- In a cricket match, a bats woman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Solution:
Total number of blls played by the batswoman = 30
Boundaries hit = 6
No. of balls in which she did not hit any numbers = 30 – 6 = 24
⸫P(she did not hit a boundary) = no. of balls in which she did not hit any boundary/total number of balls played = 24/30 = 4/5
- 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:
(i) P(a family having 2 girls) = no. of families having 2 girls/total no. of families = 475/1500 = 19/60
(ii) P ( a family having 1 girl) = no. of families having 2 girls/total no. of families = 814/1500 = 407/750
(ii) P ( a family having no girl) = no. of families having 2 girls/total no. of families = 211/1500
Sum of the probabilities in all three cases = 19/60 + 407/750 +211/1500 = 475+814+211/1500 = 1500/1500 = 1
- Refer to Example 5, Section 14.4, Chapter 14. Find the probability that a student of the class was born in August.
Solution:
Total number of students considered = 40
No. of students born in August = 6
⸫ P ( a student born in August) = no. of students born in August/total no. of students considered = 6/40 = 3/20
- Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:
| 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 tosses = 200
No. of times 2 heads occur = 72
⸫ P ( two heads coming up) = no. of times two heads occur/total no. of tosses = 72/200 = 9/25
- 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 Rs.) | Vehicles per family | |||
| 0 | 1 | 2 | Above 2 | |
| Less than 7000 | 10 | 160 | 5 | 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 Rs 10000 – 13000 per month and owning exactly 2 vehicles.
(ii) earning Rs 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than Rs 7000 per month and does not own any vehicle.
(iv) earning Rs 13000 – 16000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Solution:
Total no of families considered = 2400
(i) P ( a family earning Rs. 10000 – 13000 per month and owning exactly 2 vehicles)
= no. of families earning Rs. 10000 – 13000 per month and owning exactly 2 vehicles/total no. of families
= 29/2400
(ii) P ( a family earning Rs. 16000 or more per month and owning exactly 1 vehicle)
= no. of families earning Rs. 16000 or more per month and owning exactly 1 vehicle/total no. of families
= 579/2400 = 193/800
(iii) P ( a family earning less than Rs. 7000 per month and does not own any vehicle)
= no. of families earning Rs. 7000 per month and does not own any vehicle/total no. of families
= 10/2400 = 1/240
(iv) P ( a family earning Rs. 13000 – 16000 per month and owning more than 2 vehicles)
= no. of families earning Rs. 13000 – 16000 per month and owning more than 2 vehicles/total no. of families = 25/2400 = 1/96
(v)P(a family earning 0 vehicle or 1 vehicle) = P( a family not owning more than 1 vehicle)
= 10+0+1+2+1+160+305+535+469+579/2400 = 2062/2400 = 1031/1200
- 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.
| 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 |
Solution:
(i) Total number of students = 90
P(a student obtained less than 20%) = no. of students who obtained less than 20%/total number of students = 7/90
(ii) P(a student obtained 60 marks or above) = no. of students who obtained 60 makrs or more /total numner of students = 15+8/90 = 23/90
- 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:
(i) P(a student like statistics) = no. of students who like statistics/total number of students = 135/200 = 27/40
(i) P(a student does not like statistics) = no. of students who does not like statistics/total number of students = 65/200 = 13/40
- Refer to Q.2, Exercise 14.2.
5 3 10 2 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 1/2 km from her place of work?
Solution:
Total no. of engineers = 40
Let us arrange the data in ascending order as follows:
2, 2, 3, 3, 3, 5, 5, 6, 6, 7, 7, 7, 7, 8, 9, 9, 10, 10, 11, 11, 12, 12, 12, 12, 12, 13, 14, 15, 15, 15, 16, 17, 17, 18, 18, 19, 20, 25, 31, 32
(i)P(an engineer lives less than 7km from her place to work) = no. of engineers live less than 7km from her place to work/total no. of engineers = 9/40
(ii)P(an engineer lives more than or 7km from her place to work) = no. of engineers live more than or 7km from her place to work/total no. of engineers = 31/40
(iii)P(an engineer lives within 1/2 km from her place to work) = no. of engineers live within 1/2 km from her place to work/total no. of engineers = 0/40 = 0
1 thought on “Probability – Exercise 15.1 – Class IX”
Comments are closed.