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:
Data given in the question:
Total number of balls batswoman plays = 30
Numbers of boundary hit by batswoman = 6
To find number of time batswoman didn’t hit boundary , we will subtract
â‡’(Total number of balls batswoman plays) – (Numbers of boundary hit by batswoman)
â‡’ 30 â€“ 6 = 24
Probability of that she didn’t hit a boundary = 24/30 = 4/5
Question 2. 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
Solution:
According to question
Total numbers of families given in the question 1500
(i) Numbers of families having 2 girls = 475
Probability of chosen 2 girls = Numbers of families having 2 girls / Total numbers of families
= 475/1500 = 19/60
Probability of chosen 2 girls is 19/60(ii) Numbers of families having 1 girls = 814
Probability of chosen 1 girl = Numbers of families having 1 girl / Total numbers of families
= 814/1500 = 407/750
Probability of chosen 1 girl is 407/750(iii) Numbers of families having 2 girls = 211
Probability of chosen 0 girl = Numbers of families having 0 girls/Total numbers of families
= 211/1500Sum of the probability = (19/60)+(407/750)+(211/1500)
= (475+814+211)/1500
= 1500/1500 = 1Yes, the sum of these probabilities is 1.
Question 3. Refer to Example 5, Section 14.4, Chapter 14. Find the probability that a student of the class was born in August.
Solution:
According to questions:
Total number of students in the class in the given question = 40
Numbers of students born in August = 6
The probability that a student of the class was born in August = (Total numbers of students in the class) /
(Numbers of students born in August)
= 6/40 = 3/20
Question 4. Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:
Outcome | 3heads | 2heads | 1 head | No heads |
Frequency | 23 | 72 | 77 | 28 |
If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Solution:
Number of times 2 heads come up (in the given question) = 72
Total number of times the coins were tossed = 200
The probability of 2 heads coming up = (Number of times 2 heads come up) / (Total number of times the coins were tossed)
= 72/200 = 9/25
Question 5. 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 â‚¹) |
0 | 1 | 2 | Above 2 |
Less than 7000 | 10 | 160 | 25 | 0 |
7000-10000 | 0 | 305 | 27 | 2 |
7000-10000 | 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 â‚¹10000 â€“ 13000 per month and owning exactly 2 vehicles.
(ii) earning â‚¹16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than â‚¹7000 per month and does not own any vehicle.
(iv) earning â‚¹13000 â€“ 16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Solution:
Total number of families = 2400 (According to question)
(i) Numbers of families earning â‚¹10000 â€“13000 per month and owning exactly 2 vehicles = 29
The probability that the family chosen is earning â‚¹10000 â€“ 13000 per month and owning exactly 2 vehicles =
(Numbers of families earning â‚¹10000 â€“13000 per month and owning exactly 2 vehicles) / (Total number of families)
= 29/2400(ii) Number of families earning â‚¹16000 or more per month and owning exactly 1 vehicle = 579
The probability that the family chosen is earning â‚¹16000 or more per month and owning exactly 1 vehicle =
(Number of families earning â‚¹16000 or more per month and owning exactly 1 vehicle) / (Total number of families)
=579/2400(iii) Number of families earning less than â‚¹7000 per month and does not own any vehicle = 10
The probability that the family chosen is earning less than â‚¹7000 per month and does not own any vehicle =
(Number of families earning less than â‚¹7000 per month and does not own any vehicle)/(Total number of families)
= 10/2400 = 1/240(iv) Number of families earning â‚¹13000-16000 per month and owning more than 2 vehicles = 25
The probability that the family chosen is earning â‚¹13000 â€“ 16000 per month and owning more than 2 vehicles =
(Number of families earning â‚¹13000-16000 per month and owning more than 2 vehicles ) / (Total number of families)
= 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
The probability that the family chosen owns not more than 1 vehicle = 2062/2400 = 1031/1200
Question 6. Refer to Table 14.7, Chapter 14.
(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
(given in question)
(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 =
( Number of students who obtained less than 20% in the mathematics test)/(Total number of students)
= 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 =
(Number of students who obtained marks 60 or above ) / (Total number of students)
= 23/90
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
Opinion | Number of students |
like | 135 |
dislike | 65 |
Find the probability that a student chosen at random
(i) likes statistics, (ii) does not like it.
Solution:
Total number of students = 135+65 = 200 (According to question)
(i) Number of students who like statistics = 135
The probability that a student likes statistics = (Number of students who like statistics) / (Total number of students)
= 135/200 = 27/40(ii) Number of students who do not like statistics = 65
The probability that a student does not like statistics =
(Number of students who do not like statistics) / (Total number of students)
= 65/200 = 13/40
Question 8. Refer to Q.2, Exercise 14.2. 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 Â½ km from her place of work?
Solution:
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 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
(According to question)(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 =
(Number of engineers living less than 7 km from their place of work) / (Total numbers of engineers )
= 9/40(ii) Number of engineers living more than or equal to 7 km from their place of work = 40-9 = 31
Probability that an engineer lives more than or equal to 7 km from her place of work =
(Number of engineers living more than or equal to 7 km from their place of work ) / (Total numbers of engineers)
= 31/40(iii) Number of engineers living within Â½ km from their place of work = 0
The probability that an engineer lives within Â½ km from her place of work =
(Number of engineers living within Â½ km from their place of work) / (Total numbers of engineers)
=0/40 = 0
Question 9. Activity: Note the frequency of two-wheelers, three-wheelers and four-wheelers going past during a time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.
Solution:
The question is an activity to be performed by the students.
Question 10. Activity: Ask all the students in your class to write a 3-digit number. Choose any student from the room at random. What is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3, if the sum of its digits is divisible by 3.
Solution:
The question is an activity to be performed by the students.
Question 11. 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:
Data given in the question
Total number of bags present = 11
Number of bags containing more than 5 kg of flour = 7
The probability that any of the bags chosen at random contains more than 5 kg of flour =
(Number of bags containing more than 5 kg of flour) / (Total number of bags present)
= 7/11
Question 12. In Q.5, Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12-0.16 on any of these days.
The data obtained for 30 days is as follows:
0.03 0.08 0.08 0.09 0.04 0.17 0.16 0.05 0.02 0.06 0.18 0.20 0.11 0.08 0.12 0.13 0.22 0.07 0.08 0.01 0.10 0.06 0.09 0.18 0.11 0.07 0.05 0.07 0.01 0.04
Solution:
Total number of days in which the data was recorded = 30 days (According to the question)
Numbers of days in which sulphur dioxide was present in between the interval 0.12-0.16 = 2
The probability of the concentration of sulphur dioxide in the interval 0.12-0.16 on any of these days =
(Numbers of days in which sulphur dioxide was present in between the interval 0.12-0.16) /
(Total number of days in which the data was recorded )
= 2/30 = 1/15
Question 13. In Q.1, Exercise 14.2, 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.
Solution:
Total numbers of students = 30 (according to questions)
Number of students having blood group AB = 3
The probability that a student of this class, selected at random, has blood groupAB = (Number of students having blood group AB) / (Total numbers of students)
= 3/30 = 1/10