CBSE Important Questions Class 9 Maths Chapter 15
Important Questions Class 9 Mathematics Chapter 15 – Probability
In Chapter 15, Mathematics Class 9, students will learn how to calculate the probability that a specific experiment result will occur. The problems discussed in this exercise are based on actual events, which makes the students more interested in finding the answers.
The important questions Class 9 Mathematics Chapter 15 are made to aid students in developing a solid understanding of the subject. Students should consistently practise the important questions Class 9 Mathematics Chapter 15 in order to effectively internalise its conceptual meaning.
Our topic experts at Extramarks produced these resources for Class 9 students. These resources offer helpful hints and strategies for effectively addressing typical questions. The students may simply prepare all the concepts covered in the CBSE Syllabus in a much better and more efficient method with the help of Extramarks’ Chapter 15 Class 9 Mathematics important questions.
Students will also learn the fundamental formulas for estimating the likelihood of an event or the results of an experiment.Students will learn the probability chapter easier and better by completing the important questions Class 9 Mathematics Chapter 15.
The students are advised to evaluate each idea studied in the Class 9 Mathematics Syllabus and create an effective study plan. By only reading and memorising the principles, it is simply impossible to achieve high maths scores. Students should always strive to understand the concepts and reasoning presented in any given topic.It is advised that the students regularly practise these important questions Class 9 Mathematics Chapter 15 to have a solid understanding of fundamental mathematical ideas.
Important Questions Class 9 Mathematics Chapter 15- With Solutions
Important questions Class 9 Mathematics Chapter 15 was developed by experts at Extramarks after thorough research on each subject. This study material can be used by students to increase their confidence and prepare for the exam.
A few Mathematics Class 9 Chapter 15 important questions are provided here, along with their answers:
Question 1: In the sample study of 642 people, it is found that a total of 514 people have a high school certificate. If any person is selected at random, the probability that the selected person has a high school certificate is
(a) 0.5 (b) 0.6 (c) 0.7 (d) 0.8
Solution 1: (d)
The total number of the people in the sample study, n(S) = 642.
The number of people who have a high school certificate, n(E) = 514.
So, the probability that the person who is selected has a high school certificate
= n(E)/n(S)=514/642 = 0.8
Thus, the probability that the selected person has a high school certificate is 0.8.
Question 2: In a certain cricket match, a batswoman hits a boundary 6 times out of 30 balls she had played. Find the probability that she did not hit a boundary.
Solution 2:
As per the question,
Total number of balls = 30
Numbers of boundary = 6
Number of times the bats woman didn’t hit a boundary = 30 – 6 = 24
The probability that she did not hit boundary = 24/30 = 4/5
Question 3: 1500 families with 2 children were selected randomly, and the following data were recorded:
Number of girls in a family | 2 | 1 | 0 |
Number of families | 475 | 814 | 211 |
Compute the probability of the family, when chosen at random, having
(i) 2 girls (ii) 1 girl (iii) No girl
Also, check whether the sum of these probabilities is 1.
Solution 3:
Total number of families = 1500
(i) Numbers of families having two girls = 475
Probability = Numbers of families having two girls/Total number of families
= 475/1500 = 19/60
(ii) Numbers of families having one girl = 814
Probability = Numbers of families having one girl/Total number of families
= 814/1500 = 407/750
(iii) Numbers of families who have 0 girls = 211
Probability = Numbers of families who have 0 girls/Total number of families
= 211/1500
Sum of the probabilities = (19/60)+(407/750)+(211/1500)
= (475+814+211)/1500
= 1500/1500 = 1
Yes, the sum of the probabilities is 1.
Question 4: In a particular section of Class IX, a total of 40 students were asked about the month in which they were born, and the following graph was prepared from the data so obtained:
Find the probability that a student in the class was born in August.
Solution 4:
Image source: NCERT textbook
Total number of the students in the class = 40
Number of the students born in the month of August = 6
Probability (student of the class was born in August) = 6/40 = 3/20
Question 5: Three coins were tossed 200 times simultaneously with the following frequencies of the different outcomes:
Outcome | 3 heads | 2 heads | 1 head | No head |
Frequency | 23 | 72 | 77 | 28 |
If the three coins were simultaneously tossed again, compute the probability of 2 heads coming up.
Solution 5:
Number of times the 2 heads come up = 72
The total number of times the coins were tossed = 200
∴Probability (2 heads coming up) = 72/200 = 9/25
Question 6: An organisation which selected 2400 families at random and surveyed them to determine the relationship between the income level and the number of vehicles in the family. The information gathered is represented in the table below:
Monthly income
(in ₹) |
Vehicles per family | |||
0 | 1 | 2 | Above 2 | |
Less than 7000 | 10 | 160 | 25 | 0 |
7000-10000 | 0 | 305 | 27 | 2 |
10000-13000 | 1 | 535 | 29 | 1 |
13000-16000 | 2 | 469 | 59 | 25 |
16000 or more | 1 | 579 | 82 | 88 |
Suppose any family is selected. Find the probability that the family selected is
(i) earns ₹10000 – 13000 per month and owns exactly 2 vehicles.
