(iii) Favorable event i.e. to get an multiple of 3 are 3,6,9,12

Total number of Favorable event i.e. to get a multiple of 3 is 4

We know that PROBABILITY =

Hence probability of getting multiple of 3 =

(iv) Favorable event i.e. to get an even number are 2,4,6,8,10,12

Total number of Favorable events i.e. to get an even number is 6

We know that PROBABILITY =

Hence probability of getting an even number is

#### Page No 16.22:

#### Question 22:

In a class, there are 18 girls and 16 boys. The class teacher wants to choose one pupil for class monitor. What she does, she writes the name of each pupil on a card and puts them into a basket and mixes thoroughly. A child is asked to pick one card from the basket. What is the probability that the name written on the card is:

(i) the name of a girl

(ii) the name of a boy?

#### Answer:

GIVEN: In a class there are 18 girls and 16 boys, the class teacher wants to choose one name. The class teacher writes all pupils name on a card and puts them in basket and mixes well thoroughly .A child picks one card

TO FIND: The probability that the name written on the card is

(i) The name of a girl

(ii) The name of a boy

Total number of students in the class

(i) Total numbers of girls are 18 hence favorable cases are 18

We know that PROBABILITY =

Hence probability of getting a name of girl on the card picked =

(ii) Total numbers of boys are 16 hence favorable cases are 16

We know that PROBABILITY =

Hence probability of getting a name of boy on the card picked =

#### Page No 16.22:

#### Question 23:

Why is tossing a coin considered to be a fair way of deciding which team should choose ends in a game of cricket?

#### Answer:

When we toss a coin then the outcomes have the same probability for its occurrence they are equally likely events. So, the result of an individual coin toss is completely unpredictable.

#### Page No 16.22:

#### Question 24:

What is the probability that a number selected at random from the number 1, 2, 2, 3, 3, 3, 4, 4, 4, 4 will be their average?

#### Answer:

GIVEN: A number is selected from the numbers 1,2,2,3,3,3,4,4,4,4

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

Total numbers are 10

Average of numbers is

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 =

#### Page No 16.22:

#### Question 25:

There are 30 cards, of same size, in a bag on which numbers 1 to 30 are written. One card is taken out of the bag at random. Find the probability that the number on the selected card is not divisible by 3.

#### Answer:

GIVEN: Cards are marked with one of the numbers 1 to 30 are placed in a bag and mixed thoroughly. One card is picked at random.

TO FIND: Probability of getting a number not divisible by 3 on the picked card.

Total number of cards is 30

Cards marked number not divisible by 3 is 1,2,4,5,7,8,10,11,13,14,16,17,19,20,22,23,25,26,28,29.

Total number of cards marked numbers not divisible by 3 is 20

We know that PROBABILITY =

Hence probability of getting a number divisible by 3 on the card =

#### Page No 16.22:

#### Question 26:

A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random form the bag. Find the probability that the drawn ball is (i) red or white (ii) not black (iii) neither white nor black.

#### Answer:

GIVEN: A bag contains 5 red, 7 black and 8 white balls and a ball is drawn at random

TO FIND: Probability of getting a

(i) red or white ball

(ii) not black ball

(iii) neither white nor black

Total number of balls

(i) Total number red and white balls are

We know that PROBABILITY =

Hence probability of getting red or white ball

(ii) Total number of black balls are 7

We know that PROBABILITY =

Hence probability of getting black ball

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

Hence the probability of getting a non black ball is

(iii) Total number of neither red nor black balls i.e. red ball is 5

We know that PROBABILITY =

Hence probability of getting neither white nor black ball

#### Page No 16.22:

#### Question 27:

Find the probability that a number selected from the number 1 to 25 is not a prime number when each of the given numbers is equally likely to be selected.

#### Answer:

**GIVEN**: A number is selected from numbers 1 to 25

**TO FIND**: Probability of getting a number which is not a prime.

Total number of cards is 25.

Total number of elementary events = 25

Cards bearing non prime numbers are 1,4,6,8,9,10,12,14,15,16,18,20,21,22,24, 25

Total number of cards bearing non-prime numbers = 16

Number of favourable elementary events = 16

We know that , Probability = $\frac{\mathrm{number}\mathrm{of}\mathrm{favourable}\mathrm{elementary}\mathrm{events}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{elementary}\mathrm{events}}$

So, P(getting a card bearing a non prime number) = $\frac{16}{25}$

#### Page No 16.22:

#### Question 28:

A bag contains 8 red, 6 white and 4 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is

(i) red or white

(ii) not black

(iii) neither white nor black.

#### Answer:

GIVEN: A bag contains 8 red, 4 black and 6 white balls and a ball is drawn at random

TO FIND: Probability of getting a

(i) red or white ball

(ii) not black ball

(iii) neither white nor black

Total number of balls

(i) Total number red and white balls are

We know that PROBABILITY =

Hence probability of getting red or white ball

(ii) Total number of black balls is 4

We know that PROBABILITY =

Hence probability of getting black ball

Hence the probability of getting non black ball is

(iii) Total number of neither red nor black balls i.e. red ball is 8

We know that PROBABILITY =

Hence probability of getting neither white nor black ball

#### Page No 16.22:

#### Question 29:

Find the probability that a number selected at random from the numbers 1, 2, 3, ..., 35 is a

(i) prime number

(ii) multiple of 7

(iii) a multiple of 3 or 5

#### Answer:

GIVEN: A number is selected from numbers 1 to 35

TO FIND: Probability of getting a number

(i) which is a prime number

(ii) multiple of 7

(iii) multiple of 3 or 5

Total number of cards is 35

(i) Numbers that are primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and 31

Total prime numbers from 1 to 35 are11

We know that PROBABILITY =

Hence probability of getting a number which is a prime from 1 to 35 is equal to

(ii) Numbers that are multiple of 7 are 7,14,21,28,35

Total number that are multiple of 7 from 1 to 35 are5

We know that PROBABILITY =

Hence probability of getting number that is multiple of 7 from 1 to 35 is

(iii) Numbers that are multiple of 3 and 5 are 3, 5, 6, 8, 10, 12, 15, 18, 20, 21, 24, 25, 27, 30, 33 and 35

Total numbers that are multiple of 3 or 5 from 1 to 35 is 16

We know that PROBABILITY =

Hence probability of getting number that is multiple of 3 or 5 from 1 to 35 is equal to

#### Page No 16.22:

#### Question 30:

From a pack of 52 playing cards Jacks, queens, Kings and aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is

(i) a black queen

(ii) a red card

(iii) a black jack

(iv) a picture card (Jacks, queens and Kings are picture cards).

#### Answer:

Given: The Kings, Queens, Aces and Jacks of red color are removed from a deck of 52 playing cards and the remaining cards are shuffled and a card is drawn at random from the remaining cards

TO FIND: Probability of getting a card of

(i) A black queen

(ii) A red card

(iii) A black jack

(iv) A picture card

After removing the kings, queens, aces and the jacks of red color from the pack of 52 playing cards

Total number of cards left:

(i) Cards which are black queen is 2

We know that PROBABILITY =

Hence probability of getting a black queen =

(ii) Cards which are red are from 2 suits

Total number of red cards is

From this the kings, queens, aces and jacks of red color are taken out.

Hence total number of red cards left is

We know that PROBABILITY =

Hence probability of getting a red card is

(iii) Cards which are black jack are from 2 suits

Total number of black jack is

We know that PROBABILITY =

Hence probability of getting a black jack card

(iv) Cards which are picture cared are from 4 suits

Total number of picture cards is

From this the kings, queens, and jacks of red color are taken out.

Hence total number of picture card left is

We know that PROBABILITY =

Hence probability of getting an picture card =

#### Page No 16.22:

#### Question 31:

A bag contains lemon flavoured candles only. Malini takes out one candy without looking into the bag. what is the probability that she takes out

(i) an orange flavoured candy?

(ii) a lemon flavoured candy?

#### Answer:

GIVEN: A bag contains lemon flavored candies only. Malini takes out one candy without looking into the bag,

TO FIND: Probability that she takes out

(i) An orange flavored candy

(ii) A lemon flavored candy

(i) Probability of taking out orange flavored candy is as it is an impossible event, because the bag is filled only with lemon flavored candies

(ii) Probability of taking out lemon flavored candy is as it is a sure event, because the bag is filled only with lemon flavored candies.

#### Page No 16.23:

#### Question 32:

It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

#### Answer:

Given: Probability that 2 students should not have the same birthday

TO FIND: Probability that 2 students should have the same birthday

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

Hence the probability that should have the same birthday is

#### Page No 16.23:

#### Question 33:

A bag contains 3 red balls and 5 black balls. A ball is draw at random from the bag. What is the probability that the ball drawn is (i) red? (ii) not red?

#### Answer:

GIVEN: A bag contains 3 red and 5 black balls and a ball is drawn at random from the bag

TO FIND: Probability of getting a

(i) red ball

(ii) not red ball

Total number of balls

(i) Total number red balls are 3

We know that PROBABILITY =

Hence probability of getting red ball is

(ii) Probability of getting red ball

Hence the probability of getting not red ball

#### Page No 16.23:

#### Question 34:

A box contains 5 red marbels, 8 white marbles and 4 green marbles, One marble is taken out of the box at ramdom. What is the probability that the marble taken out will be (i) red? (ii) white?

(iii) not green?

