Important questions for Class 9 Maths Chapter 15 Probability are available with solutions here for final exam 2020. These questions are in accordance with the NCERT curriculum prescribed by CBSE board for the 2019-20 academic session. Practising these questions will help students to score good marks in this chapter. To acquire good result they can reach important questions of class 9 Maths chapter-wise here at BYJU’S.
During the exams, students get less time to revise the whole chapter. Therefore, it is necessary for them to revise atleast important concepts before they appear for the exam. Hence, here we are providing the most significant problems and solutions with respect to upcoming examination along with extra questions to give students a better practice.
Also, read:
- Important 2 Marks Questions For Cbse Class 9 Maths
- Important 3 Marks Questions For Cbse Class 9 Maths
- Important 4 Marks Questions For Cbse Class 9 Maths
Important Questions & Solutions For Class 9 Maths Chapter 15
Q.1. Compute the probability of the occurrence of an event if the probability the event not occurring is 0.56.
Solution:
Given,
P(not E) = 0.56
We know,
P(E) + P(not E) = 1
So, P(E) = 1 – P(not E)
P(E) = 1 – 0.56
Or, P(E) = 0.44
Q.2. In a factory of 364 workers, 91 are married. Find the probability of selecting a worker who is not married.
Solution:
Given,
Total workers (i.e. Sample space) = S = 364
Total married workers = P(married) = 91
Now, total workers who are not married = (E) = 364 – 91 = 273
Method 1: So, P(not married) = E/S = 273/364 = 0.75
Method 2: P(married) + P(not married) = 1
Here, P(not married) = 91/364 = 0.25
So, 0.25 + P(not married) = 1 – 0.25 = 0.75
Q. 3. From a deck of cards, 10 cards are picked at random and shuffled. The cards are as follows:
6, 5, 3, 9, 7, 6, 4, 2, 8, 2
Find the probability of picking a card having value more than 5 and find the probability of picking a card with an even number on it.
Solution:
Total number of cards = 10
Total cards having value more than 5 = 5 (6, 9, 7, 6, 8)
Total cards having an even number = 6 (6, 6, 4, 2, 8, 2)
So, the probability of picking a card having value more than 5 = 5/10 = 0.5
And, the probability of picking a card with an even number on it = 6/10 = 0.6
Q.4. From a bag of red and blue balls, the probability of picking a red ball is x/2. Find “x” if the probability of picking a blue ball is ⅔.
Solution:
Here, there are only red and blue balls.
So, the probability of picking a blue ball = probability of not picking a red ball.
P(picking a red ball) + P(not picking a red ball) = 1
x/2 + ⅔ = 1
=> 3x + 4 = 6
=> 3x = 2
Or, x = ⅔
Q.5. Two coins are tossed simultaneously for 360 times. The number of times ‘2 Tails’ appeared was three times ‘No Tail’ appeared and the number of times ‘1 tail’ appeared is double the number of times ‘No Tail’ appeared. What is the probability of getting ‘Two tails’.
Solution:
Given,
Total number of outcomes = Sample space = 360
Now, assume that the number of times ‘No Tail’ appeared to be “x”
So, the number of times ‘2 Tails’ appeared = 3x (from the question)
Also, the number of times ‘1 Tail’ appeared =2x (from the question)
As the total outcomes = 360,
x + 2x + 3x = 360
=> 6x = 360
Or, x = 60
∴ P(getting two tails) = (3 × 60)/360 = ½
Q.6: 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 a family, chosen at random, having
(i) 2 girls (ii) 1 girl (iii) No girl
Also check whether the sum of these probabilities is 1.
Solution:
Total numbers of families = 1500
(i) Numbers of families having 2 girls = 475
Probability = Numbers of families having 2 girls/Total numbers of families
P = 475/1500
P = 19/60
(ii) Numbers of families having 1 girls = 814
Probability = Numbers of families having 1 girls/Total numbers of families
P = 814/1500
P = 407/750
(iii)Numbers of families having 2 girls = 211
Probability = Numbers of families having 0 girls/Total numbers of families
= 211/1500
Sum of the probability = (19/60)+(407/750)+(211/1500)
= (475+814+211)1500 = 1500/1500 = 1
Yes, the sum of these probabilities is 1.
Q.7: A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as 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: Let E_{i} denote the event of getting the outcome i, where i = 1, 2, 3, 4, 5, 6.