(ii) earns ₹16000 or more per month and owns exactly 1 vehicle.
(iii) earns less than ₹7000 per month and does not own any vehicle.
(iv) earns ₹13000 – 16000 per month and owns more than 2 vehicles.
(v) owning not more than 1 vehicle.
Solution 6:
Total number of the families = 2400
(i) Numbers of families whose earnings are ₹10000 –13000 per month and own exactly 2 vehicles = 29
∴ probability (the family chosen has earning ₹10000 – 13000 per month and owns exactly 2 vehicles) = 29/2400
(ii) Number of families whose earning is ₹16000 or more per month and own exactly 1 vehicle = 579
∴probability (the family chosen has earning ₹16000 or more per month and owns exactly 1 vehicle) = 579/2400
(iii) Number of families whose earnings is less than ₹7000 per month and do not own any vehicle = 10
∴probability (the family chosen has earned less than ₹7000 per month and does not own any vehicle) = 10/2400 = 1/240
(iv) Number of families whose earnings is ₹13000-16000 per month and own more than 2 vehicles = 25
∴probability (the family chosen has earning ₹13000 – 16000 per month and owns more than 2 vehicles) = 25/2400 = 1/96
(v) Number of families who owns not more than 1 vehicle = 10+160+0+305+1+535+2+469+1+579
= 2062
∴probability (the selected household owns no more than one car)= 2062/2400 = 1031/1200
Question 7: A teacher who wanted to analyse the performance of two sections of students
in the mathematics test of 100 marks. Looking at the given performances, she found that a
few students got below 20 marks and a few got 70 marks or above. So she decided to
group the marks into an intervals of varying sizes as follows: 0 – 20, 20 – 30, . . ., 60 – 70,
70 – 100. Then she listed the following table:
(i) Find the probability that the student obtained less than 20% on the mathematics test.
(ii) Find the probability that the student obtained marks 60 or above.
Solution 7:
Marks | Number of students |
0 – 20 | 7 |
20 – 30 | 10 |
30 – 40 | 10 |
40 – 50 | 20 |
50 – 60 | 20 |
60 – 70 | 15 |
70 – above | 8 |
Total | 90 |
Total number of the students = 90
(i) Number of the students obtaining less than 20% in the mathematics test = 7
∴, the probability that a student obtaining less than 20% in the mathematics test = 7/90
(ii) Number of students those obtaining marks 60 or above = 15+8 = 23
∴, the probability that a student who obtained marks 60 or above = 23/90
Question 8: To know the opinion of students about the subject statistics, a survey of total 200 students was conducted. The data is recorded in the following table.
Opinion | Number of students |
like | 135 |
dislike | 65 |
Find the probability that any student is chosen at random
(i) like statistics, (ii) does not like statistics.
Solution 8:
Total number of students = 135+65 = 200
(i) Number of students those like statistics = 135
probability (student who likes statistics) = 135/200 = 27/40
(ii) Number of students who do not like statistics = 65
∴probability (student who does not like statistics) = 65/200 = 13/40
Question 9: The distance (in km) of 40 engineers from their place of residences to their place of work were represented as follows:
Find what is the empirical probability for an engineer to live:
(i) less than 7 km from her / his place of work?
(ii) more than or equal to 7 km from her / his place of work?
(iii) Within ½ km from her /his place of work?
Solution 9:
The distance (in km) of the 40 engineers from their residences to their place of work was represented as follows:
Total number of the engineers = 40
(i) Number of the engineers who are staying less than 7 km from their place of work = 9
probability (an engineer who stays less than 7 km from her / his place of work) = 9/40
(ii) Number of engineers staying in more than or equal to 7 km from their place of work = 40-9 = 31
probability (an engineer who stays more than or equal to 7 km from her / his place of work) = 31/40
(iii) Number of the engineers staying within ½ km from their place of work = 0
probability (an engineer who stays within ½ km from her / his place of work) = 0/40 = 0
Question 10: Eleven bags containing wheat flour, each marked 5 kg, whereas actually contained the following weights of flour in them(in kg):
4.97 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00 5.05 5.08
Find the probability that any of these bags, when chosen at random, may contain more than 5 kg of flour.
Solution 10:
Total number of the bags present = 11
Number of bags which have more than 5 kg of flour = 7
probability (any of the bags, when chosen at random, have more than 5 kg of flour) = 7/11
Question 11: In the survey of 364 children aged 19-36 months, it was found that about 91 children liked to eat potato chips. If any child is selected at random, the probability that the child does not like to eat potato chips is
(a) 0.80 (b) 0.50 (c) 0.75 (d) 0.25
Solution 11:
(c) the total number of children surveyed from 19-36 months, n(S) = 364. Of them, 91 liked to eat potato chips.