#### Answer:

GIVEN: A box contains 5 red, 4 green and 8 white marbles and a marble is drawn at random

TO FIND: Probability of getting a marble

(i) red

(ii) white

(iii) not green

Total number of marble:

(i) Total number red marble are 5

We know that PROBABILITY =

Hence probability of getting red marble

(ii) Total number of white marbles are 8

We know that PROBABILITY =

Hence probability of getting white marble is

(iii) Total number of green marbles is 4

We know that PROBABILITY =

Hence probability of getting green marbles

Hence the probability of getting green marbles is

#### Page No 16.23:

#### Question 35:

A lot consists of 144 ball pens of which 20 are defective and others good. Nutri will buy a pen if it is good, but will not buy if it is defective. the shopkeeper draws one pen at random and gives it to her. What is the probability that

(i) She will buy it?

(ii) She will not buy it?

#### Answer:

GIVEN: A lot consists of 144 ball pens of which 20 are defective and others good

Nuri will buy a pen if it is good but will not buy if it is defective. The shop keeper draws one pen at random and gives it to her

TO FIND: Probability that

(i) She will buy

(ii) She will not buy

Total number of bulbs is 144

(i) Total numbers of bulbs which are non defective is

We know that PROBABILITY =

Hence probabilities that she will buy a good pen which is not defective is

(ii) We know that sum of probability of occurrence of an event and probability of non occurrence of an event is 1.

Hence probabilities that she will not buy a good pen is equal to

#### Page No 16.23:

#### Question 36:

12 defective pens are accidently mixed with 132 good ones. It is not possible to just look at pen and tell whether or not it is defective. one pen is taken out at random from this lot. Determine the probability that the pen taken out is good one.

#### Answer:

GIVEN: 12 defective pens are accidently mixed with 132 good ones. One pen is taken out at random from this lot.

TO FIND: Probability that the pen taken out is good.

Total number of bulbs is

Total numbers of bulbs which are good are 132

We know that PROBABILITY =

Hence probability that pen taken out is good is

#### Page No 16.23:

#### Question 37:

Five cards − the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.

(i) What is the probability that the card is the queen?

(ii) If the queen is drawn and put a side, what is the probability that the second card picked up is (a) an ace? (b) a queen?

#### Answer:

GIVEN: Five cards-the ten, jack, queen, king and ace of diamond are well shuffled with their face downwards of cards. One card is drawn at random.

TO FIND: Probability that

(i) The card is the queen.

(ii) If the queen is drawn and left aside, what is the probability that the second card picked up is (a) an ace (b) a queen

Total number of cards is 5

(i) Total number of queens is 1

We know that PROBABILITY =

Hence probability that the card taken out is queen is equal to

(ii) When queen is drawn and left aside.

Hence Total number of cards left is 4

(a) Total number of ace card is 1

We know that PROBABILITY =

Hence probability that the card taken out is an ace is equal to

(b) Total numbers of queen is 0

We know that PROBABILITY =

Hence probabilities that card taken out is queen

#### Page No 16.23:

#### Question 38:

Harpreet tosses two different coins simultaneously (say, one is of Re 1 and other of Rs 2). What is the probability that he gets at least one head?

#### Answer:

GIVEN: Harpreet tosses two different coins simultaneously (Re1 and Rs2)

TO FIND: Probability of getting at least one head.

When two coins are tossed then the outcome will be

TT, HT, TH, HH.

Hence total number of outcome is 4.

At least one head means 1H or 2H

Hence total number of favorable outcome i.e. at least one head is 3

We know that PROBABILITY =

Hence probability of getting at least one head =

#### Page No 16.23:

#### Question 39:

Cards marked with numbers 13, 14, 15, ..., 60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that number on the card drawn is

(i) divisible by 5

(ii) a number is a perfect square

#### Answer:

GIVEN: Cards are marked with numbers 13 to 60 are placed in a box and mixed thoroughly. If one card is drawn at random from the box

TO FIND: Probability that it bears

(i) A number divisible by 5

(ii) A perfect square

Total number of cards is

(i) Cards marked with a number divisible by 5 are

15, 20, 25, 30, 35, 40, 45, 50, 55 and 60

Total numbers of cards marked numbers divisible by 5 from 13 to 60 is 10

We know that PROBABILITY =

Hence probability of getting card marked with numbers divisible by 5 from 13to60 is

(ii) Cards marked a perfect squared numbers are 16, 25, 36 and 49

Total number of disc marked with perfect square from 13 to 60 is 4

We know that PROBABILITY =

Hence probability of getting disc marked with perfect square numbers from 13 to 60 is

#### Page No 16.23:

#### Question 40:

A bag contains tickets numbered 11, 12, 13, ..., 30. A ticket is taken out from the bag at random. Find the probability that the number on the drawn ticket (i) is a multiple of 7 (ii) is greater than 15 and a multiple of 5.

#### Answer:

GIVEN: Tickets are marked with one of the numbers 11, 12, 13…30 are placed in a bag and mixed thoroughly. One ticket is picked at random.

TO FIND: Probability of getting

(i) multiple of 7

(ii) greater than15 and multiple of 5

Total number of cards is

(i) Numbers that are multiple of 7 are 7, 14, 21 and 28

Total numbers that are multiple of 7 from 11 to 30 is 4

We know that PROBABILITY =

Hence probability of getting number that is multiple of 7 from 11 to 30 is

(ii) Numbers that are greater than 15 and multiple of 5 are 20,25,30

Total numbers that are greater than 15 and multiple of 5 from 11 to 30 is 3

We know that PROBABILITY =

Hence probability of getting numbers that is greater than 15 and multiple of 5 from 11 to 30

#### Page No 16.23:

#### Question 41:

Fill in the blanks:

(i) Probability of a sure event is ...............

(ii) Probability of an impossible event is ................

(iii) The probability of an event (other than sure and impossible event) lies between .................

(iv) Every elementary event associated to a random experiment has .............. probability.

(v) Probability of an event A + Probability of event 'not A' = ................

(vi) Sum of the probabilities of each outcome in an experiment is ................

#### Answer:

Fill in the blanks:

1. Probability of sure event is 1 as it is certain that it will occur always.

2. Probability of impossible event is 0 as it will never occur.

3. Probability of an event other than sure and impossible event lies between 0 and 1.

4. Every elementary event associated to a random experiment had equal probability

5. Probability of an event A + Probability of event ‘not A’ = 1.

6. Sum of the probabilities of each outcome in an experiment is 1

#### Page No 16.23:

#### Question 42:

Examine each of the following statements and comment:

(i) If two coins are tossed at the same time, there are 3 possible outcomes−two heads, two tails, or one of each. Therefore, for each outcome, the probability of occurrence is 1/3.

(ii) If a die in thrown once, there are two possible outcomes − an odd number or an even number. Therefore, the probability of obtaining an odd number is 1/2 and the probability of obtaining an even number is 1/2.

#### Answer:

(i)

Incorrect

When two coins are tossed, the possible outcomes are (H, H), (H, T), (T, H), and (T, T). It can be observed that there can be one of each in two possible ways − (H, T), (T, H).

Therefore, the probability of getting two heads is , the probability of getting two tails is , and the probability of getting one of each is .

It can be observed that for each outcome, the probability is not .

(ii)

Correct

When a dice is thrown, the possible outcomes are 1, 2, 3, 4, 5, and 6. Out of these, 1, 3, 5 are odd and 2, 4, 6 are even numbers.

Therefore, the probability of getting an odd number is .

Similarly, the probability of getting an even number is .

#### Page No 16.23:

#### Question 43:

A box contains 100 red cards, 200 yellow cards and 50 blue cards. If a card is drawn at random from the box, then find the probability that it will be (i) a blue card (ii) not a yellow card (iii) neither yellow nor a blue card. [CBSE 2012]

#### Answer:

Total number of cards = 100 + 200 + 50 = 350

∴ Total number of outcomes = 350

(i) Number of blue cards = 50

So, the number of favourable outcomes are 50.

∴ P(drawing a blue card) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{50}{350}=\frac{1}{7}$

(ii) Number of cards which are not yellow = 100 + 50 = 150

So, the number of favourable outcomes are 150.

∴ P(drawing a non yellow card) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{150}{350}=\frac{3}{7}$

(iii) Number of cards which are neither yellow nor blue = 100

So, the number of favourable outcomes are 100.

∴ P(drawing a card which is neither yellow nor blue) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{100}{350}=\frac{2}{7}$

#### Page No 16.24:

#### Question 44:

A box contains cards numbered 3, 5, 7, 9, ..., 35, 37. A card is drawn at random form the box. Find the probability that the number on the drawn card is a prime number.

#### Answer:

The numbers 3, 5, 7, 9, ..., 35, 37 are in AP.

Here, *a* = 3 and *d* = 5 − 3 = 2

Suppose there are *n* terms in the AP.

$\therefore {a}_{n}=37\phantom{\rule{0ex}{0ex}}\Rightarrow 3+\left(n-1\right)\times 2=37\left[{a}_{n}=a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 2n+1=37\phantom{\rule{0ex}{0ex}}\Rightarrow 2n=37-1=36\phantom{\rule{0ex}{0ex}}\Rightarrow n=18$

∴ Total number of outcomes = 18

Let E be the event of drawing a card with prime number on it.

Out of the given numbers, the prime numbers are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 and 37.

So, the favourable number of outcomes are 11.

∴ Required probability = P(E) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{11}{18}$

#### Page No 16.24:

#### Question 45:

A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person form the group is selected at random. Assuming that each person is equally likely to be selected, find the probability of selecting a person who is (i) extremely patient (ii) extremely kind or honest. Which of the above you prefer more. [CBSE 2013]

#### Answer:

Number of persons in the group = 12

∴ Total number of outcomes = 12

(i) Number of persons who are extremely patient = 3

So, the favourable number of outcomes are 3.

