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 9 Maths are prepared by experts and are 100% accurate.
Page No 708:
Question 1:
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) =
(ii) Let F be the event of getting a tail.
P(getting a tail) = P (F) =
Page No 708:
Question 2:
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) =
(ii) Let F be the event of getting a tail.
P(getting a tail) = P (F) =
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
(i) P(getting 2 heads) = P(E1) =
(ii) P( getting 1 head) = P(E2) =
(iii) P( getting 0 head) = P(E3) =
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:
Question 3:
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
(i) P(getting 2 heads) = P(E1) =
(ii) P( getting 1 head) = P(E2) =
(iii) P( getting 0 head) = P(E3) =
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
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
(ii) P(getting 1 head) = P(E2) =
(iii) P(getting 0 head) = P(E3) =
(iv) P(getting 2 heads) = P(E4) =
Page No 709:
Question 4:
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
(ii) P(getting 1 head) = P(E2) =
(iii) P(getting 0 head) = P(E3) =
(iv) P(getting 2 heads) = P(E4) =
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) =
(ii) P(getting 6) = P(E2) =
(iii) P(getting 5) = P(E3) =
(iv) P(getting 1) = P(E4) =
Page No 709:
Question 5:
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) =
(ii) P(getting 6) = P(E2) =
(iii) P(getting 5) = P(E3) =
(iv) P(getting 1) = P(E4) =
Answer:
Total number of ladies = 200
Number of ladies who like coffee = 142
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) =
(ii)â P (selected lady dislikes coffee) = P(E2) = â
REMARK: In the given survey, the only possible outcomes are E1 and E2 and âP(E1) + âP(E2) = (0.71 + 0.29) = 1
Page No 709:
Question 6:
Total number of ladies = 200
Number of ladies who like coffee = 142
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) =
(ii)â P (selected lady dislikes coffee) = P(E2) = â
REMARK: In the given survey, the only possible outcomes are E1 and E2 and âP(E1) + âP(E2) = (0.71 + 0.29) = 1
Answer:
Total number of unit tests = 6
Number of tests in which the student scored more than 60% marks = 2
Page No 709:
Question 7:
Total number of unit tests = 6
Number of tests in which the student scored more than 60% marks = 2
Answer:
Total number of vehicles going past the crossing = 240
Number of two-wheelers = 84
Page No 709:
Question 8:
Total number of vehicles going past the crossing = 240
Number of two-wheelers = 84
Answer:
Total phone numbers on the directory page = 200
(i) Number of numbers with units digit 5 = 24
(ii) Number of numbers with units digit 8 = 16
Page No 709:
Question 9:
Total phone numbers on the directory page = 200
(i) Number of numbers with units digit 5 = 24
(ii) Number of numbers with units digit 8 = 16
Answer:
Total number of students = 40
(i) Number of students with blood group O = 14
(ii) Number of students with blood group AB = 6
Page No 710:
Question 10:
Total number of students = 40
(i) Number of students with blood group O = 14
(ii) Number of students with blood group AB = 6
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) =
Thus, the probability that the chosen packet contains more than 2 kg of salt is .
Page No 710:
Question 11:
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) =
Thus, the probability that the chosen packet contains more than 2 kg of salt is .
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) =
Thus, the probability that he did not hit a boundary is .
Page No 710:
Question 12:
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) =
Thus, the probability that he did not hit a boundary is .
