In the previous section we saw some problems on experimental probability. In this section we will see a few more solved examples.
Solved example 28.8
An insurance company selected 2000 drivers at random from a particular city to find a relationship between age and accidents. The data obtained is given in the table below:
Find the following:
(i) A driver is chosen at random from the city. What is the probability that, he is 18-29 years of age and has exactly 3 accidents in one year
(ii) A driver is chosen at random from the city. What is the probability that, he is 30-50 years of age and has one or more accidents in year
(iii) A driver is chosen at random from the city. What is the probability that, he has no accidents in one year
Solution:
■ The given table was prepared by adopting the following procedure:
(i) Pick a driver at random. Find two points about him:
• His age
• No.of accidents in the past one year
(ii) Based on the two answers, make a tally mark in the appropriate place in the table.
Example:
• If the age of a driver is 32, the tally mark will fall somewhere in the row marked as '30-50'
♦ If he has made 2 accidents in the past year, the tally mark will fall in the column marked as '2'
When a tally mark is placed, one trial is complete. Note that 440+160+110+ . . . + 15+9 = 2000
■ If we calculate the probabilities based on the table, those probabilities will be applicable for all the drivers in that particular city.
1. Corresponding to 18-29 age and exact 3 accidents, we have 61
• So probability = Number of trials in which the event happened⁄Total number of trials = 61⁄2000 = 0.0305
2. Corresponding to 30-50 age, we have all values in that row. But all those values are not eligible.
• The number of drivers in this age group, who made one or more accidents is 125+60+22+18 = 225
• So probability = Number of trials in which the event happened⁄Total number of trials = 225⁄2000 = 0.1125
3. In this case, age is not considered. So the no. of trials which give the favourable outcome are 440+505+360 = 1305
• So probability = Number of trials in which the event happened⁄Total number of trials = 1305⁄2000 = 0.653
Solved example 28.9
Fifty seeds were selected at random from each of 5 bags of seeds, and were kept under standardised conditions favourable to germination. After 20 days, the number of seeds which had germinated in each collection were counted and recorded as follows:
Find the following:
(i) Probability that more than 40 seeds in a bag will germinate
(ii) Probability that 49 seeds in a bag will germinate
(iii) Probability that more than 35 seeds in a bag will germinate?
Solution:
■ There are 5 bags of seeds. From each bag, 50 seeds were selected. This selection of 50 seeds should be done carefully. Because those 50 seeds should give an overall representation of the bag from which they are selected. They should not be taken from the top of the bag. Neither should they be taken from the bottom or middle. The procedure to ensure 'randomness' should be strictly followed for each of the 5 bags
■ Further more, these 5 bags are selected at random from a stack of large number of bags. 'Randomness' should be ensured while picking these 5 bags also.
■ When we calculate the probabilities, they will be applicable to a bag taken from the whole stack
■ If instead of taking 5 bags, 10 or 15 bags are tested (that is., increasing number of trials), the accuracy of the results will increase
1. No. of bags which give more than 40 seeds that would germinate = 3
• So probability = Number of trials in which the event happened⁄Total number of trials = 3⁄5 = 0.6
2. There are no bags which give 49 seeds that would germinate
• So probability = Number of trials in which the event happened⁄Total number of trials = 0⁄5 = 0
3. No. of bags which give more than 35 seeds that would germinate = 5
• So probability = Number of trials in which the event happened⁄Total number of trials = 5⁄5 = 1
Solved example 28.10
1500 families with 2 children were selected randomly, and the following data were recorded:
Compute the following:
(i) The probability that, a family chosen at random have two girls
(ii) The probability that, a family chosen at random have one girl
(iii) The probability that, a family chosen at random have no girl
(iv) Check whether the sum of all the above probabilities is 1
Solution:
■ 1500 families were selected at random from a region. There are thousands of more families in the region. When we calculate the probabilities based on the table, those probabilities, will be applicable to the whole region
1. 475 families have two girls
• So probability = Number of trials in which the event happened⁄Total number of trials = 475⁄1500 = 0.3167
2. 814 families have one girl
• So probability = Number of trials in which the event happened⁄Total number of trials = 814⁄1500 = 0.54267
3. 211 families have no girl
• So probability = Number of trials in which the event happened⁄Total number of trials = 211⁄1500 = 0.14067
4. Sum of the above probabilities = 0.3167 +0.54267 +0.14067 = 1.00004
Solved example 28.11
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
Solution:
No. of times 2 heads come up = 72
• So probability = Number of trials in which the event happened⁄Total number of trials = 72⁄200 = 9⁄25 = 0.36
Solved example 28.12
