Question-1
In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays.
Find the probability that she did not hit a boundary.
Solution:
Let E be the event of hitting the boundary.
Then, P(E) =
= = = 0.2
âˆ´ Probability of not hitting the boundary = 1 â€“ Probability of hitting the boundary
= 1 â€“ P(E)
= 1 â€“ 0.2
= 0.8
Find the probability that she did not hit a boundary.
Solution:
Let E be the event of hitting the boundary.
Then, P(E) =
= = = 0.2
âˆ´ Probability of not hitting the boundary = 1 â€“ Probability of hitting the boundary
= 1 â€“ P(E)
= 1 â€“ 0.2
= 0.8
Question-2
1500 families with 2 children were selected randomly, and the following data were recorded:
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 number of families = 475 + 814 + 211 = 1500
(i) Probability of a family, chosen at random, having 2 girls = =
(ii) Probability of a family, chosen at random, having 1 girl = =
(iii) Probability of a family, chosen at random, having no girl =
Sum of these probabilities =
=
=
= 1
Hence the sum is checked.
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 number of families = 475 + 814 + 211 = 1500
(i) Probability of a family, chosen at random, having 2 girls = =
(ii) Probability of a family, chosen at random, having 1 girl = =
(iii) Probability of a family, chosen at random, having no girl =
Sum of these probabilities =
=
=
= 1
Hence the sum is checked.
Question-3
Find the probability that a student of the class was born in August.
Solution:
Total number of students born in the year = 3 + 4 + 2 + 2 + 5 + 1 + 2 + 6 + 3 + 4 + 4 + 4
= 40
Solution:
Total number of students born in the year = 3 + 4 + 2 + 2 + 5 + 1 + 2 + 6 + 3 + 4 + 4 + 4
= 40
Number of students born in August = 6
âˆ´ Probability that a student of the class was born in August = =
Question-4
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:
If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Outcome |
3 heads |
2 heads |
1 head |
No head |
Frequency |
23 |
72 |
77 |
28 |
If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Solution:
Total number of times the three coins are tossed = 200
Number of times when 2 heads appear = 72
âˆ´ Probability of 2 heads coming up = =
Total number of times the three coins are tossed = 200
Number of times when 2 heads appear = 72
âˆ´ Probability of 2 heads coming up = =
Question-5
An organization 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:
Suppose a family is chosen. Find the probability that the family chosen is
(i) earning `10000 â€“ 13000 per month and owning exactly 2 vehicles.
(ii) Earning ` 16000 or more per month and owning exactly 1 vehicle.
(iii) Earning less than ` 7000 per month and does not own any vehicle.
(iv) Earning ` 13000 â€“ 16000 per month and owing more than 2 vehicles.
(v) Owning not more than 1 vehicle.
Solution:
Total number of families selected = 2400
(i) Number of families earning ` 10000 - 13000 per month and owning exactly
2 vehicles = 29
âˆ´ Probability that the family chosen is earning ` 10000-13000 per month and owning exactly 2 vehicles =
(ii) Number of families earning Rs.16000 or more per month and owning exactly
1 vehicle = 579 âˆ´ Probability that the family chosen is earning ` 16000 or more per month and owning exactly 1 vehicle = =
(iii) Number of families earning less than ` 7000 per month and does not own any vehicle = 10
âˆ´ Probability that the family chosen is earning less than ` 7000 per month and does not own any vehicle = = .
(iv) Number of families earning ` 13000 - 16000 per month and owning more than 2 vehicles = 25.
âˆ´ Probability that the family chosen is earning ` 13000 -16000 per month and owning more than 2 vehicles = =.
(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)
= 14 + 2048 = 2062 âˆ´ Probability that the family chosen owns not more than 1 vehicle = = .
Vehicles per family |
||||
0 | 1 | 2 | more than 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
(i) earning `10000 â€“ 13000 per month and owning exactly 2 vehicles.
(ii) Earning ` 16000 or more per month and owning exactly 1 vehicle.
(iii) Earning less than ` 7000 per month and does not own any vehicle.
(iv) Earning ` 13000 â€“ 16000 per month and owing more than 2 vehicles.
(v) Owning not more than 1 vehicle.
Solution:
Total number of families selected = 2400
(i) Number of families earning ` 10000 - 13000 per month and owning exactly
2 vehicles = 29
âˆ´ Probability that the family chosen is earning ` 10000-13000 per month and owning exactly 2 vehicles =
(ii) Number of families earning Rs.16000 or more per month and owning exactly
1 vehicle = 579 âˆ´ Probability that the family chosen is earning ` 16000 or more per month and owning exactly 1 vehicle = =
(iii) Number of families earning less than ` 7000 per month and does not own any vehicle = 10
âˆ´ Probability that the family chosen is earning less than ` 7000 per month and does not own any vehicle = = .
