Rs Aggarwal 2019 Solutions for Class 9 Math Chapter 19 Probability are provided here with simple step-by-step explanations. These solutions for Probability are extremely popular among Class 9 students for Math Probability Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 9 Math Chapter 19 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 9 Math are prepared by experts and are 100% accurate.

#### Page No 708:

#### Question 1:

A coin is tossed 500 times and we get

head: 285 times, tail: 215 times.

When a coin is tossed at random, what is the probability of getting

(i) a head?

(ii) a tail?

#### Answer:

Total number of tosses = 500

Number of heads = 285

Number of tails = 215

(i) Let E be the event of getting a head.

*P*(getting a head) = *P* (*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{heads}\mathrm{coming}\mathrm{up}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{285}{500}=0.57$

(ii) Let F be the event of getting a tail.

*P*(getting a tail) = *P* (*F*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{tails}\mathrm{coming}\mathrm{up}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{215}{500}=0.43$

#### Page No 708:

#### Question 2:

Two coins are tossed 400 times and we get

two heads: 112 times; one head: 160 times; 0 head: 128 times.

When two coins are tossed at random, what is the probability of getting

(i) 2 heads?

(ii) 1 head?

(iii) 0 head?

#### Answer:

_{ T}otal number of tosses = 400

Number of times 2 heads appear = 112

Number of times 1 head appears = 160

Number of times 0 head appears = 128

*E*

_{1},

*E*

_{2},

*E*

_{3}be the events of getting 2 heads, 1 head and 0 head, respectively. Then,

(i)

*P*(getting 2 heads) =

*P*(

*E*

_{1}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}2\mathrm{heads}\mathrm{appear}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{112}{400}=0.28$

(ii)

*P*( getting 1 head) =

*P*(

*E*

_{2}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}1\mathrm{head}\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{160}{400}=0.4$

(iii)

*P*( getting 0 head) =

*P*(

*E*

_{3}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}0\mathrm{head}\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{128}{400}=0.32$

Remark: Clearly, when two coins are tossed, the only possible outcomes are

*E*

_{1},

*E*

_{2}and

*E*

_{3}and

*P*(

*E*

_{1}) +

*P*(

*E*

_{2}) +

*P*(

*E*

_{3}) = (0.28 + 0.4 + 0.32) = 1

#### Page No 708:

#### Question 3:

Three coins are tossed 200 times and we get

three heads: 39 times; two heads: 58 times;

one head: 67 times; 0 head: 36 times.

When three coins are tossed at random, what is the probability of getting

(i) 3 heads?

(ii) 1 head?

(iii) 0 head?

(iv) 2 heads?

#### Answer:

Total number of tosses = 200

Number of times 3 heads appear = 39

Number of times 2 heads appear = 58

Number of times 1 head appears = 67

Number of times 0 head appears = 36

*E*

_{1},

*E*

_{2},

*E*

_{3}and

*E*

_{4}be the events of getting 3 heads, 2 heads, 1 head and 0 head, respectively. Then;

*P*(getting 3 heads) =

*P*(

*E*

_{1}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}3\mathrm{heads}\mathrm{appear}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{39}{200}=0.195$

(ii)

*P*(getting 1 head) =

*P*(

*E*

_{2}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}1\mathrm{head}\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{67}{200}=0.335$

(iii)

*P*(getting 0 head) =

*P*(

*E*

_{3}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}0\mathrm{head}\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{36}{200}=0.18$

(iv)

*P*(getting 2 heads) =

*P*(

*E*

_{4}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}2\mathrm{heads}\mathrm{appear}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{58}{200}=0.29$

*E*

_{1},

*E*

_{2},

*E*

_{3}

_{ and }

*E*

_{4}and

*P*(

*E*

_{1}) +

*P*(

*E*

_{2}) +

*P*(

*E*

_{3}) +

*P*(

*E*

_{4}) = (0.195 + 0.335 + 0.18 + 0.29) = 1

#### Page No 709:

#### Question 4:

A dice is thrown 300 times and the outcomes are noted as given below.

Outcome | 1 | 2 | 3 | 4 | 5 | 6 |

Frequency | 60 | 72 | 54 | 42 | 39 | 33 |

(i) 3?