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 obtained is given in a tabular form below:
Compute the following:
(i). A family chosen at random is earning 10000 - 13000 per month and owning exactly 2 vehicles
(ii). A family chosen at random is earning 16000 or more per month and owning exactly 1 vehicle
(iii). A family chosen at random is earning less than 7000 per month and does not own any vehicle
(iv). A family chosen at random is earning 13000 - 16000 per month and owning more than 2 vehicles
(v). A family chosen at random owns only 1 vehicle
Solution:
■ 2400 families were selected at random from a region. There are thousands of more families in the region. When we calculate the probabilities based on the table, those probabilities will be applicable to the whole region
1. Number of families in this income range owning exactly 2 vehicles = 29
• So probability = Number of trials in which the event happened⁄Total number of trials = 29⁄2400
2. Number of families in this income range owning exactly 1 vehicle = 579
• So probability = Number of trials in which the event happened⁄Total number of trials = 579⁄2400
3. Number of families in this income range owning no vehicles = 10
• So probability = Number of trials in which the event happened⁄Total number of trials = 10⁄2400
4. Number of families in this income range owning more than 2 vehicles = 25
• So probability = Number of trials in which the event happened⁄Total number of trials = 25⁄2400
5. In this case, income range is not required. Number of families who own exactly one vehicle is:
160 +305 +535 +469 + 579 = 2048
• So probability = Number of trials in which the event happened⁄Total number of trials = 2048⁄2400
Solved example 28.13
To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data obtained is tabulated below:
Compute the following:
(i) Probability that, a student chosen at random likes statistics
(ii) Probability that, a student chosen at random dislikes statistics
Solution:
1. Number of students who like statistics = 135
• So probability = Number of trials in which the event happened⁄Total number of trials = 135⁄200 = 27⁄40
2. Number of students who dislike statistics = 66
• So probability = Number of trials in which the event happened⁄Total number of trials = 66⁄200 = 33⁄100
Solved example 28.14
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights 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:
No. of bags which contain more than 5 kg of flour = 7
• So probability = Number of trials in which the event happened⁄Total number of trials = 7⁄11
An insurance company selected 2000 drivers at random from a particular city to find a relationship between age and accidents. The data obtained is given in the table below:
| Age of drivers (years) |
Accidents in one year | ||||
|---|---|---|---|---|---|
| 0 | 1 | 2 | 3 | over 3 | |
| 18-29 | 440 | 160 | 110 | 61 | 35 |
| 30-50 | 505 | 125 | 60 | 22 | 18 |
| Above 50 | 360 | 45 | 35 | 15 | 9 |
(i) A driver is chosen at random from the city. What is the probability that, he is 18-29 years of age and has exactly 3 accidents in one year
(ii) A driver is chosen at random from the city. What is the probability that, he is 30-50 years of age and has one or more accidents in year
(iii) A driver is chosen at random from the city. What is the probability that, he has no accidents in one year
Solution:
■ The given table was prepared by adopting the following procedure:
(i) Pick a driver at random. Find two points about him:
• His age
• No.of accidents in the past one year
(ii) Based on the two answers, make a tally mark in the appropriate place in the table.
Example:
• If the age of a driver is 32, the tally mark will fall somewhere in the row marked as '30-50'
♦ If he has made 2 accidents in the past year, the tally mark will fall in the column marked as '2'
When a tally mark is placed, one trial is complete. Note that 440+160+110+ . . . + 15+9 = 2000
■ If we calculate the probabilities based on the table, those probabilities will be applicable for all the drivers in that particular city.
1. Corresponding to 18-29 age and exact 3 accidents, we have 61
• So probability = Number of trials in which the event happened⁄Total number of trials = 61⁄2000 = 0.0305
2. Corresponding to 30-50 age, we have all values in that row. But all those values are not eligible.
• The number of drivers in this age group, who made one or more accidents is 125+60+22+18 = 225
• So probability = Number of trials in which the event happened⁄Total number of trials = 225⁄2000 = 0.1125
3. In this case, age is not considered. So the no. of trials which give the favourable outcome are 440+505+360 = 1305
• So probability = Number of trials in which the event happened⁄Total number of trials = 1305⁄2000 = 0.653
Solved example 28.9
Fifty seeds were selected at random from each of 5 bags of seeds, and were kept under standardised conditions favourable to germination. After 20 days, the number of seeds which had germinated in each collection were counted and recorded as follows:
| Bag | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Number of seeds germinated |
40 | 48 | 42 | 39 | 41 |
(i) Probability that more than 40 seeds in a bag will germinate
(ii) Probability that 49 seeds in a bag will germinate
(iii) Probability that more than 35 seeds in a bag will germinate?
