Rs Aggarwal 2021 2022 for Class 9 Maths Chapter 19 - Probability

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Rs Aggarwal 2021 2022 Solutions for Class 9 Maths Chapter 19 Probability are provided here with simple step-by-step explanations. These solutions for Probability are extremely popular among Class 9 students for Maths Probability Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2021 2022 Book of Class 9 Maths Chapter 19 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

Page No 708:

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) = $\frac{Number of heads coming up}{Total number of trials} = \frac{285}{500} = 0.57$

(ii) Let F be the event of getting a tail.
P(getting a tail) = P (F) = $\frac{Number of tails coming up}{Total number of trials} = \frac{215}{500} = 0.43$

Page No 708:

Answer:

_Total 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

In a random toss of two coins, let E₁, E₂, E₃ be the events of getting 2 heads, 1 head and 0 head, respectively. Then,

(i) P(getting 2 heads) = P(E₁) =

\frac{Number of times 2 heads appear}{Total number of trials} = \frac{112}{400} = 0.28

(ii) P( getting 1 head) = P(E₂) =

\frac{Number of times 1 head appears}{Total number of trials} = \frac{160}{400} = 0.4

(iii) P( getting 0 head) = P(E₃) =

\frac{Number of times 0 head appears}{Total number of trials} = \frac{128}{400} = 0.32

Remark: Clearly, when two coins are tossed, the only possible outcomes are E₁, E₂ and E₃ and P(E₁) + P(E₂) + P(E₃) = (0.28 + 0.4 + 0.32) = 1

Page No 708:

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

In a random toss of three coins, let E₁, E₂, E₃ and E₄ be the events of getting 3 heads, 2 heads, 1 head and 0 head, respectively. Then;

(i) P(getting 3 heads) = P(E₁) =

\frac{Number of times 3 heads appear}{Total number of trials} = \frac{39}{200} = 0.195

(ii) P(getting 1 head) = P(E₂) =

\frac{Number of times 1 head appears}{Total number of trials} = \frac{67}{200} = 0.335

(iii) P(getting 0 head) = P(E₃) =

\frac{Number of times 0 head appears}{Total number of trials} = \frac{36}{200} = 0.18

(iv) P(getting 2 heads) = P(E₄) =

\frac{Number of times 2 heads appear}{Total number of trials} = \frac{58}{200} = 0.29

Remark: Clearly, when three coins are tossed, the only possible outcomes are E₁, E₂, E₃_andE₄ and P(E₁) + P(E₂) + P(E₃) + P(E₄) = (0.195 + 0.335 + 0.18 + 0.29) = 1

Page No 709:

Answer:

Total number of throws = 300
In a random throw of a dice, let E₁, E₂, E₃, E₄, be the events of getting 3, 6, 5 and 1, respectively. Then,
(i) P(getting 3) = P(E₁) = $\frac{Number of times 3 appears}{Total number of tr ia ls} = \frac{54}{300} = 0.18$

(ii) P(getting 6) = P(E₂) = $\frac{Number of times 6 appears}{Total number of tr ia ls} = \frac{33}{300} = 0.11$

(iii) P(getting 5) = P(E₃) = $\frac{Number of times 5 appears}{Total number of tr ia ls} = \frac{39}{300} = 0.13$

(iv) P(getting 1) = P(E₄) = $\frac{Number of times 1 appears}{Total number of tr ia ls} = \frac{60}{300} = 0.20$

Page No 709:

Answer:

Total number of ladies = 200
Number of ladies who like coffee = 142

Number of ladies who dislike coffee = 58

Let E₁ and E₂ be the events that the selected lady likes and dislikes coffee, respectively.Then,
(i) P(selected lady likes coffee) = P(E₁) =

\frac{Number of ladies who like coffee}{Total number of ladies} = \frac{142}{200} = 0.71

(ii) P (selected lady dislikes coffee) = P(E₂) =

\frac{Number of ladies who dislike coffee}{Total number of ladies} = \frac{58}{200} = 0.29

REMARK: In the given survey, the only possible outcomes are E₁ and E₂and P(E₁) + P(E₂) = (0.71 + 0.29) = 1

Page No 709:

Answer:

Total number of unit tests = 6
Number of tests in which the student scored more than 60% marks = 2

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

required probability = P(E) =

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

Page No 709:

Answer:

Total number of vehicles going past the crossing = 240
Number of two-wheelers = 84

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

required probability = P(E) =

\frac{84}{240} = 0.35

Number of unit tests in which he got more than 60% marksTotal number of unit tests = 26 = 13

Page No 709:

Answer:

Total phone numbers on the directory page = 200

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

Let E₁ be the event that the units digit of selected number is 5.

∴ Required probability = P(E₁) =

\frac{24}{200} = 0.12

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

Let E₂ be the event that the units digit of selected number is 8.