(ii) 6?

(iii) 5?

(iv) 1?

#### Answer:

Total number of throws = 300

In a random throw of a dice, let E_{1}, E_{2}, E_{3}, E_{4}, be the events of getting 3, 6, 5 and 1, respectively. Then,

(i) *P*(getting 3) = *P*(*E*_{1}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}3\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{tr}\text{ia}\mathrm{ls}}=\frac{54}{300}=0.18$

(ii) *P*(getting 6) = *P*(*E*_{2}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}6\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{tr}\text{ia}\mathrm{ls}}=\frac{33}{300}=0.11$

(iii) *P*(getting 5) = *P*(*E*_{3}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}5\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{tr}\text{ia}\mathrm{ls}}=\frac{39}{300}=0.13$

(iv) *P*(getting 1) = *P*(*E*_{4}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}1\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{tr}\text{ia}\mathrm{ls}}=\frac{60}{300}=0.20$

#### Page No 709:

#### Question 5:

In a survey of 200 ladies, it was found that 142 like coffee, while 58 dislike it.

Find the probability that a lady chosen at random

(i) likes coffee

(ii) dislikes coffee

#### Answer:

Total number of ladies = 200

Number of ladies who like coffee = 142

Let

*E*

_{1}and

*E*

_{2}be the events that the selected lady likes and dislikes coffee, respectively.Then,

(i)

*P*(selected lady likes coffee) =

*P*(

*E*

_{1}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{ladies}\mathrm{who}\mathrm{like}\mathrm{coffee}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{ladies}}=\frac{142}{200}=0.71$

(ii)

*P*(selected lady dislikes coffee) =

*P*(

*E*

_{2}) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{ladies}\mathrm{who}\mathrm{dislike}\mathrm{coffee}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{ladies}}=\frac{58}{200}=0.29$

REMARK: In the given survey, the only possible outcomes are

*E*

_{1}and

*E*

_{2}

_{ }and

*P*(

*E*

_{1}) +

*P*(

*E*

_{2}) = (0.71 + 0.29) = 1

#### Page No 709:

#### Question 6:

The percentages of marks obtained by a student in six unit tests are given below.

Unit test | I | II | III | IV | V | VI |

Percentage of marks obtained | 53 | 72 | 28 | 46 | 67 | 59 |

#### Answer:

Total number of unit tests = 6

Number of tests in which the student scored more than 60% marks = 2

*E*be the event that he got more than 60% marks in the unit tests.Then,

*P*(

*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{unit}\mathrm{tests}\mathrm{in}\mathrm{which}\mathrm{he}\mathrm{got}\mathrm{more}\mathrm{than}60\%\mathrm{marks}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{unit}\mathrm{tests}}=\frac{2}{6}=\frac{1}{3}$

#### Page No 709:

#### Question 7:

On a particular day, in a city, 240 vehicles of various types going past a crossing during a time interval were observed, as shown:

Types of vehicle | Two-wheelers | Three-wheelers | Four-wheelers |

Frequency | 84 | 68 | 88 |

#### Answer:

Total number of vehicles going past the crossing = 240

Number of two-wheelers = 84

*E*be the event that the selected vehicle is a two-wheeler. Then,

*P*(

*E*) = $\frac{84}{240}=0.35$

#### Page No 709:

#### Question 8:

On one page of a telephone directory, there are 200 phone numbers. The frequency distribution of their units digits is given below:

Units digit | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

Frequency | 19 | 22 | 23 | 19 | 21 | 24 | 23 | 18 | 16 | 15 |

(i) 5?

(ii) 8?

#### Answer:

Total phone numbers on the directory page = 200

(i) Number of numbers with units digit 5 = 24

*E*

_{1}be the event that the units digit of selected number is 5.

*P*(

*E*

_{1}) = $\frac{24}{200}=0.12$

(ii) Number of numbers with units digit 8 = 16

*E*

_{2}be the event that the units digit of selected number is 8.

*P*(

*E*

_{2}) = $\frac{16}{200}=0.08$

#### Page No 709:

#### Question 9:

The following table shows the blood groups of 40 students of a class.