Solution:
■ There are 5 bags of seeds. From each bag, 50 seeds were selected. This selection of 50 seeds should be done carefully. Because those 50 seeds should give an overall representation of the bag from which they are selected. They should not be taken from the top of the bag. Neither should they be taken from the bottom or middle. The procedure to ensure 'randomness' should be strictly followed for each of the 5 bags
■ Further more, these 5 bags are selected at random from a stack of large number of bags. 'Randomness' should be ensured while picking these 5 bags also.
■ When we calculate the probabilities, they will be applicable to a bag taken from the whole stack
■ If instead of taking 5 bags, 10 or 15 bags are tested (that is., increasing number of trials), the accuracy of the results will increase
1. No. of bags which give more than 40 seeds that would germinate = 3
• So probability = Number of trials in which the event happened⁄Total number of trials = 3⁄5 = 0.6
2. There are no bags which give 49 seeds that would germinate
• So probability = Number of trials in which the event happened⁄Total number of trials = 0⁄5 = 0
3. No. of bags which give more than 35 seeds that would germinate = 5
• So probability = Number of trials in which the event happened⁄Total number of trials = 5⁄5 = 1
Solved example 28.10
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 |
(i) The probability that, a family chosen at random have two girls
(ii) The probability that, a family chosen at random have one girl
(iii) The probability that, a family chosen at random have no girl
(iv) Check whether the sum of all the above probabilities is 1
Solution:
■ 1500 families were selected at random from a region. There are thousands of more families in the region. When we calculate the probabilities based on the table, those probabilities, will be applicable to the whole region
1. 475 families have two girls
• So probability = Number of trials in which the event happened⁄Total number of trials = 475⁄1500 = 0.3167
2. 814 families have one girl
• So probability = Number of trials in which the event happened⁄Total number of trials = 814⁄1500 = 0.54267
3. 211 families have no girl
• So probability = Number of trials in which the event happened⁄Total number of trials = 211⁄1500 = 0.14067
4. Sum of the above probabilities = 0.3167 +0.54267 +0.14067 = 1.00004
Solved example 28.11
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 |
Solution:
No. of times 2 heads come up = 72
• So probability = Number of trials in which the event happened⁄Total number of trials = 72⁄200 = 9⁄25 = 0.36
Solved example 28.12
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 obtained is given in a tabular form below:
| Monthly income (Rs) |
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 |
(i). A family chosen at random is earning 10000 - 13000 per month and owning exactly 2 vehicles
(ii). A family chosen at random is earning 16000 or more per month and owning exactly 1 vehicle
(iii). A family chosen at random is earning less than 7000 per month and does not own any vehicle
(iv). A family chosen at random is earning 13000 - 16000 per month and owning more than 2 vehicles
(v). A family chosen at random owns only 1 vehicle
Solution:
■ 2400 families were selected at random from a region. There are thousands of more families in the region. When we calculate the probabilities based on the table, those probabilities will be applicable to the whole region
1. Number of families in this income range owning exactly 2 vehicles = 29
• So probability = Number of trials in which the event happened⁄Total number of trials = 29⁄2400
2. Number of families in this income range owning exactly 1 vehicle = 579
• So probability = Number of trials in which the event happened⁄Total number of trials = 579⁄2400
3. Number of families in this income range owning no vehicles = 10
• So probability = Number of trials in which the event happened⁄Total number of trials = 10⁄2400
4. Number of families in this income range owning more than 2 vehicles = 25
• So probability = Number of trials in which the event happened⁄Total number of trials = 25⁄2400
5. In this case, income range is not required. Number of families who own exactly one vehicle is:
160 +305 +535 +469 + 579 = 2048
• So probability = Number of trials in which the event happened⁄Total number of trials = 2048⁄2400
Solved example 28.13
To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data obtained is tabulated below:
| Opinion | Number of students |
|---|---|
| Like | 135 |
| Dislike | 66 |
(i) Probability that, a student chosen at random likes statistics
(ii) Probability that, a student chosen at random dislikes statistics
Solution:
1. Number of students who like statistics = 135
• So probability = Number of trials in which the event happened⁄Total number of trials = 135⁄200 = 27⁄40
2. Number of students who dislike statistics = 66
• So probability = Number of trials in which the event happened⁄Total number of trials = 66⁄200 = 33⁄100
Solved example 28.14
Eleven bags of wheat flour, each marked 5 kg, actually contained the following weights 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:
No. of bags which contain more than 5 kg of flour = 7
• So probability = Number of trials in which the event happened⁄Total number of trials = 7⁄11
We have completed this discussion on experimental probability. Part III of this discussion can be seen in chapter 36.
In the next section, we will see Quadratic equations.
In the next section, we will see Quadratic equations.
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