∴ P(selecting a person who is extremely patient) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{3}{12}=\frac{1}{4}$

(ii) Number of persons who are extremely honest = 6

Number of persons who are extremely kind = 12 − (3 + 6) = 3

∴ Number of persons who are extremely kind or honest = 6 + 3 = 9

So, the favourable number of outcomes are 9.

∴ P(selecting a person who is extremely kind or honest) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{9}{12}=\frac{3}{4}$

#### Page No 16.24:

#### Question 46:

Cards numbered 1 to 30 are put in a bag. A card is drawn at random from this bag. Find the probability that the number on the drawn card is

(i) not divisible by 3

(ii) a prime number greater than 7

(iii) not a perfect square number [CBSE 2014]

#### Answer:

Total number of outcomes = 30

(i) There are 10 numbers divisible by 3.

∴ Number of numbers not divisible by 3 = 30 − 10 = 20

So, the favourable number of outcomes are 20.

∴ P(drawing a card with number not divisible by 3) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{20}{30}=\frac{2}{3}$

(ii) Prime numbers greater than 7 are 11, 13, 17, 19, 23 and 29.

So, the favourable number of outcomes are 6.

∴ P(drawing a card with prime number greater than 7) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{6}{30}=\frac{1}{5}$

(iii) The perfect squares are 1, 4, 9, 16, 25. These are 5 in number.

∴ Number of numbers which are not perfect squares = 30 − 5 = 25

So, the favourable number of outcomes are 25.

∴ P(drawing a card with not a perfect square number) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{25}{30}=\frac{5}{6}$

#### Page No 16.24:

#### Question 47:

A piggy bank contains hundred 50 paise coins, fifity â‚ą1 coins, twenty â‚ą2 coins and ten â‚ą5 coins. If it is equally likely that one of the coins will fall out when the bank is turned upside down, find the probability that the coin which fell

(i) will be a 50 paise coin

(ii) will be of value more than â‚ą1

(iii) will be of value less than â‚ą5

(iv) will be a â‚ą1 or â‚ą2 coin [CBSE 2014]

#### Answer:

Total number of coins = 100 + 50 + 20 + 10 = 180

So, the total number of outcomes is 180.

(i) Number of 50 paise coins = 100

So, the favourable number of outcomes are 100.

∴ P(the coin which fell is a 50 paise coin) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{100}{180}=\frac{5}{9}$

(ii) A coin of value more than â‚ą1 can be â‚ą2 or â‚ą5 coin.

Number of â‚ą2 or â‚ą5 coins = 20 + 10 = 30

So, the favourable number of outcomes are 30.

∴ P(the coin which fell has value more than â‚ą1 ) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{30}{180}=\frac{1}{6}$

(iii) A coin of value less than â‚ą5 can be 50 paise or â‚ą1 or â‚ą2 coin.

Number of 50 paise or â‚ą1 or â‚ą2 coins = 100 + 50 + 20 = 170

So, the favourable number of outcomes are 170.

∴ P(the coin which fell has value less than â‚ą5 ) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{170}{180}=\frac{17}{18}$

(iv) Number of â‚ą1 or â‚ą2 coins = 50 + 20 = 70

So, the favourable number of outcomes are 70.

∴ P(the coin which fell will be a â‚ą1 or â‚ą2 coin ) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{70}{180}=\frac{7}{18}$

#### Page No 16.24:

#### Question 48:

A bag contains cards numbered from 1 to 49. A card is drawn from the bag at random, after mixing the card thoroughly. Find the probability that the number on the drawn card is

(i) an odd number

(ii) a multiple of 5

(iii) a perfect square

(iv) an even prime number

#### Answer:

The cards are numbered from 1 to 49.

∴ Total number of outcomes = 49

(i) The odd numbers from 1 to 49 are 1, 3, 5,..., 49.

The numbers 1, 3, 5,..., 49 are in AP.

Here, *a* = 1 and *d* = 2.

Suppose there are *n* terms in the AP.

$\therefore {a}_{n}=49\phantom{\rule{0ex}{0ex}}\Rightarrow 1+\left(n-1\right)\times 2=49\left[{a}_{n}=a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 2n-1=49\phantom{\rule{0ex}{0ex}}\Rightarrow 2n=49+1=50\phantom{\rule{0ex}{0ex}}\Rightarrow n=25$

So, the favourable number of outcomes are 25.

∴ P(number on the drawn card is odd) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{25}{49}$

(ii) The numbers from 1 to 49 which are multiple of 5 are 5, 10, 15, 20, 25, 30, 35, 40 and 45.

So, the favourable number of outcomes are 9.

∴ P(number on the drawn card is a multiple of 5) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{9}{49}$

(iii) The perfect square numbers from 1 to 49 are 1, 4, 9, 16, 25, 36 and 49.

So, the favourable number of outcomes are 7.

∴ P(number on the drawn card is a perfect square) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{7}{49}=\frac{1}{7}$

(iv) An even prime number is 2.

So, the favourable number of outcomes is 1.

∴ P(number on the drawn card is an even prime number) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{1}{49}$

**Note**: 2 is the only even prime number.

#### Page No 16.24:

#### Question 49:

A box contains 20 cards numbered from 1 to 20. A card is drawn at random from the box. Find the probability that the number on the drawn card is

(i) divisible by 2 or 3 (ii) a prime number [CBSE 2015]

#### Answer:

Total number of outcomes = 20

(i) The numbers from 1 to 20 which are divisible by 2 or 3 are 2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18 and 20.

So, the favourable number of outcomes are 13.

∴ P(number on the drawn card is divisible by 2 or 3) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{13}{20}$

(ii) The prime numbers from 1 to 20 are 2, 3, 5, 7, 11, 13, 17 and 19.

So, the favourable number of outcomes are 8.

∴ P(number on the drawn card is a prime number) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{8}{20}=\frac{2}{5}$

#### Page No 16.24:

#### Question 50:

In a simultaneous throw of pair of dice, find the probability of getting:

(i) 8 as the sum

(ii) a doublet

(iii) a doublet of prime numbers

(iv) a doublet of odd numbers

(v) a sum greater than 9

(vi) an even number on first

(vii) an even number on one and a multiple of 3 on the other

(viii) neither 9 nor 11 as the sum of the numbers on the faces

(ix) a sum less than 6

(x) a sum less than 7

(xi) a sum more than 7

(xii) at least once

(xiii) a number other than 5 on any dice.

(xiv) even number on each die [CBSE 2014, 2015]

(xv) 5 as the sum [CBSE 2014, 2015]

(xvi) 2 will come up at least once [CBSE 2015]

(xvii) 2 will not come either time [CBSE 2015]

#### Answer:

GIVEN: A pair of dice is thrown

TO FIND: Probability of the following:

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

(i) Favourable events i.e. getting the sum of numbers on the dice equal to 8

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

Hence total number of favourable events i.e. the sum of numbers on the dice equal to 8 is 5

We know that PROBABILITY =

Hence probability of getting the sum of numbers on the dice equal to 8

(ii) Favourable events i.e. getting the same number on both the dice

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

Hence total number of favourable events i.e. the same number on both the dice is 6

We know that PROBABILITY =

Hence probability of getting the same number on both the dice=

(iii) Favourable events i.e. getting the same prime number on both the dice are

(2, 2), (3, 3) and (5, 5)

Hence total number of favourable events i.e. the same prime number on both the dice is 3

We know that PROBABILITY =

Hence probability of getting the same prime number on both the dice=

(iv) Favourable events i.e. getting the same odd number on both the dice are

(1, 1), (3, 3) and (5, 5)

Hence total number of favourable events i.e. the same odd number on both the dice is 3

We know that PROBABILITY =

Hence probability of getting the same odd number on both the dice is

(v) Favourable events i.e. getting the total of numbers on the dice is greater than 9

(5,5), (5,6), (6,4), (4,6), (6,5), (6,6)

Hence total number of favourable events i.e. getting the total of numbers on the dice is greater than 9 is 6

We know that PROBABILITY =

Hence probability of getting the total of numbers on the dice greater than 9 is

(vi) Favourable events i.e. getting an even number on the first dice

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

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

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

Hence total number of favourable events i.e. getting an even number on the first dice=18

We know that PROBABILITY =

Hence probability of getting the an even number on the first dice=

(vii) Favourable event that an even number on first dice and multiple of 3 on the other dice are (2,3), (2,6),(4,3), (4,6),(6,3), (6,6), (3,2), (6,2), (3,4), (6,4) and (3,6)

Hence total number of favourable events i.e. getting that an even number on first dice and multiple of 3 on the other dice is 11

We know that PROBABILITY =

Hence probability of getting that an even number on one dice and multiple of 3 on the other dice is equal to

(viii) Favourable event neither 9 nor 11 as the sum of the number of faces 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),

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

(5,1), (5,2), (5,3), (5,5),

(6,1), (6,2),), (6,4), (6,6),

Hence total number of favourable events i.e. getting neither 9 nor 11 as the sum of the number of faces is 30

We know that PROBABILITY =

Hence probability of getting neither 9 nor 11 as the sum of the number of faces is equal to

(ix) Favourable event i.e. sum less than 6

(1,1), (1,2), (1,3), (1,4),

(2,1), (2,2), (2,3)

(3,1), (3,2), (4,1),

Hence total number of favourable events i.e. sum less than 6 is 10

We know that PROBABILITY =

Hence probability of getting sum less than 6 is equal to

(x) Favourable event a sum less than 7

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

(2,1), (2,2), (2,3), (2,4),

(3,1), (3,2), (3,3),

(4,1), (4,2), (5,1),

Hence total number of favourable events i.e. sum less than 7 is 15

We know that PROBABILITY =

Hence probability of getting sum less than 7 is equal to

(xi) Favourable event a sum less than 7

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

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

Hence total number of favourable events i.e. sum more than 7 is 15

We know that PROBABILITY =

Hence probability of getting sum more than is equal to

(xii) Favourable event at least once

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

Hence total number of favourable events i.e. at least once is 11

We know that PROBABILITY =

Hence probability of getting at least once is equal to

(xiii) Favourable event other than 5 on any dice

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

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

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

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

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

Hence total number of favourable events i.e. other than 5 on any dice is 25

We know that PROBABILITY =

Hence probability of getting other than 5 on any dice is equal to

(xiv) Favourable outcomes for getting even number on each die are

(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)

Hence, the number of favourable outcomes are 9.