Blood group | A | B | O | AB |

Number of students | 11 | 9 | 14 | 6 |

(i) O?

(ii) AB?

#### Answer:

Total number of students = 40

(i) Number of students with blood group O = 14

*E*

_{1}be the event that the selected student's blood group is O.

*P*(

*E*

_{1}) = $\frac{14}{40}=0.35$

(ii) Number of students with blood group AB = 6

*E*

_{2}be the event that the selected student's blood group is AB.

*P*(

*E*

_{2}) = $\frac{6}{40}=0.15$

#### Page No 710:

#### Question 10:

12 Packets of salt, each marked 2 kg, actually contained the following weights (in kg) of salt:

1.950, 2.020, 2.060, 1.980, 2.030, 1.970,

2.040, 1.990, 1.985, 2.025, 2.000, 1.980.

Out of these packets, one packet is chosen at random.

What is the probability that the chosen packet contains more than 2 kg of salt?

#### Answer:

Total number of salt packets = 12

Number of packets which contains more than 2 kg of salt = 5

∴ P(Chosen packet contains more than 2 kg of salt) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{packets}\mathrm{which}\mathrm{contains}\mathrm{more}\mathrm{than}2\mathrm{kg}\mathrm{of}\mathrm{salt}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{salt}\mathrm{packets}}=\frac{5}{12}$

Thus, the probability that the chosen packet contains more than 2 kg of salt is $\frac{5}{12}$.

#### Page No 710:

#### Question 11:

In a circket match, a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that he did not hit a boundary.

#### Answer:

Number of balls played by the batsman = 30

Number of balls in which he hits boundaries = 6

∴ Number of balls in which he did not hit a boundary = 30 − 6 = 24

P(Batsman did not hit a boundary) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{balls}\mathrm{in}\mathrm{which}\mathrm{he}\mathrm{did}\mathrm{not}\mathrm{hit}\mathrm{a}\mathrm{boundary}}{\mathrm{Number}\mathrm{of}\mathrm{balls}\mathrm{played}\mathrm{by}\mathrm{the}\mathrm{batsman}}=\frac{24}{30}=\frac{4}{5}$

Thus, the probability that he did not hit a boundary is $\frac{4}{5}$.

#### Page No 710:

#### Question 12:

An organisation selected 2400 families at random and surveyed them to determine a relationship between the income level and the number of vehicles in a family. The information gathered is listed in the table below:

Monthly income (in ₹) |
Number of vehicles per family | |||

0 | 1 | 2 | 3 or more | |

Less than ₹ 25000 | 10 | 160 | 25 | 0 |

₹ 25000 – ₹ 30000 | 0 | 305 | 27 | 2 |

₹ 30000 – ₹ 35000 | 1 | 535 | 29 | 1 |

₹ 35000 – ₹ 40000 | 2 | 469 | 59 | 25 |

₹ 40000 or more | 1 | 579 | 82 | 88 |

Suppose a family is chosen at random. Find the probability that the family chosen is

(i) earning ₹ 25000 – ₹ 30000 per month and owning exactly 2 vehicles.

(ii) earning ₹ 40000 or more per month and owning exactly 1 vehicle.

(iii) earning less than ₹ 25000 per month and not owning any vehicle.

(iv) earning ₹ 35000 – ₹ 40000 per month and owning 2 or more vehicles.

(v) owning not more than 1 vehicle.

#### Answer:

Number of families surveyed = 2400

(i) Number of families earning ₹ 25000 – ₹ 30000 per month and owning exactly 2 vehicles = 27

∴ P(Family chosen is earning ₹ 25000 – ₹ 30000 per month and owning exactly 2 vehicles)

= Number of families earning ₹ 25000 – ₹ 30000 per month and owning exactly 2 vehicles / Number of families surveyed

$=\frac{27}{2400}\phantom{\rule{0ex}{0ex}}=\frac{9}{800}$

(ii) Number of families earning ₹ 40000 or more per month and owning exactly 1 vehicle = 579

∴ P(Family chosen is earning ₹ 40000 or more per month and owning exactly 1 vehicle)