Then
Probability of the outcome 1 = P(E_{1})
= Frequency of 1/Total number of times the die is thrown
= 179/1000
= 0.179
Similarly,
P(E_{2} ) = 150/1000 = 0.15,
P(E_{3}) = 157/1000 = 0.157,
P(E_{4}) = 149/1000 = 0.149,
P(E_{5}) = 175/1000 = 0.175
and P(E_{6}) = 190/1000 = 0.19
You can check: P(E_{1}) + P(E_{2}) + P(E_{3}) + P(E_{4}) + P(E_{5}) + P(E_{6}) = 1
Q.8: An organisation selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed 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 a family is chosen. Find the probability that the family chosen is
- earning ₹10000 – 13000 per month and owning exactly 2 vehicles.
- earning ₹16000 or more per month and owning exactly 1 vehicle.
- earning less than ₹7000 per month and does not own any vehicle.
- earning ₹13000 – 16000 per month and owning more than 2 vehicles.
- owning not more than 1 vehicle.
Solution:
Total number of families = 2400
(i) Numbers of families earning ₹10000 –13000 per month and owning exactly 2 vehicles = 29
Therefore, the probability that the family earning between ₹10000 – 13000 per month and owning exactly 2 vehicle = 29/2400
(ii) Number of families earning ₹16000 or more per month and owning exactly 1 vehicle = 579
Therefore, the probability that the family earning between ₹16000 or more per month and owning exactly 1 vehicle = 579/2400
(iii) Number of families earning less than ₹7000 per month and does not own any vehicle = 10
Therefore, the probability that the family earning less than ₹7000 per month and does not own any vehicle = 10/2400 = 1/240
(iv) Number of families earning ₹13000-16000 per month and owning more than 2 vehicles = 25
Therefore, the probability that the family earning between ₹13000 – 16000 per month and owning more than 2 vehicles = 25/2400 = 1/96
(v) Number of families owning not more than 1 vehicle = 10+160+0+305+1+535+2+469+1+579 = 2062
Therefore, the probability that the family owns not more than 1 vehicle = 2062/2400 = 1031/1200
Q.9: Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights of flour (in kg): 4.97 5.05 5.08 5.03 5.00 5.06 5.08 4.98 5.04 5.07 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg of flour.
Solution: Total number of bags present = 11
Number of bags containing more than 5 kg of flour = 7
Therefore,
the probability that any of the bags chosen at random contains more than 5 kg of flour = 7/11
Q.10: The distance (in km) of 40 engineers from their residence to their place of work were found as follows: 5 3 10 20 25 11 13 7 12 31 19 10 12 17 18 11 32 17 16 2 7 9 7 8 3 5 12 15 18 3 12 14 2 9 6 15 15 7 6 12.
What is the empirical probability that an engineer lives:
(i) less than 7 km from her place of work?
(ii) more than or equal to 7 km from her place of work?
(iii) within km from her place of work?
Solution:
The distance (in km) of 40 engineers from their residence to their place of work was found as follows:
5 3 10 20 25 11 13 7 12 31 19 10 12 17 18 11 3 2 17 16 2 7 9 7 8 3 5 12 15 18 3 12 14 2 9 6 15 15 7 6 12
Total numbers of engineers = 40
(i) Number of engineers living less than 7 km from their place of work = 9
the probability that an engineer lives less than 7 km from her place of work =9/40
(ii) Number of engineers living more than or equal to 7 km from their place of work
= 40 – 9 = 31
The probability that an engineer lives more than or equal to 7 km from her place of work
= 31/40
(iii) Number of engineers living within 1/2 km from their place of work = 0
The probability that an engineer lives within 1/2km from her place of work = 0/40 = 0
Q.11: Refer to the table below:
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 |
(i) Find the probability that a student obtained less than 20% in the mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.
Solution: Total number of students = 90
(i) Number of students who obtained less than 20% in the mathematics test = 7
The probability that a student obtained less than 20% in the mathematics test = 7/90
(ii) Number of students who obtained marks 60 or above = 15+8 = 23 the probability that a student obtained marks 60 or above =23/90
Extra Questions For Class 9 Maths Chapter 15 (Probability)
- Can the experimental probability of an event be a negative number? If not, why?
- Can the experimental probability of an event be greater than 1? Justify your answer.
- 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.
- A company selected 4000 households at random and surveyed them to find out a relationship between the level and the number of television sets in a home. The information so obtained is listed in the following table.
Monthly income
(in Rs.) |
Number of television/household | |||
0 | 1 | 2 | Above 2 | |
<10000 | 20 | 80 | 10 | 0 |
10000-14999 | 10 | 240 | 60 | 0 |
15000-19999 | 0 | 380 | 120 | 30 |
20000-24999 | 0 | 520 | 370 | 80 |
25000 and above | 0 | 1100 | 760 | 220 |
Find the probability:
(i) of a household earning Rs.10000-14999 per year and having exactly one television.
(ii) of a household earning Rs.25000 and more per year and owning two televisions
(iii) of a household having no televisions.