∴ Number of children who did not like to eat potato chips, n(E) = 364 – 91 = 273
∴ Probability (the child does not like to eat potato chips) = n(E)/n(S) = 273/364 = 0.75
Therefore, the probability that the child does not like to eat potato chips is 0.75.
Question 12: Two coins when tossed 1000 times, and the outcomes are recorded below
Image source: NCERT textbook
Based on this information, the probability of at most one head is
- 15 (B) 14 (C) 45 (D) 34
Solution 12:
(c) The total number of the coins tossed, n(S) = 1000
Number of the outcomes where at most one head occurs, n(E) = 550 + 250 = 800
=n(E)/n(S) = 800/1000 = 4/5
Therefore, the probability of at most one head occurring is 4/5.
Question 13: 80 bulbs are selected randomly from a lot, and their lifetime (in hours) in the form of a frequency table is recorded and given below
Image source: NCERT textbook
One bulb is selected at random from the lot. The probability that its life is 1150
hours are
- 180 (B) 716 (C) 0 (D) 1
Solution 13:
(c) Total number of the bulb in a lot, n (S) = 80
Number of the bulbs whose lifetime is 1150, n(E) = 0
Probability that its lifetime is 1150 h = n(E)/n(S) = 0/80 = 0
Hence, the probability that its lifetime is 1150 is 0.
Question 14: Refer to the above question. The probability that the bulbs selected randomly from the lot has a life of less than 900 h is
(a) 7/16 (b) 5/16 (c) 11/40 (d) 9/16
Solution 14:
(d) Total number of the bulbs in a lot, n(S) = 80
Number of the bulbs whose life is less than 900 h, n(E) = 10 + 12 + 23 = 45
Probability that the bulbs have lifetime less than 900 h =n(E)/n(S) = 45/80 = 9/16
Hence, the probability that the bulb has a lifetime of less than 900 is 9/16.
Question 15: Can the experimental probability of any event be a negative number? If not, why?
Solution 15:
No, as the number of trials when the event occurs cannot be negative, and hence the total number of trials is always positive.
Question 16: Can the experimental probability of any event be greater than 1? Justify your answer.
Solution 16:
No, since the number of trials when the event can occur can never be greater than the total number of trials.
Question 17: As the number of tosses of a coin increase, the ratio of the number of heads to the total number of tosses will be ½. Is it correct? If not, write the correct one.
Solution 17:
No, since the number of coins increases, the ratio of the number of the heads to the total number of tosses will be nearer to ½ but not exactly ½.
Question 18: A company selected 4000 households at random and surveyed them to find out the relationship between their income level and the number of television sets in their home. The information obtained is listed in the following table.
Image source: NCERT textbook
Find the probability
(i) of a household which earns Rs. 10000 – Rs. 14999 per year and have exactly one television.
(ii) a household which earns Rs. 25000 and more per year owns 2 televisions.
(iii) of a household which do not have any television.
Solution 18:
The total number of households selected by the company,
n(S) = 4000
(i) Number of households earning Rs. 10000 – Rs. 14999 per year and who have exactly one television, n(E1) = 240
Therefore, required probability = n(E1)n(S) = 2404000 = 6100 = 350= 0.06
Hence, the probability of the number of households earning Rs. 10000 – Rs. 14999 per year and who have exactly one television is 0.06.
(ii) Number of households earning Rs. 25000 and more per year owning 2 televisions, n(E2) = 760
Therefore, required probability = n(E2)n(S) = 7604000 = 0.19
Hence, the probability of the number of households earning Rs. 25000 and more per year owning 2 televisions is 0.19.
(iii) Number of households not having any television, n(E3) = 30
Therefore, required probability = n(E3)n(S) = 304000 = 3400
Hence, the probability of the number of households not having any television is 3400.
Question 19: Two dice when thrown simultaneously 500 times. Each time the sum of the two numbers appearing on their tops is noted and recorded in the following table as given .
Image source: NCERT textbook
If the dice are thrown once more, then what is the probability of getting a sum of
(i) 3?
(ii) more than 10?
(iii) less than or equal to 5?
(iv) between 8 and 12?
Solution 19:
Total number of times when two dice are thrown simultaneously, n(S) = 500
(i) Number of the times of getting a sum 3, n(E) = 30
Therefore, Probability of getting a sum 3 = n(E)n(S) = 30500 = 350= 0.06
probability (getting a sum of 3) is 0.06.
(ii) Number of the times of getting a sum of more than 10, n(E1) = 28 + 15 = 43
Therefore, the probability of getting a sum more than 10 = n(E1)n(S) = 43500 = 0.086
probability (getting a sum of more than 10) is 0.086.
(iii) Number of the times of getting a sum of less than or equal to 5, n(E2) = 55 + 42 + 30+ 14 = 141
So, probability of getting a sum of less than or equal to 5 = n(E2)n(S) = 141500 = 0.282
probability (getting a sum of less than or equal to 5) is 0.282.