= Number of families earning ₹ 40000 or more per month and owning exactly 1 vehicle / Number of families surveyed

$=\frac{579}{2400}\phantom{\rule{0ex}{0ex}}=\frac{193}{800}$

(iii) Number of families earning less than ₹ 25000 per month and not owning any vehicle = 10

∴ P(Family chosen is earning less than ₹ 25000 per month and not owning any vehicle)

= Number of families earning less than ₹ 25000 per month and not owning any vehicle / Number of families surveyed

$=\frac{10}{2400}\phantom{\rule{0ex}{0ex}}=\frac{1}{240}$

(iv) Number of families earning ₹ 35000 – ₹ 40000 per month and owning 2 or more vehicles = 59 + 25 = 84

∴ P(Family chosen is earning ₹ 35000 – ₹ 40000 per month and owning 2 or more vehicles)

= Number of families earning ₹ 35000 – ₹ 40000 per month and owning 2 or more vehicles / Number of families surveyed

$=\frac{84}{2400}\phantom{\rule{0ex}{0ex}}=\frac{7}{200}$

(v) Number of families owning not more than 1 vehicle

= Number of families owning 0 vehicle + Number of families owning 1 vehicle

= 10 + 0 + 1 + 2 + 1 + 160 + 305 + 535 + 469 + 579 = 2062

∴ P(Family chosen is owning not more than 1 vehicle)

= Number of families owning not more than 1 vehicle / Number of families surveyed

$=\frac{2062}{2400}\phantom{\rule{0ex}{0ex}}=\frac{1031}{1200}$

#### Page No 710:

#### Question 13:

The table given below shows the marks obtained by 30 students in a test.

Marks (Class interval) |
1 – 10 | 11 – 20 | 21 – 30 | 31 – 40 | 41 – 50 |

Number of students (Frequency) |
7 | 10 | 6 | 4 | 3 |

Out of these students, one is chosen at random. What is the probability that the marks of the chosen student

(i) are 30 or less?

(ii) are 31 or more?

(iii) lie in the interval 21 – 30?

#### Answer:

Total number of students = 30

(i) Number of students whose marks are 30 or less = 7 + 10 + 6 = 23

∴ P(Marks of the chosen student are 30 or less) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{whose}\mathrm{marks}\mathrm{are}30\mathrm{or}\mathrm{less}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{students}}=\frac{23}{30}$

(ii) Number of students whose marks are 31 or more = 4 + 3 = 7

∴ P(Marks of the chosen student are 31 or more) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{whose}\mathrm{marks}\mathrm{are}31\mathrm{or}\mathrm{more}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{students}}=\frac{7}{30}$

(iii) Number of students whose marks lie in the interval 21–30 = 6

∴ P(Marks of the chosen student lie in the interval 21–30) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{whose}\mathrm{marks}\mathrm{lie}\mathrm{in}\mathrm{the}\mathrm{interval}21-30}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{students}}=\frac{6}{30}=\frac{1}{5}$

#### Page No 711:

#### Question 14:

Age (in years) | 18 – 29 | 30 – 39 | 40 – 49 | 50 – 59 |

Number of teachers | 3 | 27 | 37 | 8 |

(i) 40 or more than 40 years old?

(ii) of an age lying between 30 – 39 years (including both)?

(iii) 18 years or more and 49 years or less?

(iv) 18 years or more old?

(v) above 60 years of age?

Note Here 18 – 29 means 18 or more but less than or equal to 29.

#### Answer:

Total number of teachers = 75

(i) Number of teachers who are 40 or more than 40 years old = 37 + 8 = 45

∴ P(Selected teacher is 40 or more than 40 years old) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{teachers}\mathrm{who}\mathrm{are}40\mathrm{or}\mathrm{more}\mathrm{than}40\mathrm{years}\mathrm{old}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{teachers}}=\frac{45}{75}=\frac{3}{5}$

(ii) Number of teachers of an age lying between 30 – 39 years (including both) = 27

∴ P(Selected teacher is of an age lying between 30 – 39 years (including both))