∴ P(even number on each die) $=\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{9}{36}=\frac{1}{4}$

(xv) Favourable outcomes for getting 5 as the sum are

(1, 4), (2, 3), (3, 2), (4, 1)

Hence, the number of favourable outcomes are 4.

∴ P(getting 5 as the sum) $=\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{4}{36}=\frac{1}{9}$

(xvi) Favourable outcomes for 2 coming up atleast once are

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

Hence, the number of favourable outcomes are 11.

∴ P(2 will come up at least once) $=\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{11}{36}$

(xvii) Favourable outcomes for 2 not coming either time are

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

Hence, the number of favourable outcomes are 25.

∴ P(2 will not come up either time) $=\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{25}{36}$

#### Page No 16.24:

#### Question 51:

What is the probability that an ordinary year has 53 Sundays?

#### Answer:

GIVEN: An ordinary year

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

Total number of days in an ordinary year is 365days

Hence number of weeks in an ordinary year is

In an ordinary 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 is 7

Favorable day I.e. Sunday is 1

We know that PROBABILITY =

Hence probability that an ordinary year has 53 Sundays is equal to

#### Page No 16.24:

#### Question 52:

What is the probability that a leap year has 53 Tuesdays and 53 Mondays?

#### Answer:

GIVEN: A leap year

TO FIND: Probability that a leap year has 53 Tuesdays and 53 Mondays

Total number of days in a non leap year is 366days

Hence number of weeks in a non leap year is

In a non 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 53 Tuesdays and 53 Mondays the additional days should include Monday and Tuesday

Hence total number of pairs of days is 7

Favorable day i.e. in which one Tuesday and one Monday is there is only 1

We know that PROBABILITY =

Hence probability that a leap year has 53 Tuesdays and 53 Mondays is equal to#### Page No 16.24:

#### Question 53:

A black die and a white die are thrown at the same time. Write all the possible outcomes. What is the probability?

(i) that the sum of the two numbers that turn up is 7?

(ii) of obtaining a total of 6?

(iii) of obtaining a total of 10?

(iv) of obtaining the same number on both dice?

(v) of obtaining a total more than 9?

(vi) that the sum of the two numbers appearing on the top of the dice is 13?

(vii) that the sum of the numbers appearing on the top of the dice is less than or equal to 12?

(viii) that the product of numbers appearing on the top of the dice is less than 9. [CBSE 2014]

(ix) that the difference of the numbers appearing on the top of two dice is 2. [CBSE 2014]

(x) that the numbers obtained have a product less then 16.

#### Answer:

GIVEN: A pair of dice is thrown

TO FIND: Probability of the following:

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

(i) Favorable events i.e. getting the sum of numbers on the dice equal to 8

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

Hence total number of favorable events i.e. the sum of numbers on the dice equal to 8 is 5

We know that PROBABILITY =

Hence probability of getting the sum of numbers on the dice equal to 8

(ii) Favorable events i.e. getting the sum of numbers on the dice equal to 6

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

Hence total number of favorable events i.e. the sum of numbers on the dice equal to 6 is 5

We know that PROBABILITY =

Hence probability of getting the sum of numbers on the dice equal to 6 is

(iii) Favorable events i.e. getting the sum of numbers on the dice equal to10 is (4, 6), (5, 5) and (6, 4)

Hence total number of favorable events i.e. the sum of numbers on the dice equal to 6 is 3

We know that PROBABILITY =

Hence probability of getting the sum of numbers on the dice equal to10 is

(iv) Favorable events i.e. getting the same number on both the dice

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

Hence total number of favorable events i.e. the same number on both the dice is 6

We know that PROBABILITY =

Hence probability of getting the same number on both the dice=

(v) Favorable events i.e. getting the sum of numbers on the dice is greater than 10

is (5, 5), (5, 6), (6, 4), (6, 5) and (6, 6)

Hence total number of favorable events i.e. getting the total of numbers on the dice greater than 9 is 6

We know that PROBABILITY =

Hence probability of getting the total of numbers on the dice greater than 9 is

(vi) Favorable events i.e. getting the sum of both numbers appearing on the top of the dice is 13 is 0 since the highest sum of score we can get is 12

Hence probability of getting the sum of both numbers appearing on the top of the dice 13 is equal to

(vii) Favorable events i.e. getting the sum of both numbers appearing on the top of the dice less than or equal to 12 is a sure event

Hence probability of getting the sum of both numbers appearing on the top of the dice less than or equal to 12 is equal to

(viii) Favourable outcomes for getting the product of numbers less than 9 are

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

Thus, the number of favourable outcomes are 16.

∴ P(getting the product of numbers less than 9) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{16}{36}=\frac{4}{9}$

(ix) Favourable outcomes for getting the difference of the numbers as 2 are

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

Thus, the number of favourable outcomes are 8.

∴ P(getting the difference of the numbers as 2) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{8}{36}=\frac{2}{9}$

(x) Let *B* be the event of getting the numbers whose product is less than 16.

The outcomes in favour of event *B* 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), (4,1), (4,2), (4,3), (5,1), (5,2), (5,3), (6,1) and (6,2).

Number of favourable outcomes = 25

∴ P(*B* ) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{favourable}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{25}{36}$

#### Page No 16.25:

#### Question 54:

A bag contains cards which are numbered form 2 to 90. A card is drawn at random from the bag. Find the probability that it bears

(i) a two digit number

(ii) a number which is a perfect square.

#### Answer:

GIVEN: Cards are marked with one of the numbers 2 to 90 are placed in a bag and mixed thoroughly. One card is picked at random.

TO FIND: Probability of getting

(i) a two digit number

(ii) a number which is a perfect square

Total number of cards is.

(i) Cards marked two digit starts from 10

Total number of cards marked two digits from 10 to 90 is

We know that PROBABILITY =

Hence probability of getting a two digit card =

(ii) Cards which are perfect square form 2 to 90 are 4,9,16,25,36,49,64,81

Total number of cards marked perfect square from 2 to 90 are 8

We know that PROBABILITY =

Hence probability of getting perfect square card =

#### Page No 16.25:

#### Question 55:

The faces of a red cube and a yellow cube are numbered from 1 to 6. Both cubes are rolled. What is the probability that the top face of each cube will have the same number?

#### Answer:

GIVEN: The face of red cube and a yellow cube are marked 1 to 6

TO FIND: Probability of getting the same number on both the cubes

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 same number on both the dice are

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

Hence total number of favorable events i.e. the same number on both the cube is 6

We know that PROBABILITY =

Hence probability of getting the same number on both the cube=

#### Page No 16.25:

#### Question 56:

The probability of selecting a green marble at random from a jar that contains only green, white and yellow marbles is 1/4. The probability of selecting a white marble at random from the same jar is 1/3. If this jar contains 10 yellow marbles. What is the total number of marbles in the jar?

#### Answer:

GIVEN: A bag contains green, white and yellow marbles.

(i) Probability of selecting green marbles =

(ii) Probability of selecting white marbles =

(iii) The jar contains 10 yellow marbles.

TO FIND: Total number of marbles in the same jar

We know that sum of probabilities of all elementary events is 1.

Hence,

We know that PROBABILITY =

Hence

#### Page No 16.25:

#### Question 57:

(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot . What is the probability that this bulb is defective?

(ii) Suppose the bulb drawn in (i) is not defective and not replaced . Now bulb is drawn at random from the rest . What is the probability that this bulb is not defective ?

#### Answer:

(i) Total number of bulbs = 20

Total number of defective bulbs = 4

P (getting a defective bulb)

(ii) Remaining total number of bulbs = 19

Remaining total number of non-defective bulbs = 16 − 1 = 15

P (getting a not defective bulb)

#### Page No 16.25:

#### Question 58:

A box contains 90 discs which are numbered from 1 to 90 . If one discs is drawn at random from the box , find the probability that it bears

(i) a two digit number

(ii) a perfect square number

(iii) a number divisible by 5.

#### Answer:

Total number of discs = 90

(i) Total number of two-digit numbers between 1 and 90 = 81

P (getting a two-digit number)

(ii) Perfect squares between 1 and 90 are 1, 4, 9, 16, 25, 36, 49, 64, and 81. Therefore, total number of perfect squares between 1 and 90 is 9.

P (getting a perfect square)

(iii) Numbers that are between 1 and 90 and divisible by 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, and 90. Therefore, total numbers divisible by 5 = 18

Probability of getting a number divisible by 5

#### Page No 16.25:

#### Question 59:

Two dice, one blue and one grey, are thrown at the same time. Complete the following table:

Event: 'Sum on two dice' |
2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

Probability |

From the above table a student argues that there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability $\frac{1}{11}.$ Do you agree with this argument?