(iv) Number of families earning ` 13000 - 16000 per month and owning more than 2 vehicles = 25.
âˆ´ Probability that the family chosen is earning ` 13000 -16000 per month and owning more than 2 vehicles = =.
(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)
= 14 + 2048 = 2062 âˆ´ Probability that the family chosen owns not more than 1 vehicle = = .
Question-6
In a mathematics test, the marks 90 students of class IX are listed in the table below:
(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.
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 obtaining less than 20% in the mathematics test = 7 âˆ´ Probability that a student obtained less than 20% in mathematics test =
(ii) Number of students obtaining marks 60 or above = 15 + 8 = 23 âˆ´ Probability that a student obtained marks 60 or above =
Total number of students = 90
(i) Number of students obtaining less than 20% in the mathematics test = 7 âˆ´ Probability that a student obtained less than 20% in mathematics test =
(ii) Number of students obtaining marks 60 or above = 15 + 8 = 23 âˆ´ Probability that a student obtained marks 60 or above =
Question-7
To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table.
Find the probability that a student chosen at random
(i) like statistics, (ii) does not like it.
Opinion |
Number of students |
Like |
135 |
Dislike |
65 |
Find the probability that a student chosen at random
(i) like statistics, (ii) does not like it.
Solution:
Total number of students = 200
(i) Number of students who like statistics = 135
âˆ´ Probability that a student chosen at random likes statistics = =
(ii) Number of students who do not like statistics = 65
Probability that a student chosen at random does not like it = =
Aliter:
Probability that a student chosen at random does like statistics
= 1 - probability that a student chosen at random likes statistics
= 1 - =
Total number of students = 200
(i) Number of students who like statistics = 135
âˆ´ Probability that a student chosen at random likes statistics = =
(ii) Number of students who do not like statistics = 65
Probability that a student chosen at random does not like it = =
Aliter:
Probability that a student chosen at random does like statistics
= 1 - probability that a student chosen at random likes statistics
= 1 - =
Question-8
The distance (in km) of 40 female 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:
a. Less than 7 km from her place of work?
b. More than or equal to 7 km from her place of work?
c. Within km from her place of work?
Solution:
Total number of female engineers = 40
(a) Number of female engineers whose distance (in km) from their residence to their place of work is less than 7 km = 9. âˆ´ Probability that an engineer lives less than 7 km from her place of work =
(b) Number of female engineers whose distance (in km) from their residence to their place of work is more than or equal to 7 km = 31 âˆ´ Probability that an engineer lives more than or equal to 7 km from her place of residence =
Aliter
Probability that an engineer lives more than or equal to 7 km from her place of residence
= 1 - probability that an engineer lives less than 7 km from her place of work
= 1 - =
(c) Number of female engineers whose distance (in km) from their residence to their place of work is within km = 0. âˆ´ Probability that an engineer lives within km from her place of work = = 0.
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:
a. Less than 7 km from her place of work?
b. More than or equal to 7 km from her place of work?
c. Within km from her place of work?
Solution:
Total number of female engineers = 40
(a) Number of female engineers whose distance (in km) from their residence to their place of work is less than 7 km = 9. âˆ´ Probability that an engineer lives less than 7 km from her place of work =
(b) Number of female engineers whose distance (in km) from their residence to their place of work is more than or equal to 7 km = 31 âˆ´ Probability that an engineer lives more than or equal to 7 km from her place of residence =
Aliter
Probability that an engineer lives more than or equal to 7 km from her place of residence
= 1 - probability that an engineer lives less than 7 km from her place of work
= 1 - =
(c) Number of female engineers whose distance (in km) from their residence to their place of work is within km = 0. âˆ´ Probability that an engineer lives within km from her place of work = = 0.
Question-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 of wheat flour = 11,
Number of bags of wheat flour containing more than 5 kg of flour = 7.
âˆ´ Probability that any of the bags, chosen at random,
contains more than 5 kg of flour = .
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 of wheat flour = 11,
Number of bags of wheat flour containing more than 5 kg of flour = 7.
âˆ´ Probability that any of the bags, chosen at random,
contains more than 5 kg of flour = .
Question-10
You were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.
"The blood groups of 30 students of Class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
Represent this data in the form of a frequency distribution table.
Use this table to determine the probability that a student of this class, selected random, has blood group AB.
Solution:
Total number of students = 30
Number of students having blood groups AB = 3
âˆ´ Probability that a student of this class, selected at random, has blood group AB = = = 0.1
"The blood groups of 30 students of Class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.
Represent this data in the form of a frequency distribution table.
Use this table to determine the probability that a student of this class, selected random, has blood group AB.
Solution:
Total number of students = 30
Number of students having blood groups AB = 3
âˆ´ Probability that a student of this class, selected at random, has blood group AB = = = 0.1