= $\frac{\mathrm{Number}\mathrm{of}\mathrm{teachers}\mathrm{of}\mathrm{an}\mathrm{age}\mathrm{lying}\mathrm{between}30-39\mathrm{years}\left(\mathrm{including}\mathrm{both}\right)}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{teachers}}=\frac{27}{75}=\frac{9}{25}$

(iii) Number of teachers 18 years or more and 49 years or less = 3 + 27 + 37 = 67

∴ P(Selected teacher is 18 years or more and 49 years or less) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{teachers}18\mathrm{years}\mathrm{or}\mathrm{more}\mathrm{and}49\mathrm{years}\mathrm{or}\mathrm{less}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{teachers}}=\frac{67}{75}$

(iv) Number of teachers 18 years or more old = 3 + 27 + 37 + 8 = 75

∴ P(Selected teacher is 18 years or more old) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{teachers}18\mathrm{years}\mathrm{or}\mathrm{more}\mathrm{old}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{teachers}}=\frac{75}{75}=1$

(v) Number of teachers above 60 years of age = 0

∴ P(Selected teacher is above 60 years of age) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{teachers}\mathrm{above}60\mathrm{years}\mathrm{of}\mathrm{age}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{teachers}}=\frac{0}{75}=0$

#### Page No 711:

#### Question 15:

Age (in years) | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 |

Number of patients | 90 | 50 | 60 | 80 | 50 | 30 |

What is the probability that his age is

(i) 30 years or more but less than 40 years?

(ii) 50 years or more but less than 70 years?

(iii) 10 years or more but less than 40 years?

(iv) 10 years or more?

(v) less than 10 years?

#### Answer:

Total number of patients = 360

(i) Number of patients whose age is 30 years or more but less than 40 years = 60

*E*

_{1}be the event that the selected patient's age is in between 30 - 40.

*P*(patient's age 30 is years or more but less than 40 years) =

*P*(

*E*

_{1}) = $\frac{60}{360}=\frac{1}{6}$

(ii) Number of patients whose age is 50 years or more but less than 70 years = (50 +30) = 80

*E*

_{2}be the event that the selected patient's age is in between 50 - 70.

∴

*P*(patient's age is 50 years or more but less than 70 years) =

*P*(

*E*

_{2}) = $\frac{80}{360}=\frac{2}{9}$

(iii) Let

*E*

_{3 }be the event that the selected patient is 10 years or more but less than 40 years.

Number of patients whose age is 10 years or more but less than 40 years = 90 + 50 + 60 = 200

*E*

_{3}) = $\frac{200}{360}=\frac{5}{9}$

(iv) Number of patients whose age is 10 years or more = 90 + 50 + 60 + 80 + 50 + 30 = 360

*E*

_{4}be the event that the selected patient's age is 10 years or more. Then

∴

*P*(patient's age is 10 years or more) =

*P*(

*E*

_{4}) = $\frac{360}{360}=1$

(v) Number of patients whose age is less than 10 years = 0

*E*

_{5}be the event that the selected patient's age is less than 0.

*P*(patient's age is less than 10 years)=

*P*(

*E*

_{5}) = $\frac{0}{360}=0$

#### Page No 711:

#### Question 16:

Marks | 0 – 20 | 20 – 30 | 30 – 40 | 40 – 50 | 50 – 60 | 60 – 70 | 70 and above |

Number of students | 7 | 8 | 12 | 25 | 19 | 10 | 9 |

What is the probability that the chosen student

(i) gets 20% or less marks?

(ii) gets 60% or more marks?

#### Answer:

Total number of students = 90

(i) Number of students who gets 20% or less marks = Number of students who gets 20 or less marks = 7

∴ P(Student gets 20% or less marks) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{who}\mathrm{gets}20\%\mathrm{or}\mathrm{less}\mathrm{marks}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{students}}=\frac{7}{90}$

(ii) Number of students who gets 60% or more marks = Number of students who gets 60 or more marks = 10 + 9 = 19

∴ P(Student gets 60% or more marks) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{who}\mathrm{gets}60\%\mathrm{or}\mathrm{more}\mathrm{marks}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{students}}=\frac{19}{90}$

#### Page No 711:

#### Question 17:

It is known that a box of 800 electric bulbs contains 36 defective bulbs.One bulb is taken at random out of the box. What is the probability that the bulb chosen is non-defective?