#### Answer:

GIVEN: Two dice are thrown

TO FIND: Probability of the following:

Event: sum of two dice | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

Probability |

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

(1) Favorable events i.e. getting the sum of numbers on the dice equal to 2 is

(1, 1),

Hence total number of favorable events i.e. the sum of numbers on the dice equal to 2 is 1

We know that PROBABILITY =

Hence probability of getting the sum of numbers on the dice equal to2

(2) Favorable events i.e. getting the sum of numbers on the dice equal to 3

are (1,2) and (2,1)

Hence total number of favorable events i.e. the sum of numbers on the dice equal to 3 is 2

We know that PROBABILITY =

Hence probability of getting the sum of numbers on the dice equal to 3 =

(3) Favorable events i.e. getting the sum of numbers on the dice equal to 4

are (1,3), (2,2) and (3,1)

Hence total number of favorable events i.e. the sum of numbers on the dice equal to 4 is 3

We know that PROBABILITY =

Hence probability of getting the sum of numbers on the dice equal to4 is

(4) Favorable events i.e. getting the sum of numbers on the dice equal to 5

are (1,4), (2,3), (3,2) and (4,1)

Hence total number of favorable events i.e. getting the sum of numbers on the dice equal to 5 is 4

We know that PROBABILITY =

Hence probability of getting the sum of numbers on the dice equal to 5 is

(5) Favorable events i.e. getting the total of numbers on the dice equal to 6 are

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

Hence total number of favorable events i.e. getting the total of numbers on the dice equal to 6 is 5

We know that PROBABILITY =

Hence probability of getting the sum of numbers on the dice equal to 6 is

(6) Favorable events i.e. getting the sum of both numbers equal to 7 are

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

Hence total number of favorable events i.e. getting the total of numbers on the dice equal to 7 is 6

We know that PROBABILITY =

Hence probability of getting the total of numbers on the dice equal to 7 is

(7) Favorable events i.e. getting the total of numbers on the dice equal to 8 are

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

Hence total number of favorable events i.e. getting the total of numbers on the dice equal to 8 is 5

We know that PROBABILITY =

Hence probability of getting the total of numbers on the dice equal to 8 is

(8) Favorable events i.e. getting the sum of numbers on the dice equal to 9 are

(3,6), (4,5) ,(5,4) and (6,3)

Hence total number of favorable events i.e. getting the sum of numbers on the dice equal to 9 is 4

We know that PROBABILITY =

Hence probability of getting the sum of numbers on the dice equal to9 is

(9) Favorable events i.e. getting the sum of numbers on the dice equal to 10 are

(4,6), (5,5) and (6,4)

Hence total number of favorable events i.e. the sum of numbers on the dice equal to 10 is 3

We know that PROBABILITY =

Hence probability of getting the sum of numbers on the dice equal to10=

(10) Favorable events i.e. getting the sum of numbers on the dice equal to 11 are (5, 6) and (6,5),

Hence total number of favorable events i.e. the sum of numbers on the dice equal to 11 is 2

We know that PROBABILITY =

Hence probability of getting the sum of numbers on the dice equal to11 is

(11) Favorable events i.e. getting the sum of numbers on the dice equal to 12 is

(6, 6)

Hence total number of favorable events i.e. the sum of numbers on the dice equal to 12 is 1

We know that PROBABILITY =

Hence probability of getting the sum of numbers on the dice equal to 12

The complete table is as follows

Event: Sum of two dices | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

Probability |

From the above table it is clear that each event does not have same probability

#### Page No 16.25:

#### Question 60:

A bag contains 6 red balls and some blue balls. if the probability of a drawing a blue ball from the bag is twice that of a red ball, find the number of blue balls in the bag.

#### Answer:

GIVEN: A bag contains 6 red balls and some blue balls. If the probability of drawing a blue ball from the bag is twice that of a red ball,

TO FIND: the number of blue balls in the bag.

Let the probability of getting a red ball be

The probability of not getting a red ball or getting a blue ball be

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

Hence the probability of getting a red ball is

We know that PROBABILITY =

Hence total number of blue balls = total number of balls −red balls

Hence total number of blue balls is

#### Page No 16.25:

#### Question 61:

The king, queen and jack of clubs are removed form a deck of 52 playing cards and the remaining cards are shuffled. A card is drawn from the remaining cards. Find the probability of getting a card of (i) heart (ii) queen (iii) clubs (iv) a face card (v) a queen of diamond.

#### Answer:

Given: King, Queen and Jack of Clubs are removed from a deck of 52 playing cards and the remaining cards are shuffled and a card is drawn at random from the remaining cards

TO FIND: Probability of getting a card

(i) Heart

(ii) Queen

(iii) Clubs

After removing the king, queen and the jack of clubs from the pack of 52 playing cards

Total number of cards left:

(i) Cards which are heart

Total number of heart cards is

We know that PROBABILITY =

Hence probability of getting a heart card is equal to

(ii) Cards which are queen

Total number of queen cards is

From this 4 queen cards one queen of club is taken out.

Hence total number of queen cards left is

We know that PROBABILITY =

Hence probability of getting an queen card

(iii) Cards which are clubs

Total number of club cards is

The king, queen and jack of clubs are removed

Hence total number of club cards left is

We know that PROBABILITY =

Hence probability of getting a club card is equal to

(iv) a face card

Total face cards = 3 × 4 = 12

king, queen and jack of clubs are removed from the pack

So, face cards left = 12 − 3 = 9

Total cards left = 52 − 3 = 49

Probability of getting a face card = =$=\frac{9}{49}$

(v) a queen of diamond

After removing king, queen and jack of clubs we are left with 52 − 3 = 49 cards total

Probability of getting queen of diamond = $\frac{1}{49}$

#### Page No 16.25:

#### Question 62:

Two dice are thrown simultaneously. What is the probability that:

(i) 5 will not come up on either of them?

(ii) 5 will come up on at least one?

(iii) 5 will come up at both dice?

#### Answer:

GIVEN: Two dice are thrown

TO FIND: Probability of the following:

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

(i) Favorable events i.e. 5 will not come up on either of them

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

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

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

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

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

Hence total number of favorable events i.e. 5 will not come up on either of them is 25

We know that PROBABILITY =

Hence probability that 5 will not come up on either of them is equal to

(ii) Favorable events i.e. 5 will come on at least once

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

Hence total number of favorable events i.e.5 will not come on at least once is 11

We know that PROBABILITY =

Hence probability of getting i.e. 5 will come on at least once is equal to

(iii) Favorable events i.e. 5 will come on both side

is (5, 5)

Hence total number of favorable events i.e. 5 will come on both side is 1

We know that PROBABILITY =

Hence probability of getting 5 will come on both side is equal to

#### Page No 16.26:

#### Question 63:

A number is selected at random from first 50 natural numbers. Find the probability that it is a multiple of 3 and 4.

#### Answer:

Total number of outcomes = 50

Let E be the event of getting a number which is a multiple of 3 and 4.

Now, the common multiples of 3 and 4 among first 50 natural numbers are 12, 24, 36 and 48.

So, the favourable number of outcomes are 4.

∴ Required probability = P(E) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{4}{50}=\frac{2}{25}$

#### Page No 16.26:

#### Question 64:

A dice is rolled twice. Find the probability that

(i) 5 will not come up either time

(ii) 5 will come up exactly one time [CBSE 2014]

#### Answer:

When a dice is rolled twice, 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

(i) The outcomes where 5 will not come up either time are

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

So, the number of favourable outcomes are 25.

∴ P(5 will not come up either time) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{25}{36}$

(ii) The outcomes where 5 will come up exactly one time are

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

So, the number of favourable outcomes are 10.

∴ P(5 will come up exactly one time) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{10}{36}=\frac{5}{18}$

#### Page No 16.26:

#### Question 65:

All the black face cards are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is drawn at random. Find the probability of getting

(i) face card

(ii) red card

(iii) black card

(iv) king

#### Answer:

In a pack of 52 playing cards, there are 2 black jacks, 2 black queens and 2 black kings.

All the black face cards are removed from the pack. Then,

Number of remaining cards = 52 − 6 = 46

∴ Total number of outcomes = 46

(i) There are 6 face cards (2 red jacks, 2 red queens and 2 red kings) in the remaining pack of cards.

So, the favourable number of outcomes are 6.

∴ P(getting a face card) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{6}{46}=\frac{3}{23}$

(ii) There are 26 red cards in the remaining pack of cards.

So, the favourable number of outcomes are 26.

∴ P(getting a red card) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{26}{46}=\frac{13}{23}$

(iii) Number of black cards in the remaining pack of cards = 26 − 6 = 20

So, the favourable number of outcomes are 20.

∴ P(getting a black card) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{20}{46}=\frac{10}{23}$

(iv) There are 2 red king cards in the remaining pack of cards.

So, the favourable number of outcomes are 2.

∴ P(getting a king) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{2}{46}=\frac{1}{23}$

#### Page No 16.26:

#### Question 66:

Cards numbered from 11 to 60 are kept in a box. If a card is drawn at random from the box, find the probability that the number on the drawn card is

(i) an odd number

(ii) a perfect square number

(iii) divisible by 5

(iv) a prime number less than 20 [CBSE 2014]

#### Answer:

The cards are numbered from 11 to 60.

∴ Total number of outcomes = 60 − 10 = 50

(i) The odd numbers from 11 to 60 are 11, 13, 15,..., 59.