(iv) Number of the times of getting a sum between 8 and 12, n(E3) = 53 + 46 + 28 = 127
Therefore, the Probability of getting a sum between 8 and 12 = n(E3)n(S) = 127500 = 0.254
Hence, the probability of getting a sum between 8 and 12 is 0.254.
Question 20: Bulbs are packed in cartons, each containing 40 bulbs, seven hundred cartons were examined for defective bulbs, and the results are given in the following table.
Image source: NCERT textbook
One carton was selected at random. What is the probability that it has
(i) no defective bulb?
(ii) defective bulbs from 2-6?
(iii) defective bulbs less than 4?
Solution 20:
Total number of cartoons, n(S) = 700
(i) Number of cartoons which has no defective bulb, n(E1) =400
Therefore, the probability that no defective bulb = n(E1)n(S) = 400700= 47
Hence, the probability that there are no defective bulbs is 47.
(ii) Number of cartoons which has defective bulbs from 2 to 6, n(E2) =48 + 41 + 18+8+3 = 118
Therefore, the probability that the defective bulb from 2 to 6 = n(E2)n(S) = 118700= 59350
Hence, the probability that no defective bulbs are 59350.
(iii) Number of cartoons which has defective bulbs less than 4, n(E3) =400 + 180 + 48+ 41 = 669
Therefore, the probability that the defective bulb less than 4 = n(E3)n(S) = 669700
Hence, the probability that the defective bulbs are less than 4 is 669700.
Question 21: Over the past 200 working days, the number of defective parts produced by a machine is represented in the following table
Image source: NCERT textbook
Determine the probability that tomorrow’s output will have
(i) no defective part.
(ii) at least one defective part,
(iii) not more than 5 defective parts,
(iv) more than 13 defective parts.
Solution 21:
Total number of working days, n(S) = 200
(i) Number of days when no defective part is, n(E1) = 50
Probability that no defective part = n(E1)/n(S) = 50/200 = ¼ = 0.25
(ii) Number of days in which at least one defective part is
n(E2)= 32 + 22+ 18+ 12+ 10+ 10+8+6=6+2+2 = 150
Therefore, the probability that at least one defective part = n(E2)n(S) = 150200 = 34 = 0.75
Hence, the probability that at least one defective part is 0.75.
(iii) Number of days in which more than 5 defective days,
n(E3) = 50 + 32+ 22+18+12+12 = 146
Therefore, the probability that not more than 5 defective parts
= n(E3)n(S) = 146200 = 0.73
Hence, the probability that not more than 5 defective parts is 0.73
(iv) Number of the days when more than 13 defective parts, n (E4) = 0
Therefore, the probability that more than 13 defective parts = n (E4)n(S) = 0200 =0
Hence, the probability that there are more than 13 defective parts is 0.
Question 22: A recent survey found that the ages of workers in a factory are as follows.
Image source: NCERT textbook
If any person is selected at random, find the probability that the person is
(i) 40 yr or more. (ii) under 40 yr.
(iii) having age from 30-39 years. (iv) under 60 but over 39 years.
Solution 22:
Total number of the workers in a factory,
n(S) = 38 + 27 + 86 + 46 + 3 = 200
(i) Number of persons selected at the age of 40 years or more,
n(E1) = 86 + 46 + 3 = 135
Therefore, the probability that the persons selected at the age of 40 years or more,
P(E1) = n(E1)n(S) = 135200 = 0.675
Hence, the probability that the person who i selected at the age of 40 years or more is 0.675.
(ii) Number of persons selected under the age of 40 years,
n(E2) = 38 + 27 = 65
Therefore, the probability that the persons selected under the age of 40 years,
P(E2) = n(E2)n(S) = 65200 = 0.325
Hence, the probability that the person selected is under the age of 40 years is 0.325.
(iii) Number of persons selected having age from 30 to 39 years, n(E3) = 27
Therefore, the probability that the persons selected from 30 to 39 years,
P(E3) = n(E3)n(S) = 27200 = 0.135
Hence, the probability that the person selected from 30 to 39 years is 0.135.
(iv) Number of persons selected having age from 60 but over 39 year, n(E4) = 86 + 46 = 132
Therefore, the probability that the persons selected having aged 60 but over 39 years,
P(E4) = n(E4)n(S) = 132200 = 0.66
Hence, the probability that the person selected has aged from 60 but is over 39 years is 0.66.
Question 23: Compute the probability of the occurrence of an event if the probability of the event not occurring is 0.56.
Solution 23:
Given,
P(not E) = 0.56
We know,
P(E) + P(not E) = 1
Thus, P(E) = 1 – P(not E)
P(E) = 1 – 0.56
Or, P(E) = 0.44
Question 24: From a deck of cards, 10 cards were picked at random and then shuffled. The cards are as follows:
6, 5, 3, 9, 7, 6, 4, 2, 8, 2
Find the probability of picking a card having a value of more than 5 and find the probability of picking a card with an even number appearing on it.