#### Answer:

Total number of electric bulbs in the box = 800

Number of defective electric bulbs in the box = 36

∴ Number of non-defective electric bulbs in the box = 800 − 36 = 764

P(Bulb chosen is non-defective) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{non}\mathrm{defective}\mathrm{electric}\mathrm{bulbs}\mathrm{in}\mathrm{the}\mathrm{box}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{electric}\mathrm{bulbs}\mathrm{in}\mathrm{the}\mathrm{box}}=\frac{764}{800}=\frac{191}{200}$

Thus, the probability that the bulb chosen is non-defective is $\frac{191}{200}$.

#### Page No 712:

#### Question 18:

Fill in the blanks.

(i) Probability of an impossible event = ........ .

(ii) Probability of a sure event = ........ .

(iii) Let *E* be an event. Then*, P*(not *E*) = ......... .

(iv) *P*(*E*) + *P*(not* E*) = ........ .

(v) ....... ≤ *P*(*E*) ≤ ....... .

#### Answer:

(i) Probability of an impossible event = __ 0 __

(ii) Probability of a sure event = __ 1 __

(iii) Let *E* be an event. Then*, P*(not *E*) = __ ____1 − P( E) __ .

(iv)

*P*(

*E*) +

*P*(not

*E*) =

__1__

(v)

__0__≤

*P*(

*E*) ≤

__1__

#### Page No 712:

#### Question 1:

In a sample survey of 645 people, it was found that 516 people have a high school certificate. If a person is chosen at random, what is the probability that he / she has a high school certificate?

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

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

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

(d) $\frac{4}{5}$

#### Answer:

Total number of people surveyed = 645

Number of people who have a high school certificate = 516

∴ P(Person has a high school certificate) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{people}\mathrm{who}\mathrm{have}\mathrm{a}\mathrm{high}\mathrm{school}\mathrm{certificate}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{people}\mathrm{surveyed}}=\frac{516}{645}=\frac{4}{5}$

Hence, the correct answer is option (d).

#### Page No 712:

#### Question 2:

In a medical examination of students of a class, the following blood groups are recorded:

Blood group | A | B | AB | O |

Number of students | 11 | 15 | 8 | 6 |

(a) $\frac{13}{20}$

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

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

(d) $\frac{11}{40}$

#### Answer:

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

Explanation:

Total number of students = 40

Number of students with blood group AB = 8

*E*be the event that the selected student's blood group is AB.

*P*(

*E*) = $\frac{8}{40}=\frac{1}{5}$

#### Page No 713:

#### Question 3:

80 bulbs are selected at random from a lot and their lifetime is hours is recorded as under.

Lifetime (in hours) | 300 | 500 | 700 | 900 | 1100 |

Frequency | 10 | 12 | 23 | 25 | 10 |

(a) $\frac{1}{80}$

(b) $\frac{7}{16}$

(c) 1

(d) 0

#### Answer:

(d) 0

Maximum lifetime a bulb has is 1100 hours. There is no bulb with lifetime 1150 hours.

#### Page No 713:

#### Question 4:

In a survey of 364 children aged 19 – 36 months, it was found that 91 liked to eat potato chips. If a child is selected at random, the probability that he / she does not like to eat potato chips is

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

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

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

(d) $\frac{4}{5}$

#### Answer:

Total number of children surveyed = 364

Number of children who liked to eat potato chips = 91

Number of children who do not liked to eat potato chips = 364 − 91 = 273

∴ P(Child does not like to eat potato chips) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{children}\mathrm{who}\mathrm{do}\mathrm{not}\mathrm{liked}\mathrm{to}\mathrm{eat}\mathrm{potato}\mathrm{chips}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{children}\mathrm{surveyed}}=\frac{273}{364}=\frac{3}{4}$

Hence, the correct answer is option (c).