Answer:
Number of families surveyed = 2400
(i) Number of families earning â¹ 25000 – â¹ 30000 per month and owning exactly 2 vehicles = 27
∴ P(Family chosen is earning â¹ 25000 – â¹ 30000 per month and owning exactly 2 vehicles)
= Number of families earning â¹ 25000 – â¹ 30000 per month and owning exactly 2 vehicles / Number of families surveyed
(ii) Number of families earning â¹ 40000 or more per month and owning exactly 1 vehicle = 579
∴ P(Family chosen is earning â¹ 40000 or more per month and owning exactly 1 vehicle)
= Number of families earning â¹ 40000 or more per month and owning exactly 1 vehicle / Number of families surveyed
(iii) Number of families earning less than â¹ 25000 per month and not owning any vehicle = 10
∴ P(Family chosen is earning less than â¹ 25000 per month and not owning any vehicle)
= Number of families earning less than â¹ 25000 per month and not owning any vehicle / Number of families surveyed
(iv) Number of families earning â¹ 35000 – â¹ 40000 per month and owning 2 or more vehicles = 59 + 25 = 84
∴ P(Family chosen is earning â¹ 35000 – â¹ 40000 per month and owning 2 or more vehicles)
= Number of families earning â¹ 35000 – â¹ 40000 per month and owning 2 or more vehicles / Number of families surveyed
(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
Page No 710:
Question 13:
Number of families surveyed = 2400
(i) Number of families earning â¹ 25000 – â¹ 30000 per month and owning exactly 2 vehicles = 27
∴ P(Family chosen is earning â¹ 25000 – â¹ 30000 per month and owning exactly 2 vehicles)
= Number of families earning â¹ 25000 – â¹ 30000 per month and owning exactly 2 vehicles / Number of families surveyed
(ii) Number of families earning â¹ 40000 or more per month and owning exactly 1 vehicle = 579
∴ P(Family chosen is earning â¹ 40000 or more per month and owning exactly 1 vehicle)
= Number of families earning â¹ 40000 or more per month and owning exactly 1 vehicle / Number of families surveyed
(iii) Number of families earning less than â¹ 25000 per month and not owning any vehicle = 10
∴ P(Family chosen is earning less than â¹ 25000 per month and not owning any vehicle)
= Number of families earning less than â¹ 25000 per month and not owning any vehicle / Number of families surveyed
(iv) Number of families earning â¹ 35000 – â¹ 40000 per month and owning 2 or more vehicles = 59 + 25 = 84
∴ P(Family chosen is earning â¹ 35000 – â¹ 40000 per month and owning 2 or more vehicles)
= Number of families earning â¹ 35000 – â¹ 40000 per month and owning 2 or more vehicles / Number of families surveyed
(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
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) =
(ii) Number of students whose marks are 31 or more = 4 + 3 = 7
∴ P(Marks of the chosen student are 31 or more) =
(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) =
Page No 711:
Question 14:
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) =
(ii) Number of students whose marks are 31 or more = 4 + 3 = 7
∴ P(Marks of the chosen student are 31 or more) =
(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) =
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) =
(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))
=
(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) =
(iv) Number of teachers 18 years or more old = 3 + 27 + 37 + 8 = 75
∴ P(Selected teacher is 18 years or more old) =
(v) Number of teachers above 60 years of age = 0
∴ P(Selected teacher is above 60 years of age) =
Page No 711:
Question 15:
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) =
(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))
=
(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) =
(iv) Number of teachers 18 years or more old = 3 + 27 + 37 + 8 = 75
∴ P(Selected teacher is 18 years or more old) =
(v) Number of teachers above 60 years of age = 0
∴ P(Selected teacher is above 60 years of age) =
Answer:
Total number of patients = 360
(i) Number of patients whose age is 30 years or more but less than 40 years = 60
(ii) Number of patients whose age is 50 years or more but less than 70 years = (50 +30) = 80
∴ P(patient's age is 50 years or more but less than 70 years) = P(E2) =
(iii) Let E3 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
(iv) Number of patients whose age is 10 years or more = 90 + 50 + 60 + 80 + 50 + 30 = 360
∴ P(patient's age is 10 years or more) = P(E4) =
(v) Number of patients whose age is less than 10 years = 0
Page No 711:
Question 16:
Total number of patients = 360
(i) Number of patients whose age is 30 years or more but less than 40 years = 60
(ii) Number of patients whose age is 50 years or more but less than 70 years = (50 +30) = 80
∴ P(patient's age is 50 years or more but less than 70 years) = P(E2) =
(iii) Let E3 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
(iv) Number of patients whose age is 10 years or more = 90 + 50 + 60 + 80 + 50 + 30 = 360
∴ P(patient's age is 10 years or more) = P(E4) =
(v) Number of patients whose age is less than 10 years = 0
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) =
(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) =
Page No 711:
Question 17:
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) =
(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) =
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) =
Thus, the probability that the bulb chosen is non-defective is .