Solution 24:
Total number of the cards = 10
Total number of cards having a value of more than 5 = 5
i.e. {6, 9, 7, 6, 8}
Total number of cards having an even number = 6
i.e. {6, 6, 4, 2, 8, 2}
Thus, the probability of picking a card having the value of more than 5 = 5/10 = 0.5
And the probability of picking a card having an even number on it = 6/10 = 0.6
Question 25: From a bag of the red and blue balls, the probability of picking a red ball is x/2. Find “x” if the probability of picking the blue ball is ⅔.
Solution 25:
Here, we have only red and blue balls.
P(picking any red ball) + P(picking any blue ball) = 1
x/2 + ⅔ = 1
=> 3x + 4 = 6
=> 3x = 2
Or, x = ⅔
Question 26: Two coins are tossed simultaneously 360 times. The number of times ‘2 Tails’ appeared was three times of ‘No Tail’ appeared and the number of times ‘1 tail’ appeared was double the number of times ‘No Tail’ appeared. What is the probability of getting the ‘Two tails’?
Solution 26:
Given,
Total number of the outcomes = Sample space = 360
Now, assuming that the number of times ‘No Tail’ appeared to be “x.”
So, the number of times the ‘2 Tails’ appeared = 3x (from the question)
Also, the number of times the ‘1 Tail’ appeared =2x (from the question)
As the total number of outcomes = 360,
x + 2x + 3x = 360
=> 6x = 360
Or, x = 60
∴ P(getting two tails) = (3 × 60)/360 = ½
Question 27: A die when thrown 1000 times with the frequencies for the outcomes to be 1, 2, 3, 4, 5 and 6 and are given in the following table:
Outcome | 1 | 2 | 3 | 4 | 5 | 6 |
Frequency | 179 | 150 | 157 | 149 | 175 | 190 |
Find the probability of getting each outcome.
Solution 27: Let us assume Ei to be denoting an event of getting an outcome which is i, where i = 1, 2, 3, 4, 5, or 6.
Then
The probability of the outcome 1 can be shown as P(E1)
= Frequency of 1 / the total number of times the die was thrown
= 179/1000
= 0.179
In such similar cases we will have,
P(E2 ) = 150/1000 = 0.15
P(E3) = 157/1000 = 0.157
P(E4) = 149/1000 = 0.149
P(E5) = 175/1000 = 0.175
and P(E6) = 190/1000 = 0.19
You can check as: P(E1) + P(E2) + P(E3) + P(E4) + P(E5) + P(E6) = 1
Question 28: The blood groups of some of the students of Class IX were surveyed and recorded as below :
Blood Group | A | B | AB | O |
No. of students | 19 | 6 | 13 | 12 |
If any student at random is chosen, find the probability that he/she is having blood group A or AB.
Solution 28:
Here,
total number of the students = 19 + 6 + 13 + 12 = 50
Number of the students having blood group A or AB = 19 + 13 = 32
Required probability = 3850 = 1925
Question 29: A group of a total of 80 students studying in Class X are selected and were asked for their choice of subject to be taken in Class XI, which is represented as below :
Stream | PCM | PCB | Commerce | Humanities | Total |
No. of students | 29 | 18 | 21 | 12 | 80 |
If any student is chosen at random, find the probability that he/she is a student of either commerce or humanities stream.
Solution 29:
Here, the total number of students = 80
Total number of students in Commerce or Humanities stream = 33
Required probability = 3380
Question 30: A dice is rolled a number of times, and its outcomes are recorded as below:
Outcome | 1 | 2 | 3 | 4 | 5 | 6 |
Frequency | 35 | 45 | 50 | 38 | 53 | 29 |
Find the probability of getting an odd number.
Solution 30:
Total number of the outcomes = 250
Total number of the outcomes of getting odd numbers = 35 + 50 + 53 = 138
∴ P(getting an odd number) = 138250 = 69125
Question 31: The probability of guessing the correct answer to a certain question is x2. If probability of not guessing the correct answer is 23, then find x.
Solution 31:
Here, the probability of guessing the correct answer = x2
And the probability of not guessing the correct answer = 23
Now,
x2 + 23 = 1
⇒ 3x + 4 = 6
⇒ 3x = 2
⇒ x = 23
Question 32: A bag which contains x white, y red and z blue balls. If a ball is drawn at random, then what is the probability of drawing a blue ball?
Solution 32:
Number of blue balls = Z
Total balls = x + y + Z
∴ P(a blue ball) = zx+y+z
Question 33: 750 families with 3 children were selected randomly, and the following data were recorded:
Number of girls ina family | 0 | 1 | 2 | 3 |
Number of families | 120 | 220 | 310 | 100 |
If any family member is chosen at random, compute the probability that it has :
(i) no boy child
(ii) no girl child
Solution 33:
(i) P(no boy child) = 100750 = 215
and P (no girl child) = 120750 = 425
Question 34: Suppose the probability of winning a race an athlete is 16, less than twice the probability of him losing the race. Find the probability of winning the race.