#### Page No 713:

#### Question 5:

Two coins are tossed 1000 times and the outcomes are recorded as given below:

Number of heads | 2 | 1 | 0 |

Frequency | 200 | 550 | 250 |

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

(b) $\frac{4}{5}$

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

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

#### Answer:

Total number of times two coins are tossed = 1000

Number of times of getting at most one head = Number of times of getting 0 heads + Number of times of getting 1 head = 250 + 550 = 800

∴ P(Getting at most one head) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{of}\mathrm{getting}\mathrm{at}\mathrm{most}\mathrm{one}\mathrm{head}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{times}\mathrm{two}\mathrm{coins}\mathrm{are}\mathrm{tossed}}=\frac{800}{1000}=\frac{4}{5}$

Hence, the correct answer is option (b).

#### Page No 713:

#### Question 6:

80 bulbs are selected at random from a lot and their lifetime in hours is recorded as under.

Lifetime (in hours) | 300 | 500 | 700 | 900 | 1100 |

Frequency | 10 | 12 | 23 | 25 | 10 |

(a) $\frac{27}{40}$

(b) $\frac{29}{40}$

(c) $\frac{5}{16}$

(d) $\frac{11}{40}$

#### Answer:

(b) $\frac{29}{40}$

Explanation:

Total number of bulbs in the lot = 80

Number of bulbs with life time of more than 500 hours = (23 + 25 + 10) = 58

*E*be the event that the chosen bulb's life time is more than 500 hours.

*P*(

*E*) = $\frac{58}{80}=\frac{29}{40}$

#### Page No 713:

#### Question 7:

To know the opinion of the students about the subject Sanskrit, a survey of 200 students was conducted. The data is recorded as under.

Opinion | like | dislike |

Number of students | 135 | 65 |

(a) $\frac{13}{27}$

(b) $\frac{27}{40}$

(c) $\frac{13}{40}$

(d) $\frac{27}{13}$

#### Answer:

Total number of students surveyed = 200

Number of students who does not like the subject Sanskrit = 65

∴ P(Student chosen at random does not like the subject Sanskrit) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{who}\mathrm{does}\mathrm{not}\mathrm{like}\mathrm{the}\mathrm{subject}\mathrm{Sanskrit}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{students}\mathrm{surveyed}}=\frac{65}{200}=\frac{13}{40}$

Hence, the correct answer is option (c).

#### Page No 714:

#### Question 8:

A coin is tossed 60 times and the tail appears 35 times. What is the probability of getting a head?

(a) $\frac{7}{12}$

(b) $\frac{12}{7}$

(c) $\frac{5}{12}$

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

#### Answer:

(c) $\frac{5}{12}$

Explanation:

Total number of trials = 60

Number of times tail appears = 35

*E*be the event of getting a head.

∴

*P*(getting a head) =

*P*(

*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{head}\mathrm{appears}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{trials}}=\frac{25}{60}=\frac{5}{12}$

#### Page No 714:

#### Question 9:

It is given that the probability of winning a game is 0.7. What is the probability of losing the game?

(a) 0.8

(b) 0.3

(c) 0.35

(d) 0.15

#### Answer:

(b) 0.3

Explanation:

Let E be the event of winning the game. Then,

*P*(E) = 0.7

*P*(not E) = *P*(losing the game) = 1− *P*(E) ⇒ 1− 0.7 = 0.3

#### Page No 714:

#### Question 10:

In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays. What is the probability that in a given delivery, the ball does not hit the boundary?

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

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

(c) $\frac{4}{5}$

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

#### Answer:

(c) $\frac{4}{5}$

Explanation:

Total number of balls faced = 30

Number of times the ball hits the boundary = 6

Number of times the ball does not hit the boundary = (30 − 6 )= 24

*E*be the event that the ball does not hit the boundary. Then,

*P*(

*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{ball}\mathrm{does}\mathrm{not}\mathrm{hit}\mathrm{the}\mathrm{boundary}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{balls}}=\frac{24}{30}=\frac{4}{5}$

#### Page No 714:

#### Question 11:

A bag contains 16 cards bearing numbers 1, 2, 3, ..., 16 respectively. One card is chosen at random. What is the probability that the chosen card bears a number divisible by 3?