Page No 712:
Question 18:
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) =
Thus, the probability that the bulb chosen is non-defective is .
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) = â1 − P(E)â .
(iv) P(E) + P(not E) = 1
(v) 0 ≤ P(E) ≤ 1
Page No 712:
Question 1:
(i) Probability of an impossible event = 0
(ii) Probability of a sure event = 1
(iii) Let E be an event. Then, P(not E) = â1 − P(E)â .
(iv) P(E) + P(not E) = 1
(v) 0 ≤ P(E) ≤ 1
Answer:
Total number of people surveyed = 645
Number of people who have a high school certificate = 516
∴ P(Person has a high school certificate) =
Hence, the correct answer is option (d).
Page No 712:
Question 2:
Total number of people surveyed = 645
Number of people who have a high school certificate = 516
∴ P(Person has a high school certificate) =
Hence, the correct answer is option (d).
Answer:
(c)
Explanation:
Total number of students = 40
Number of students with blood group AB = 8
Page No 713:
Question 3:
(c)
Explanation:
Total number of students = 40
Number of students with blood group AB = 8
Answer:
(d) 0
Maximum lifetime a bulb has is 1100 hours. There is no bulb with lifetime 1150 hours.
Page No 713:
Question 4:
(d) 0
Maximum lifetime a bulb has is 1100 hours. There is no bulb with lifetime 1150 hours.
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) =
Hence, the correct answer is option (c).
Page No 713:
Question 5:
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) =
Hence, the correct answer is option (c).
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) =
Hence, the correct answer is option (b).
Page No 713:
Question 6:
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) =
Hence, the correct answer is option (b).
Answer:
(b)
Explanation:
Total number of bulbs in the lot = 80
Number of bulbs with life time of more than 500 hours = (23 + 25 + 10) = 58
Page No 713:
Question 7:
(b)
Explanation:
Total number of bulbs in the lot = 80
Number of bulbs with life time of more than 500 hours = (23 + 25 + 10) = 58
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) =
Hence, the correct answer is option (c).
Page No 714:
Question 8:
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) =
Hence, the correct answer is option (c).
Answer:
(c)
â
Explanation:
Total number of trials = 60
Number of times tail appears = 35
∴ P(getting a head) = P (E) =
Page No 714:
Question 9:
(c)
â
Explanation:
Total number of trials = 60
Number of times tail appears = 35
∴ P(getting a head) = P (E) =
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− âP(E) ⇒ 1− 0.7 = 0.3
Page No 714:
Question 10:
(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− âP(E) ⇒ 1− 0.7 = 0.3
Answer:
(c)
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
P(E) =
Page No 714:
Question 11:
(c)
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
P(E) =
Answer:
(b)
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.
Page No 714:
Question 12:
(b)
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.
Answer:
(b)
Explanation:
Total number of balls in the bag = 5 + 8 + 7 = 20
Number of black balls = 8
Page No 714:
Question 13:
(b)
Explanation:
Total number of balls in the bag = 5 + 8 + 7 = 20
Number of black balls = 8
Answer:
(c)
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) =
Page No 714:
Question 14:
(c)
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) =
Answer:
(a)
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) =
Page No 715:
Question 15:
(a)
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) =
Answer:
(d)
Explanation:
Total number of students = 36
Number of students born in October = 3
P(E) =
Page No 715:
Question 16:
(d)
Explanation:
Total number of students = 36
Number of students born in October = 3
P(E) =
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) =
Hence, the correct answer is option (c).
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