Solution 34:
Let the probability of the athlete winning the race be p
∴ Probability of the athlete losing the race = 1 – p
As per the question, we have
p = 2 (1 – p) – 16
⇒ 6p = 12 – 12p – 1
⇒ 18p = 11
⇒ p = 1118
Hence, the probability of winning the race is 1118.
Question 35:
Three coins were tossed simultaneously 150 times with the following frequencies of different outcomes :
Number of tails | 0 | 1 | 2 | 3 |
Frequency | 25 | 30 | 32 | 63 |
Compute the probability of getting :
(i) At least 2 tails
(ii) Exactly one tail
Solution 35:
Here, the total number of the chances = 150
(i) Total number of the chances having at least 2 tails = 32 + 63 = 95
∴ Required probability = 95150 = 1930
(ii) Total number of the chances having exactly one tail = 30
∴ Required probability = 30150 = 15
Question 36: Find the probability that the student will get 70% or more in the next unit test. Also, the probability that students get less than 70%.
Solution 36:
Here, the marks are out of 50, so we will first find its percentage (i.e., out of 100)
Unit Test | I | II | III | IV | V |
Marks (Out of 100) | 68 | 70 | 72 | 68 | 74 |
Total number of outcomes = 5
Probability of getting 70% or more marks = 35
Probability of getting less than 70% = 25
Question 37: 100 plants were sown in six different colonies A, B, C, D, E, and F. After 31 days, the number of plants survived as follows :
Colonies | A | B | C | D | E | F |
No. of plants survived | 80 | 90 | 84 | 76 | 82 | 92 |
What is the probability of :
(i) more than 80 plants surviving in a colony?
(ii) less than 82 plants surviving in a colony?
Solution 37:
From the question, we have the total number of colonies = 6
(i) Number of the colonies in which more than 80 plants survived = 4 (i.e., B, C, E and F)
∴ P(more than 80 plants survived in the colony) = 46 = 23
(ii) Number of the colonies in which less than 82 plants survived = 2 (i.e., A and D)
∴ P (less than 82 plants survived in the colony) = 26 = 13
Question 38: A die was rolled 100 times, and the number of times 6 came up was noted. If the experimental probability thus calculated from this information is 25, then how many times 6 came up? Justify your answer.
Solution 38:
Here, the total number of trials = 100
Let x be the number of times occurring 6.
We know the probability of an event = Frequency of the event occurring in the total number of trials
⇒ x100 = 25 [∵ Probability is given)
⇒ x = 40
Question 39:
Three coins were tossed simultaneously 250 times. The distribution of the various outcomes is listed below :
(i) Three tails: 30,
(ii) Two tails: 70,
(iii) One tail: 90,
(iv) No tail: 60
Find the respective probability of each of the event and check that the sum of all probabilities is 1.
Solution 39:
Here, the total number of outcomes = 250
(i) Total number of three tails = 30
Therefore, P(of three tails) = 30250 = 325
(ii) Total number of two tails = 70
Therefore, P(of two tails) = 70250 = 725
(iii) Total number of one tail = 90
Therefore, P(of one tails) = 90250 = 925
(iv) Total number of no tail = 60
Therefore, P(of no tail) = 60250 = 625
Now, sum of all probabilities = 325 + 725 + 925 + 625 = 2525 =1
Question 40: A travel company has 100 drivers to drive buses to various tourist destinations. Given
below is the table showing the resting time of the drivers after they covers a certain distance (in km).
Distance (in km) | After 80 km | After 115 km | After 155 km | After 200 km |
No. of drivers | 13 | 47 | 30 | 10 |
What is the probability that the driver was chosen at random :
(a) takes halt after covering 80 km?
(b) takes halt after covering 115 km?
(c) takes halt after covering 155 km?
(d) takes halt after crossing 200 km?
Solution 40:
Total number of the drivers = 100
(a) P (takes halt after covering 80 km) = 13100
(b) P (takes halt after covering 115 km) = 60100= 35
(c) P (takes halt after covering 155 km) = 90100= 910
(d) P (takes halt after crossing 200 km) = 10100 = 110
Question 41: A company which selected 2300 families at random and surveyed them to determine the relationship between income level and the number of vehicles in their home. The information gathered is listed in the table below :
If a family is chosen at random, find the probability that the family is :
(i) earning ₹7000 – ₹13000 per month and owns exactly 1 vehicle.
(ii) owns not more than one vehicle. (iii) earning more than ₹13000 and owns 2 or more than 2 vehicles. (iv) owns no vehicle
Solution 41:
From the question, we have the total number of the families = 2300
(i) Number of the families earning ₹7000 to ₹13000 per month and owning exactly 1 vehicle = 295 + 525 = 820
Therefore, required probability = 8202300 = 41115
(ii) Number of families owning not more than one vehicle = 1962
Therefore, required probability = 19622300 = 9811150
(iii) Number of families earning more than Rs. 13000 and owns 2 or more than 2 vehicles =224
Therefore, required probability = 2242300 = 56575
(iv) Number of families owning no vehicle =14
Therefore, required probability = 142300 = 71150
Question 42: A survey of 2000 people of different age groups was conducted to find out their preference for watching different types of movies :
Type I + Family
Type II → Comedy and Family
Type III → Romantic, Comedy, and Family 242.