(a) $\frac{3}{16}$

(b) $\frac{5}{16}$

(c) $\frac{11}{16}$

(d) $\frac{13}{16}$

#### Answer:

(b) $\frac{5}{16}$

Explanation:

Total number of cards in the bag = 16

Numbers on the cards that are divisible by 3 are 3, 6, 9, 12 and 15.

*E*be the event that the chosen card bears a number divisible by 3.

*P*(

*E*) = $\frac{5}{16}$

#### Page No 714:

#### Question 12:

A bag contains 5 red, 8 black and 7 white balls. One ball is chosen at random. What is the probability that the chosen ball is black?

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

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

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

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

#### Answer:

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

Explanation:

Total number of balls in the bag = 5 + 8 + 7 = 20

Number of black balls = 8

*E*be the event that the chosen ball is black.

*P*(

*E*) = $\frac{8}{20}=\frac{2}{5}$

#### Page No 714:

#### Question 13:

The outcomes of 65 throws of a dice were noted as shown below:

Outcome | 1 | 2 | 3 | 4 | 5 | 6 |

Number of times | 8 | 10 | 12 | 16 | 9 | 10 |

(a) $\frac{3}{35}$

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

(c) $\frac{31}{65}$

(d) $\frac{36}{65}$

#### Answer:

(c) $\frac{31}{65}$

Explanation:

Total number of throws = 65

Let E be the event of getting a prime number.

Then, E contains 2, 3 and 5, i.e. three numbers.

∴ *P*(getting a prime number) = *P*(*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{prime}\mathrm{number}\text{s}\mathrm{occur}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{throws}}=\frac{(10+12+9)}{65}=\frac{31}{65}$

#### Page No 714:

#### Question 14:

In 50 throws of a dice, the outcomes were noted as shown below:

Outcome | 1 | 2 | 3 | 4 | 5 | 6 |

Number of times | 8 | 9 | 6 | 7 | 12 | 8 |

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

(b) $\frac{3}{50}$

(c) $\frac{1}{8}$

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

#### Answer:

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

Explanation:

Total number of trials = 50

Let E be the event of getting an even number.

Then, E contains 2, 4 and 6, i.e. 3 even numbers.

∴ *P*(getting an even number) = *P*(*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{even}\mathrm{number}s\mathrm{appear}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{throws}}=\frac{(9+7+8)}{50}=\frac{24}{50}=\frac{12}{25}$

#### Page No 715:

#### Question 15:

The table given below shows the month of birth of 36 students of a class.

Month of birth | Jan | Feb | Mar | Apr | May | June | July | Aug | Sept | Oct | Nov | Dec |

No. of students | 4 | 3 | 5 | 0 | 1 | 6 | 1 | 3 | 4 | 3 | 4 | 2 |

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

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

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

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

#### Answer:

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

Explanation:

Total number of students = 36

Number of students born in October = 3

*E*be the event that the chosen student was born in October. Then,

*P*(

*E*) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{students}\mathrm{born}\mathrm{in}\mathrm{October}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{students}}=\frac{3}{36}=\frac{1}{12}$

#### Page No 715:

#### Question 16:

Two coins are tossed simultaneously 600 times to get 2 heads : 234 times, 1 head : 206 times, 0 head : 160 times.

If two coins are tossed at random, what is the probability of getting at least one head?

(a) $\frac{103}{300}$

(b) $\frac{39}{100}$

(c) $\frac{11}{15}$

(d) $\frac{4}{15}$

#### Answer:

Number of times two coins are tossed simultaneously = 600

Number of times of getting at least one head = Number of times of getting 1 head + Number of times of getting 2 heads = 206 + 234 = 440

∴ P(Getting at least one head) = $\frac{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{of}\mathrm{getting}\mathrm{at}\mathrm{least}\mathrm{one}\mathrm{head}}{\mathrm{Number}\mathrm{of}\mathrm{times}\mathrm{two}\mathrm{coins}\mathrm{are}\mathrm{tossed}\mathrm{simultaneously}}=\frac{440}{600}=\frac{11}{15}$

Hence, the correct answer is option (c).

View NCERT Solutions for all chapters of Class 9