- 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.