Type IV → Action, Romantic, Comedy and Family
Age group | Type I | Type II | Type III | Type IV | All |
18-29
30-50 Above 50 |
440
505 360 |
160
125 45 |
110
60 35 |
61
22 15 |
35
18 9 |
Find the probability that any person chosen at random is :
(a) in 18-29 years of age and they like type II movies
(b) above 50 years of age and they like all types of movies
(c) in 30-50 years and likes type I movies. :
Solution 42:
(a) Let E1 be the event of the age group (18 – 29) years and liking type II movies
Favourable outcomes to event E1 = 160
∴ P(E1) = 1602000 = 225
(b) Let E2 be the event of age group above 50 years and liking all types of movies
Favourable number of outcomes to event E2 = 9
∴ P(E2) = 92000
(c) Let E3 be the event between the age group (30 – 50) years and liking type I movies
Favourable outcomes to event E3 = 505
∴ P(E3) = 5052000 = 101400
Question 43: In a kitchen, there are total of 108 utensils consisting of bowls, plates, and glasses. The ratio of the bowls, plates and the glasses is 4:2:3. A utensil is picked at random. Find the probability that :
(i) it is a plate.
(ii) it is not a bowl.
Solution 43:
Total number of utensils in the kitchen = 108
Let the number of bowls be 4x, the number of plates be 2x, and the number of glasses be 3x
∴ 4x + 2x + 3x = 108
9x = 108
x = 1089 = 12
Thus, number of bowls = 4 × 12 = 48
Number of plates = 2 × 12 = 24
Number of glasses = 3 × 12 =
(i)P (a plate) = 24108 = 29
(ii) P (not a bowl) = 24+36108 = 60108 = 59
Question 44:
A bag which contains 20 balls, out of which x are white.
(a) If one of the ball is drawn at random, find the probability that it is white.
(b) If 10 more white balls are again put in the bag, the probability of drawing a white ball out of it will be double that in part (a), find x.
Solution 44:
Here, the total no. of balls = 20
No. of white balls = x
∴ P(white ball) = x20
Now, 10 more white balls are added
∴ Total no. of balls = 20 + 10 = 30
Total no. of white balls = x + 10
According to the statement of the question, we have
x+1030 = 2 × x20
⇒ x+103 = x
⇒ x + 10 = 3x
⇒ 2x = 10
⇒ x = 5
Question 45:
Here is a representation from a mortality table.
(i) Based on the given information, what is the probability of the person aged 60′ of dying within a year?
(ii) What is the probability that a person aged 61′ will be living for 4 years?
Solution 45:
(i)16090 persons of age 60 (16090 – 11490), i.e., 4600, who died before reaching their 61st birthday.
Therefore, P (a person aged 60 dies within a year) = 460016090 = 4601609
(ii) Number of persons of age 61 years = 11490
Number of the persons surviving for 4 years = 2320
P (a person aged 61 will be living for 4 years) = 232011490 = 2321149
Question 46: An insurance company which selected 2000 drivers at random (i.e., without any preference for one driver over another) in a particular city to find the relationship between their age and accidents. The data obtained are represented in the following table :
Age group of drivers (in years) | Number of accidents in 1 year | ||||
0 | 1 | 2 | 3 | More than 3 | |
18-28
29-50 Above 50 |
395
520 390 |
155
118 57 |
105
65 35 |
70
20 8 |
29
21 12 |
Find the probability of the following events for the driver selected at random from the city :
(i) being 18 – 29 years of age and have exactly 3 accidents in one year.
(ii) being 30 – 50 years of age and have one or more accidents in a year.
(iii) had no accident in one year.
Solution 46:
Here, the total number of the drivers = 2000
(i) The number of the drivers in the age group 18 – 29 who have exactly 3 accidents = 70
So, P (driver in the age group 18 – 29 having exactly 3 accidents in one year) = 702000 = 0.035
(ii). The number of the drivers in the age group 30 – 50 and who have one or more than one accidents in one year = 118 + 65 + 20 + 21 = 224
P (driver in the age group 30 – 50 and who have one or more accidents in one year) = 2242000 = 0.112
(iii) The number of the drivers who have no accident in one year = 395 + 520 + 390 = 1305
So, P (driver having no accident) = 13052000 = 0.6525
Benefits of Solving Important Questions Class 9 Mathematics Chapter 15
These Class 9 Mathematics Chapter 15 important questions provide an overview of the problems that will probably be asked in the board exams. Learning about these can also give you more assurance as you take the examinations. Therefore, the important questions Class 9 Mathematics Chapter 15 of CBSE will aid students in achieving high marks in the board exams.