∴ Required probability = P(E₂) =

\frac{16}{200} = 0.08

Page No 709:

Answer:

Total number of students = 40
(i) Number of students with blood group O = 14

Let E₁ be the event that the selected student's blood group is O.

∴ Required probability = P(E₁) =

\frac{14}{40} = 0.35

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

Let E₂ be the event that the selected student's blood group is AB.

∴ Required probability = P(E₂) =

\frac{6}{40} = 0.15

Page No 710:

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) = $\frac{Number of packets which contains more than 2 kg of salt}{Total number of salt packets} = \frac{5}{12}$

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

Page No 710:

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

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

Page No 710:

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

$= \frac{27}{2400} = \frac{9}{800}$
(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

$= \frac{579}{2400} = \frac{193}{800}$
(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

$= \frac{10}{2400} = \frac{1}{240}$
(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

$= \frac{84}{2400} = \frac{7}{200}$
(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

$= \frac{2062}{2400} = \frac{1031}{1200}$

Page No 710:

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) = $\frac{Number of students whose marks are 30 or less}{Total number of students} = \frac{23}{30}$

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

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

(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) = $\frac{Number of students whose marks lie in the interval 21 - 30}{Total number of students} = \frac{6}{30} = \frac{1}{5}$

Page No 711:

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) = $\frac{Number of teachers who are 40 or more than 40 years old}{Total number of teachers} = \frac{45}{75} = \frac{3}{5}$

(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))

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

(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) = $\frac{Number of teachers 18 years or more and 49 years or less}{Total number of teachers} = \frac{67}{75}$

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

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

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

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

Page No 711:

Answer:

Total number of patients = 360
(i) Number of patients whose age is 30 years or more but less than 40 years = 60

Let E₁ 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(E₁) =

\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

Let E₂ 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(E₂) =

\frac{80}{360} = \frac{2}{9}

(iii) Let E₃be 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

∴ P(Patient's age is 10 years or more but less than 40 years) = P(E₃) =

\frac{200}{360} = \frac{5}{9}

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

Let E₄ 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(E₄) =

\frac{360}{360} = 1

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

Let E₅ be the event that the selected patient's age is less than 0.

∴ P(patient's age is less than 10 years)= P(E₅) =

\frac{0}{360} = 0

Page No 711:

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

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

Page No 711:

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) = $\frac{Number of non defective electric bulbs in the box}{Total number of electric bulbs in the box} = \frac{764}{800} = \frac{191}{200}$

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

Page No 712:

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:

Answer:

Total number of people surveyed = 645

Number of people who have a high school certificate = 516

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

Hence, the correct answer is option (d).

Page No 712:

Answer:

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

∴ Required probability = P(E) =

\frac{8}{40} = \frac{1}{5}

1440 = 0.35

Page No 713:

Answer:

(d) 0
Maximum lifetime a bulb has is 1100 hours. There is no bulb with lifetime 1150 hours.

Page No 713:

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) = $\frac{Number of children who do not liked to eat potato chips}{Total number of children surveyed} = \frac{273}{364} = \frac{3}{4}$

Hence, the correct answer is option (c).

Page No 713:

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

Hence, the correct answer is option (b).

Page No 713:

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

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

∴ Required probability = P(E) =

\frac{58}{80} = \frac{29}{40}

Page No 713:

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) = $\frac{Number of students who does not like the subject Sanskrit}{Total number of students surveyed} = \frac{65}{200} = \frac{13}{40}$

Hence, the correct answer is option (c).

Page No 714:

Answer:

Number of times head appears = 60 − 35 = 25

Let E be the event of getting a head.
∴ P(getting a head) = P (E) =

\frac{Number of times head appears}{Total number of trials} = \frac{25}{60} = \frac{5}{12}

Page No 714:

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:

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

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

P(E) =

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

Page No 714:

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.

Number of cards with numbers divisible by 3 = 5

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

∴ Required probability = P(E) =

\frac{5}{16}

Page No 714:

Answer:

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

Explanation:
Total number of balls in the bag = 5 + 8 + 7 = 20
Number of black balls = 8

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

∴ Required probability = P(E) =

\frac{8}{20} = \frac{2}{5}

Page No 714:

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) = $\frac{Number of times prime number s occur}{Total number of throws} = \frac{(10 + 12 + 9)}{65} = \frac{31}{65}$

Page No 714:

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) = $\frac{Number of times even number s appear}{Total number of throws} = \frac{(9 + 7 + 8)}{50} = \frac{24}{50} = \frac{12}{25}$

Page No 715:

Answer:

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

Explanation:
Total number of students = 36
Number of students born in October = 3

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

P(E) =

\frac{Number of students born in October}{Total number of students} = \frac{3}{36} = \frac{1}{12}

Page No 715:

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

Hence, the correct answer is option (c).

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