The numbers 11, 13, 15,..., 59 are in AP.

Here, *a* = 11 and *d* = 2.

Suppose there are *n* terms in the AP.

$\therefore {a}_{n}=59\phantom{\rule{0ex}{0ex}}\Rightarrow 11+\left(n-1\right)\times 2=59\left[{a}_{n}=a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 2n+9=59\phantom{\rule{0ex}{0ex}}\Rightarrow 2n=59-9=50\phantom{\rule{0ex}{0ex}}\Rightarrow n=25$

So, the favourable number of outcomes are 25.

∴ P(number on the drawn card is odd) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{25}{50}=\frac{1}{2}$

(ii) The perfect square numbers from 11 to 60 are 16, 25, 36 and 49.

So, the favourable number of outcomes are 4.

∴ P(number on the drawn card is a perfect square) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{4}{50}=\frac{2}{25}$

(iii) The numbers from 11 to 60 divisible by 5 are 15, 20, 25, 30, 35, 40, 45, 50, 55 and 60.

So, the favourable number of outcomes are 10.

∴ P(number on the drawn card is divisible by 5) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{10}{50}=\frac{1}{5}$

(iv) Prime numbers less than 20 from 11 to 60 are 11, 13, 17 and 19.

So, the favourable number of outcomes are 4.

∴ P(number on the drawn card is a prime number less than 20) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{4}{50}=\frac{2}{25}$

#### Page No 16.26:

#### Question 67:

All kings and queens are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is randomly drawn from it. Find the probability that this card is (i) a red face card (ii) a black card. [CBSE 2014]

#### Answer:

In a pack of 52 playing cards, there are 4 queen and 4 king cards.

All the king and queen cards are removed from the pack. Then,

Number of remaining cards = 52 − (4 + 4) = 44

∴ Total number of outcomes = 44

(i) There are 2 red face cards (2 red jacks) in the remaining pack of cards.

So, the favourable number of outcomes are 2.

∴ P(getting a red face card) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{2}{44}=\frac{1}{22}$

(ii) There are 2 black kings and 2 black queens among the 4 queen and 4 king cards removed from the pack of cards.

Number of black cards in the remaining pack of cards = 26 − 4 = 22

So, the favourable number of outcomes are 22.

∴ P(getting a black card) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{22}{44}=\frac{1}{2}$

#### Page No 16.26:

#### Question 68:

All jacks, queens and kings are removed from a pack of 52 cards. The remaining cards are well-shuffled and then a card is randomly drawn from it. Find the probability that this card is (i) a black face card (ii) a red card. [CBSE 2014]

#### Answer:

In a pack of 52 playing cards, there are 4 jack, 4 queen and 4 king cards.

All jack, queen and king cards are removed from the pack. Then,

Number of remaining cards = 52 − 12 = 40

∴ Total number of outcomes = 40

(i) There are 0 face cards in the remaining pack of cards.

So, the favourable number of outcomes are 0.

∴ P(drawing a black face card) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{0}{40}=0$

(ii) There are 2 red jacks, 2 red queens and 2 red kings among 4 jack, 4 queen and 4 king cards removed from the pack.

Number of red cards in the remaining pack of cards = 26 − (2 + 2 + 2) = 20

So, the favourable number of outcomes are 20.

∴ P(drawing a red card) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{20}{40}=\frac{1}{2}$

#### Page No 16.26:

#### Question 69:

Red queens and black jacks are removed from a pack of 52 playing cards. A card is drawn at random from the remaining cards, after reshuffling them. Find the probability that the card drawn is

(i) a king (ii) of red colour (iii) a face card (iv) a queen [CBSE 2014]

#### Answer:

In a pack of 52 playing cards, there are 2 red queens and 2 black jacks.

Red queens and black jacks are removed from the pack. Then,

Number of remaining cards = 52 − 4 = 48

∴ Total number of outcomes = 48

(i) There are 4 king cards in the remaining pack of cards.

So, the favourable number of outcomes are 4.

∴ P(drawing a king) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{4}{48}=\frac{1}{12}$

(ii) Two red queens are removed from the pack of cards.

Number of red cards in the remaining pack of cards = 26 − 2 = 24

So, the favourable number of outcomes are 24.

∴ P(drawing a red card) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{24}{48}=\frac{1}{2}$

(iii) There are 12 face cards in the pack, out of which 2 red queens and 2 black jacks are removed.

Number of face cards in the remaining pack of cards = 12 − (2 + 2) = 8

So, the favourable number of outcomes are 8.

∴ P(drawing a face card) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{8}{48}=\frac{1}{6}$

(iv) There are 4 queen cards in the pack, out of which 2 red queens are removed.

Number of queens in the remaining pack of cards = 4 − 2 = 2

So, the favourable number of outcomes are 2.

∴ P(getting a queen) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{2}{48}=\frac{1}{24}$

#### Page No 16.26:

#### Question 70:

In a bag there are 44 identical cards with figure of circle or square on them. There are 24 circles, of which 9 are blue and rest are green and 20 squares of which 11 are blue and rest are green. One card is drawn from the bag at random. Find the probability that it has the figure of

(i) square (ii) green colour (iii) blue circle and (iv) green square. [CBSE 2015]

#### Answer:

There are 44 identical cards with figure of circle or square in the bag.

∴ Total number of outcomes = 44

(i) There are 20 squares on the cards.

So, the favourable number of outcomes is 20.

∴ P(card drawn has the figure of square) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{20}{44}=\frac{5}{11}$

(ii) Number of cards with green circle = 24 − 9 = 15

Number of cards with green square = 20 − 11 = 9

∴ Number of cards with green colour figure = 15 + 9 = 24

So, the favourable number of outcomes is 24.

∴ P(card drawn has the figure of green colour) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{24}{44}=\frac{6}{11}$

(iii) There are 9 cards with circle of blue colour.

So, the favourable number of outcomes is 9.

∴ P(card drawn has the figure of blue circle) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{9}{44}$

(iv) Number of cards with green square = 20 − 11 = 9

So, the favourable number of outcomes is 9.

∴ P(card drawn has the figure of green square) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{9}{44}$

#### Page No 16.26:

#### Question 71:

All red face cards are removed from a pack of playing cards. The remaining cards are well shuffled and then a card is drawn at random from them. Find the probability that the drawn card is (i) a red card (ii) a face card and (iii) a card of clubs. [CBSE 2015]

#### Answer:

In a pack of 52 playing cards, there are 2 red jacks, 2 red queens and 2 red kings.

All the red face cards are removed from the pack. Then,

Number of remaining cards = 52 − 6 = 46

∴ Total number of outcomes = 46

(i) Number of red cards in the remaining pack of cards = 26 − 6 = 20

So, the favourable number of outcomes are 20.

∴ P(drawing a red card) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{20}{46}=\frac{10}{23}$

(ii) There are 6 face cards (2 black jacks, 2 black queens and 2 black kings) in the remaining pack of cards.

So, the favourable number of outcomes are 6.

∴ P(drawing a face card) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{6}{46}=\frac{3}{23}$

(iii) There are 13 cards of club in the remaining pack of cards.

So, the favourable number of outcomes are 13.

∴ P(drawing a card of club) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{13}{46}$

#### Page No 16.26:

#### Question 72:

Two customers are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another. What is the probability that both will visit the shop on:

(i) the same day?

(ii) Different days?

(iii) consecutive days?

#### Answer:

GIVEN: Two customers are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any one day as on another.

TO FIND: Probability that both will visit the shop on:

(i) The same day

(ii) Different days

(iii) Consecutive days

Two customers can visit the shop on two days in ways.

Hence total number of ways =36

(i) Two customer can visit the shop on any day of the week i.e.

MONDAY, TUESDAY, WEDNESDAY, THURSDAY, FRIDAY, SATURDAY

Favorable number of ways = 6

We know that PROBABILITY =

Hence probability of the customer visiting the shop on the same day =

(ii) We know that sum of probability of occurrence of an event and probability of non occurrence of an event is 1.

Hence probability of the customer visiting the shop on the different day is

(iii) Two costumer can visit the shop in two consecutive days in the following ways:

(MONDAY, TUESDAY), (WEDNESDAY, THURSDAY), (THURSDAY, FRIDAY) (FRIDAY, SATURDAY)

Favorable number of ways =5

We know that PROBABILITY =

Hence probability of the customer visiting the shop on two consecutive days =

#### Page No 16.33:

#### Question 1:

Suppose you drop a tie at random on the rectangular region shown in the given figure. What is the probability that it will land inside the circle with diameter 1 m?

#### Answer:

**Given:** Suppose you drop a tie at random on the rectangular region shown in figure

**To find: **Probability that it will land in inside the circle of diameter 1m

Total area of circle with diameter 1 m

We know that PROBABILITY =

Hence probability that the tie will land in the circle is .

#### Page No 16.33:

#### Question 2:

In the accompanying diagram a fair spinner is placed at the centre *O* of the circle Diameter *AOB* and radius *OC* divide the circle into three regions labelled *X*, *Y* and *Z*. It ∠BOC = 45°. What is the probability that the spinner will land in the region *X*? (in the given figure).

#### Answer:

**Given:** A fair spinner is placed at the centre O of the circle. Diameter AOB and radius OC divide the circle into three regions labeled X, Y and Z and angle

**To find: **Probability that the spinner will land in X region?

Total angle of circle is 360°.