Below are a few benefits for students to practise from our set of Important Questions Class 9 Mathematics Chapter 15:
- Students can rapidly clear any doubts by gaining access to these solutions. For every chapter of CBSE 9th Class Mathematics, specific exercise solutions are also available.
- The thorough explanations aid students in conceptual learning and the improvement of their problem-solving abilities.
- For students’ benefit, the solutions are provided in simple steps.
- In order to help students visualise the problems and solutions, diagrams are also provided.
Revision notes, which include a thorough explanation, key formulas, and time-saving advice, are also offered to students to assist them in getting a quick review of all the topics. For better understanding, students can also get access to some other resources by clicking on the links below.
- CBSE revision notes
- CBSE syllabus
- CBSE sample papers
- CBSE past years’ question papers
- CBSE extra questions and solutions
Q.1 Two coins are tossed, and the outcomes are as follows:
Number of Heads |
2 |
1 |
0 |
Frequency |
200 |
550 |
250 |
What is the probability of getting at most one head
Marks:1
Ans
$\begin{array}{l}\text{Here, the total number of trials = 200 + 550 + 250 = 1000}\\ \text{Let E be the event of getting at most one head i.e., 0 head or}\\ \text{1 head.}\\ \text{Total number of outcomes related to event E = 250 + 550}\\ \text{= 800}\\ \text{Required probability, P(E) =}\frac{800}{1000}=\frac{4}{5}\\ \text{So, the probability of getting at most one head =}\frac{4}{5}.\end{array}$
Q.2 A total 210 batteries are placed in a workshop. The recorded data for their life-time (in days) is given as:
Life-time (in days) | 100 | 150 | 200 | 250 | 300 |
Number of batteries | 25 | 30 | 60 | 55 | 40 |
If a battery is selected at random from the workshop, determine the probability that its life-time is more than 200 days.
Marks:1
Ans
$\begin{array}{l}\text{Here, the total number of batteries is given by:}\\ \mathrm{n}\left(\mathrm{S}\right)=210\\ \text{Now, the number of batteries with life more than}\\ \text{200 days is given by:}\\ \mathrm{n}\left(\mathrm{E}\right)=55+40\\ \text{}=95\\ \mathrm{So},\text{}\mathrm{the}\text{}\mathrm{probability}\text{}\mathrm{that}\text{a randomly selected battery}\mathrm{has}\\ \text{life more than 200 days is given by:}\\ \mathrm{P}\left(\mathrm{E}\right)=\frac{\left(\begin{array}{l}\text{Number of batteries with life}\\ \text{more than 200 days}\end{array}\right)\text{}}{\text{Total number of batteries}}\\ \text{}=\frac{\mathrm{n}\left(\mathrm{E}\right)}{\mathrm{n}\left(\mathrm{S}\right)}\\ \text{}=\frac{95}{210}\\ \text{}=\frac{19}{42}\\ \mathrm{Hence},\text{}\mathrm{the}\text{}\mathrm{probability}\text{}\mathrm{that}\text{a randomly selected battery}\mathrm{has}\\ \text{life more than 200 days is}\frac{19}{42}\text{.}\end{array}$
Q.3 In an activity ‘Save the nature’, 100 schools participated in which each school planted 100 plants. After one month, the number of plants that survived were recorded as in the table given below:
No. of plants survived |
less than 25 |
26-50 |
51-60 |
61-70 |
more than 70 |
total no. of schools |
No. of schools (frequencies) |
15 |
20 |
30 |
30 |
5 |
100 |
If a school is selected at random for inspection, what is the probability that
(i) more than 25 plants survived in the school
(ii) less than 61 plants survived in the school
Marks:4
Ans
(i) Total number of school in which plants were planted = 100
No. of schools in which more than 25 plants survived = 20 + 30 + 30 + 5 = 85
P (more than 25 plants survived in the school) =
$\frac{85}{100}$
= 0.85.
(ii) Total number of schools in which plants were planted = 100
No. of school in which less than 61 plants survived = 15 + 20 + 30
= 65
P (less than 61 plants survived) =
$\frac{65}{100}$
= 0.65.
Q.4 A cycle manufacturing company keeps a record of replacement of tyres and the distance covered by them before replacement. Table given below shows the record of 100 tyres:
Distance (in km) | 0-2000 | 2000-5000 | 5000-7000 | 7000-10000 |
Frequency | 30 | 50 | 10 | 10 |
What is the probability to replace a tyre covering less than 5000 km
Marks:3
Ans
Total number of tyres = 100
Tyres which covered distance less than 5000 = 30 + 50 = 80
Probability to replace a tyre which covered distance less than 5000 km =
$\frac{80}{100}$
= 0.8
Q.5 Two coins are tossed simultaneously 100 times and the frequencies of two outcomes are recorded as follows:
(i) One head = 20 (ii) Two heads = 50
Find the probability of getting no head.
Marks:2
Ans
Number of times in which we get no heads = 100 (20 + 50)
= 30
Therefore,
the required probability =
$\frac{30}{100}$
= 0.3
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