We know that PROBABILITY =

Hence probability of “spinner will land in X region” is

#### Page No 16.33:

#### Question 3:

A target shown in the given figure consists of three concentric circle of radii 3, 7 and 9 cm respectively. A dart is thrown and lands on the target. What is the probability that the dart will land on the shaded region?

#### Answer:

**Given:** A target is shown in figure consists of three concentric circles of radius 3, 7, and 9 cm. A dart is thrown and lands on the target

**To find: **Probability that the dart will land in shaded region?

Total area of circle with radius 9 cm

We know that Probability =

Hence probability of the spinner will land in shaded region is

#### Page No 16.33:

#### Question 4:

In the given figure, points *A*, *B*, *C* and *D* are the centres of four circles that each have a radius of length one unit. If a point is selected at random from the interior of square *ABCD.* What is the probability that the point will be chosen from the shaded region.

#### Answer:

**Given:** A, B, C, and D are the centers of four circles that each have a radius of length one unit. If a point is selected at random from the interior of square ABCD

**To find:** Probability that the point will be chosen from the shaded region,

In the figure we can see 4 circles of radius 1 unit.

Area of quarter circle with centre A:

Since all the circles are of the same radius, hence the area of quarter with centre B, C, D will be same as the area of circle of quarter of circle with centre A.

Hence total area covered by 4 quarter circle will be

Side of the square will be 2 units

Area of square ABCD = 4 unit^{2}

Area of the shaded portion

We know that PROBABILITY

Hence probability of the shaded region is.

#### Page No 16.33:

#### Question 5:

In the given figure, *JKLM *is a square with sides of length 6 units. Points *A* and *B* are the mid points of sides *KL* and *LM* respectively. If a point is selected at random from the interior of the square. What is the probability that the point will be chosen from the interior of Δ*JAB*?

#### Answer:

**Given:** JKLM is a square with sides of length 6units. Points A* *and B are the midpoints of sides KL and ML respectively. If a point is selected at random from the interior of the square

**To find:** Probability that the point will be chosen from the interior of ΔJAB.

We the following figure

Area of square JLKM is equal to

Now we have

Now area of the triangle AJB

We know that Probability

=

Hence the Probability that the point will be chosen from the interior of ΔAJB is .

#### Page No 16.34:

#### Question 6:

In the given figure, a square dart board is shown. The length of a side of the larger square is 1.5 times the length of a side of the smaller square. If a dart is thrown and lands on the larger square. What is the probability that it will land in the interior of the smaller square?

Figure

#### Answer:

**Given:** A square dart board is shown. The length of a side of the larger square is 1.5 times the length of a side of the smaller square. If a dart is thrown and lands on the larger square

**To find:** Probability that it will land in the interior of the smaller square

Let the length of smaller square is *x* cm

Therefore the length of side of bigger square will be 1.5*x* cm

Figure

We know that Probability =

Hence probability that the dart will land in the interior of the smaller square is equal to =.

#### Page No 16.34:

#### Question 1:

Cards each marked with one of the numbers 4, 5, 6, ..., 20 are placed in a box and mixed thoroughly. One card is drawn at random from the box. What is the probability of getting an even number?

#### Answer:

GIVEN: Cards are marked with one of the numbers 4, 5, 6…20 are placed in a box and mixed thoroughly. One card is picked at random.

TO FIND: Probability of getting an even number on the picked card.

Total number of cards is 20-3=17

Cards marked even number are 4,6,8,10,12,14,16,18,20

Total number of cards marked even is 9

We know that PROBABILITY =

Hence probability of getting an even number on the card =

#### Page No 16.34:

#### Question 2:

One card is drawn from a well shuffled deck of 52 playing cards. What is the probability of getting a non-face card?

#### Answer:

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

TO FIND: Probability of getting a non-face card

Total number of cards are52.

Cards which are non-face are cards of numbers from 1 to 10 of each suit and there are four suits

Total number of non-face cards is

We know that PROBABILITY =

Hence probability of getting non-faced card =

Hence probability of getting a non-face card =

#### Page No 16.34:

#### Question 3:

A bag contains 5 red, 8 green and 7 white balls, One ball is drawn at random from the bag. What is the probability of getting a white ball or a green ball?

#### Answer:

GIVEN: A bag contains5 red, 8 green and 7 white balls

TO FIND: Probability of getting a white ball or a green ball.

Total number of balls

Total number of green or white balls

We know that PROBABILITY =

Hence probability of getting a green or a white ball =

Hence probability of getting an green or white ball =

#### Page No 16.34:

#### Question 4:

A die is thrown once. What is the probability of getting a prime number?

#### Answer:

GIVEN: A dice is thrown once

TO FIND: Probability of getting a prime number.

Total number on a dice is 6.

Prime numbers on a dice are 2, 3 and 5

Total number of prime numbers on dice is 3

We know that PROBABILITY =

Hence probability of getting a prime number =

Hence probability of getting a prime number =

#### Page No 16.34:

#### Question 5:

A die thrown once. What is the probability of getting a number lying between 2 and 6?

#### Answer:

GIVEN: A dice is thrown once

TO FIND: Probability of getting a number lying between 2 and 6.

Total number on a dice is 6.

Number lying between 2 and 6 are 3, 4 and 5

Total number of numbers lying between 2 and 6 is 3

We know that PROBABILITY =

Hence probability of getting a number lying between 2 and 6 =

Hence probability of getting a number lying between 2 and 6 =

#### Page No 16.34:

#### Question 6:

A die is thrown once. What is the probability of getting an odd number?

#### Answer:

GIVEN: A dice is thrown once

TO FIND: Probability of getting an odd number.

Total number on a dice is 6.

Odd numbers on a dice are 1, 3 and 5

Total number of odd numbers on dice is 3

We know that PROBABILITY =

Hence probability of an odd number =

Hence probability of getting an odd number =

#### Page No 16.35:

#### Question 7:

If $\overline{E}$ denote the complement or negation of an even E, what is the value of P(E) + P($\overline{E}$) ?

#### Answer:

Given: denotes the complement or negation of an event *E*.

TO FIND:

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

Hence

Hence the result is

#### Page No 16.35:

#### Question 8:

One card is drawn at random from a well shuffled deck of 52 cards. What is the probability of getting an ace?

#### 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 are52.

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=

Hence probability of getting an Ace =

#### Page No 16.35:

#### Question 9:

Two coins are tossed simultaneously. What is the probability of getting at least one head?

#### Answer:

GIVEN: Two coins are tossed simultaneously.

TO FIND: Probability of getting at least one head.

When two coins are tossed then the outcome will be

TT, HT, TH, HH.

Hence total number of outcome is 4.

At least one head means 1H or 2H.

Hence total number of favorable outcome i.e. at least one head is 3

We know that PROBABILITY =

Hence probability of getting at least one head =

#### Page No 16.35:

#### Question 10:

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn bears a number which is a multiple of 3?

#### Answer:

GIVEN: Tickets are marked from 1 to 20 are mixed up. One ticket is picked at random.

TO FIND: Probability that the ticket bears a multiple of 3

Total number of cards is 20

Cards marked multiple of 3 number are 3,6,9,12,15,18

Total number of cards marked multiple of 3 are 6

We know that PROBABILITY =

Hence probability of getting a multiple of 3 on the ticket =

Hence probability of getting a multiple of 3 on the ticket

#### Page No 16.35:

#### Question 11:

From a well shuffled pack of cards, a card is drawn at random. Find the probability of getting a black queen.

#### Answer:

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

TO FIND: Probability of getting a black queen

Total number of cards are52.

Cards which are black queen are 1 from 2 black suits

Total number of black queen cards is

We know that PROBABILITY =

Hence probability of getting an black queen=

Hence probability of getting an black queen =

#### Page No 16.35:

#### Question 12:

A die is thrown once. Find the probability of getting a number less than 3.

#### Answer:

GIVEN: A dice is thrown once

TO FIND: Probability of getting a number less than 3

Total number on a dice is 6.

Number less than 3 are1 and 2

Total number of numbers less than 3 is 2

We know that PROBABILITY =

Hence probability of getting a number less than 3 is equal to

Hence probability of getting a number less than 3 =

#### Page No 16.35:

#### Question 13:

Two coins are tossed simultaneously. Find the probability of getting exactly one head.

#### Answer:

GIVEN: Two coins are tossed simultaneously.

TO FIND: Probability of getting exactly one head.

When two coins are tossed then the outcome will be

TT, HT, TH, HH.

Hence total number of outcome is 4.

Exactly one head we get 2 times

Hence total number of favorable outcome i.e. exactly one head is 2

We know that PROBABILITY =

Hence probability of getting exactly one head =

Hence probability of getting exactly one head =

#### Page No 16.35:

#### Question 14:

A die is thrown once. What is the probability of getting a number greater than 4?

#### Answer:

GIVEN: A dice is thrown once

TO FIND: Probability of getting a number greater than 4

Total number on a dice is 6.

Numbers greater than 4 are 5 and 6

Total number of numbers greater than 4 is 2

We know that PROBABILITY =

Hence probability of getting a number greater than 4 is equal to

Hence probability of getting a number greater than 4 is

#### Page No 16.35:

#### Question 15:

What is the probability that a number selected at random from the numbers 3, 4, 5, ....9 is a multiple of 4?

#### Answer:

GIVEN: numbers are 3, 4, 5, 6…9

TO FIND: Probability of Getting multiple of 4

Total number is 9−3+1=7

Numbers which are multiple of 4 between 3 and 9 are 4 and 8

Total number which are multiple of 4 between 3 to 9 is 4 and 8 is 2

We know that PROBABILITY =

Hence probability of getting a number which is a multiple of 4 is

Hence probability of getting a number which is a multiple of 4 is equal to

#### Page No 16.35:

#### Question 16:

A letter of English alphabet is chosen at random. Determine the probability that the chosen letter is a consonant. [CBSE 2015]

#### Answer:

There are 26 letters of English alphabet.

Total number of outcomes = 26

Out of 26 letters, 21 are consonants and 5 are vowels.

Favourable number of outcomes = 21

∴ P(chosen letter is a consonant) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{21}{26}$

#### Page No 16.35:

#### Question 17:

A bag contains 3 red and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is not red.

#### Answer:

Bag contains 3 red balls and 5 black balls

Total balls in the bag = 3 + 5 = 8

Probability of getting a ball which is not red = 1 − Probability of getting red ball

$=1-\frac{3}{8}=\frac{5}{8}$

#### Page No 16.35:

#### Question 18:

A number is chosen at random from the number –3, –2, –1, 0, 1, 2, 3. What will be the probability that square of this number is less then or equal to 1?

#### Answer:

S = {−3, −2, −1, 0, 1, 2, 3}

Let E be the event of getting a number whose square is less than or equal to 1.

So, E = {−1, 1, 0}

$P\left(E\right)=\frac{3}{7}$.

Hence, the probability of getting a number whose square is less than or equal to is $\frac{3}{7}$.

#### Page No 16.35:

#### 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) $\frac{4}{9}$

(b) $\frac{5}{9}$

(c) $\frac{1}{9}$

(d) $\frac{2}{3}$

#### 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 16.35:

#### Question 2:

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

(a) $\frac{4}{9}$

(b) $\frac{5}{9}$

(c) $\frac{1}{9}$

(d) $\frac{2}{3}$

#### 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 16.36:

#### Question 3:

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

(a) $\frac{1}{3}$

(b) $\frac{2}{3}$

(c) $\frac{1}{9}$

(d) $\frac{2}{9}$

#### 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 16.36:

#### Question 4:

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

(a) $\frac{1}{4}$

(b) $\frac{3}{8}$

(c) $\frac{1}{2}$

(d) $\frac{1}{4}$

#### 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 16.36:

#### Question 5:

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

(a) $\frac{1}{2}$

(b) $\frac{1}{3}$

(c) $\frac{1}{6}$

(d) $\frac{2}{3}$

#### 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 16.36:

#### Question 6:

The probability of guessing the correct answer to a certain test questions is $\frac{x}{12}$. If the probability of not guessing the correct answer to this question is $\frac{2}{3}$, 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 16.36:

#### 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) $\frac{1}{4}$

(b) $\frac{1}{3}$

(c) $\frac{4}{9}$

(d) $\frac{7}{9}$

#### 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 16.36:

#### 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) $\frac{1}{10}$

(b) $\frac{3}{10}$

(c) $\frac{7}{10}$

(d) $\frac{9}{10}$

#### 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 16.36:

#### Question 9:

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

(a) $\frac{3}{5}$

(b) $\frac{2}{5}$

(c) $\frac{2}{3}$

(d) $\frac{1}{3}$

#### 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 16.36:

#### Question 10:

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

(a) $\frac{1}{4}$

(b) $\frac{1}{13}$

(c) $\frac{1}{52}$

(d) $\frac{12}{13}$

#### 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 16.36:

#### Question 11:

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

(a) $\frac{2}{3}$

(b) $\frac{1}{6}$

(c) $\frac{1}{3}$

(d) $\frac{5}{6}$

#### 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 16.36:

#### Question 12:

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

(a) $\frac{2}{3}$

(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 16.36:

#### 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 16.36:

#### 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 16.36:

#### 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 16.37:

#### 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 16.37:

#### 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) $\frac{19}{20}$

(b) $\frac{1}{25}$

(c) $\frac{1}{20}$

(d) $\frac{17}{20}$

#### 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 16.37:

#### 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) $\frac{13}{25}$

(b) $\frac{21}{50}$

(c) $\frac{12}{25}$

(d) $\frac{23}{50}$

#### 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 16.37:

#### Question 19:

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

(a) $\frac{1}{12}$

(b) $\frac{1}{6}$

(c) $\frac{3}{4}$

(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 16.37:

#### Question 20:

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

(a) $\frac{1}{3}$

(b) $\frac{3}{7}$

(c) $\frac{1}{6}$

(d) $\frac{1}{2}$

#### 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 16.37:

#### 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) $\frac{2}{3}$ (b) $\frac{1}{3}$ (c) $\frac{1}{2}$ (d) $\frac{1}{6}$ [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) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{3}{6}=\frac{1}{2}$

Hence, the correct answer is option C.

#### Page No 16.37:

#### 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) $\frac{1}{2}$ (b) $\frac{1}{3}$ (c) $\frac{1}{6}$ (d) $\frac{5}{6}$ [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) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{3}{6}=\frac{1}{2}$

Hence, the correct answer is option A.

#### Page No 16.37:

#### 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) $\frac{7}{90}$ (b) $\frac{10}{90}$ (c) $\frac{4}{45}$ (d) $\frac{9}{89}$ [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) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{8}{90}=\frac{4}{45}$

Hence, the correct answer is option C.

#### Page No 16.37:

#### 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) $\frac{4}{15}$ (b) $\frac{2}{15}$ (c) $\frac{1}{5}$ (d) $\frac{1}{3}$ [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) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{3}{15}=\frac{1}{5}$

Hence, the correct answer is option C.

#### Page No 16.37:

#### 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) $\frac{1}{4}$ (b) $\frac{1}{8}$ (c) $\frac{3}{4}$ (d) $\frac{7}{8}$ [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) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{3}{4}$

Hence, the correct answer is option C.

#### Page No 16.37:

#### 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) $\frac{1}{36}$ (b) $\frac{1}{2}$ (c) $\frac{1}{6}$ (d) $\frac{1}{4}$ [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) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{9}{36}=\frac{1}{4}$

Hence, the correct answer is option D.

#### Page No 16.37:

#### 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) $\frac{2}{3}$ (b) $\frac{1}{6}$ (c) $\frac{1}{3}$ (d) $\frac{11}{30}$ [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) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{10}{30}=\frac{1}{3}$

Hence, the correct answer is option C.

#### Page No 16.38:

#### 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) $\frac{1}{13}$ (b) $\frac{9}{13}$ (c) $\frac{4}{13}$ (d) $\frac{12}{13}$ [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) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{48}{52}=\frac{12}{13}$

Hence, the correct answer is option D.

#### Page No 16.38:

#### Question 29:

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

(a) $\frac{5}{7}$

(b) $\frac{2}{7}$

(c) $\frac{3}{7}$

(d) $\frac{1}{7}$

#### 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 16.38:

#### Question 30:

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) $\frac{7}{9}$

(b) $\frac{5}{9}$

(c) $\frac{2}{3}$

(d) $\frac{1}{9}$

#### 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 16.38:

#### Question 31:

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

(a) $\frac{2}{7}$

(b) $\frac{5}{7}$

(c) $\frac{6}{7}$

(d) $\frac{1}{7}$

#### 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 16.38:

#### Question 32:

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

(a) $\frac{1}{18}$

(b) $\frac{7}{36}$

(c) $\frac{1}{6}$

(d) $\frac{2}{9}$

#### 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 16.38:

#### Question 33:

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

(a) $\frac{6}{7}$

(b) $\frac{1}{7}$

(c) $\frac{5}{7}$

(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 16.38:

#### Question 34:

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

(a) $\frac{17}{45}$

(b) $\frac{1}{5}$

(c) $\frac{17}{90}$

(d) $\frac{8}{45}$

#### 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 16.38:

#### Question 35:

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

(a) $\frac{2}{3}$

(b) $\frac{1}{6}$

(c) $\frac{1}{3}$

(d) $\frac{5}{6}$

#### 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 16.38:

#### Question 36:

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

(a) $\frac{2}{7}$

(b) $\frac{4}{7}$

(c) $\frac{5}{7}$

(d) $\frac{6}{7}$

#### 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 16.38:

#### Question 37:

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

(a) $\frac{3}{10}$

(b) $\frac{29}{100}$

(c) $\frac{1}{3}$

(d) $\frac{7}{25}$

#### 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 16.38:

#### Question 38:

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) $\frac{1}{2}$ (b) $\frac{1}{3}$ (c) $\frac{1}{6}$ (d) $\frac{1}{12}$ [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) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{6}{36}=\frac{1}{6}$

Hence, the correct answer is option C.

#### Page No 16.39:

#### Question 39:

Mark the correct alternative in each of the following:

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

(a) $\frac{7}{8}$ (b) $\frac{1}{8}$ (c) $\frac{5}{8}$ (d) $\frac{3}{4}$ [CBSE 2014]

#### Answer:

The possible outcomes are BBB, BBG, BGB, GBB, BGG, GBG, GGB, GGG.

∴ Total number of outcomes = 8

The favourable outcomes are BBB, BBG, BGB, GBB, BGG, GBG, GGB.

So, the favourable number of outcomes are 7.

∴ P(at least one boy) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{7}{8}$

Hence, the correct answer is option A.

#### Page No 16.39:

#### Question 40:

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) $\frac{1}{5}$ (b) $\frac{3}{25}$ (c) $\frac{4}{25}$ (d) $\frac{2}{25}$ [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) = $\frac{\mathrm{Favourable}\mathrm{number}\mathrm{of}\mathrm{outcomes}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{outcomes}}=\frac{4}{25}$

Hence, the correct answer is option C.

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