In the previous section we saw the results when the number of times 'two coins are tossed simultaneously' is increased. In this section we will see how those findings can be used to solve practical problems. Before that, we will just recall what the following terms mean:
1. Trial 2. Outcome 3. Favourable outcome 4. Event
1. Trial:
• In experiment I, each toss of the coin is a trial
• In experiment II, each roll of the die is a trial
• In experiment III, each 'simultaneous toss of two coins' is a trial
2. Outcome:
• In experiment I, consider any one trial. It has two possible outcomes:
H or T
• In experiment II, consider any one trial. It has six possible outcomes:
1, 2, 3, 4, 5 or 6
• In experiment III, consider any one trial. It has three possible outcomes:
TT, HT or HH
3. Favourable outcome:
■ In experiment I, consider any one trial. It has two possible outcomes:
H or T
• Let the outcome in that trial be H.
♦ If we already wanted 'H' before the beginning of the trial, then it is a favourable outcome
♦ If what we wanted was 'T' before the beginning of the trial, then it is not a favourable outcome
• Let the outcome in that trial be T.
♦ If we already wanted 'T' before the beginning of the trial, then it is a favourable outcome
♦ If what we wanted was 'H' before the beginning of the trial, then it is not a favourable outcome
■ In experiment II, consider any one trial. It has six possible outcomes:
1, 2, 3, 4, 5 or 6
• Let the outcome in that trial be 5.
♦ If we already wanted '5' before the beginning of the trial, then it is a favourable outcome
♦ If what we wanted was a number other than '5' before the beginning of the trial, then it is not a favourable outcome
• Let the outcome in that trial be 4.
♦ If we already wanted 'an even number' before the beginning of the trial, then it is a favourable outcome. (In that case, the out comes 2 and 6 will also be favourable)
♦ If what we wanted was an odd number, before the beginning of the trial, then it is not a favourable outcome
■ In experiment III, consider any one trial.It has three possible outcomes:
TT, HT or HH
• Let the outcome in that trial be HT.
♦ If we already wanted 'HT' before the beginning of the trial, then it is a favourable outcome
♦ If what we wanted was TT or HH before the beginning of the trial, then it is not a favourable outcome
4. Event:
■ In experiment I, consider any one trial.
• At the end of that trial, we will get an outcome. This outcome may or may not be favourable
• If what we get is a favourable outcome, we say: An event has occurred
Example:
Suppose we decided that we want H. This decision was made before carrying out the trial.
At the end of the trial, the outcome is H
Then an event has occurred
■ In experiment II, consider any one trial.
Suppose we decided that we want 2. This decision was made before carrying out the trial.
At the end of the trial, the outcome is 2
Then an event has occurred
If at the end of the trial the outcome is 1, 3, 4, 5 or 6, we don't have an event
Another example:
Suppose we decided that we want an even number. This decision was made before carrying out the trial.
At the end of the trial, the outcome is 6
Then an event has occurred
If at the end of the trial the outcome is 2, 4 or 6, we have an event
If at the end of the trial the outcome is 1, 3 or 5, we don't have an event
■ In experiment III, consider any one trial.
Suppose we decided that we want HT. This decision was made before carrying out the trial.
At the end of the trial, the outcome is HT
Then an event has occurred
If at the end of the trial the outcome is HH or TT, we don't have an event
• We know that Probability = Number of favourable outcomes⁄Total number of all possible outcomes.
We have seen the details in an earlier chapter 1.5.
■ In the present chapter we did some experiments
• In each experiment, we did trials
♦ At the end of each trial, we got an outcome
♦ This outcome may be favourable or unfavourable
• If it is favourable, we have an event
■ So in this chapter we will be dealing with 'Experimental probability'.
• 'Experimental probability' is also called 'Empirical probability'
• Empirical probability of an event E to occur is denoted as P(E)
• It is given by the ratio: Number of trials in which the event happened⁄Total number of trials.
(In this chapter wherever we write 'probability', it would mean 'empirical probability'. This shortening is for convenience)
An example:
1. In the experiment ID, we obtained some values related to heads. They are: 0.2, 0.333, 0.356, . . . (see table 28.2 in the first section of this chapter)
2. Each of them are probability values. That is., we can write:
P(E) = 0.2
P(E) = 0.333
P(E) = 0.356
Where E is the event of 'obtaining head' in 'tossing a coin'.
3. So we find that P(E) changes when the number of trials changes.
4. Suppose someone asks us the following question:
What is the empirical probability of obtaining head when a coin is tossed once?
Ans:
If we have done 15 trials we would be able to say 0.2
If we have done 30 trials we would be able to say 0.333
If we have done 45 trials we would be able to say 0.356
Another example:
1. In the experiment IID:
P(E) = 0.2
P(E) = 0.125
P(E) = 0.16
Where E is the event of 'obtaining 4' in 'rolling a die'. (see table 28.4 in the second section of this chapter)
2. Suppose someone asks us the following question:
What is the empirical probability of obtaining 4 when a die is rolled once?
Ans:
If we have done 20 trials we would be able to say 0.2
If we have done 40 trials we would be able to say 0.125
If we have done 60 trials we would be able to say 0.16
Now we will see some solved examples:
Solved example 28.1
A coin is tossed 1000 times. Head was obtained 455 times and tail was obtained 545 times. If E is the event of getting a head and F, the event of getting a tail, then Compute P(E) and P(F)
Solution:
1. The coin is tossed 1000 times. So total number of trials = 1000
2. In each trial there are two possible outcomes: H and T
3. Consider the probability for H.
• When we consider the probability for H, obtaining H in a trial is a favourable outcome.
• If a favourable outcome is obtained in a trial, that trial gives us an event.
• So when H is considered, 455 trials give us an event.
• In other words, when H is considered, we have 455 events
• Thus P(E) = Number of trials in which the event happened⁄Total number of trials = 455⁄1000. = 0.455
4. Consider the probability for T.
• When we consider the probability for T, obtaining T in a trial is a favourable outcome.
• If a favourable outcome is obtained in a trial, that trial gives us an event.
• So when T is considered, 545 trials give us an event.
• In other words, when T is considered, we have 545 events
• Thus P(E) = Number of trials in which the event happened⁄Total number of trials = 545⁄1000 = 0.545
5. Note that P(E) + P(F) = 0.455 + 0.545 = 1
This is because H and T are the only possible outcomes. If H does not occur, T will obviously occur and vice versa
Solved example 28.2
Two coins are tossed simultaneously 500 times. The results are:
HH 105 times, HT 275 times and TT 120 times
Find the probability for HH, HT and TT
Solution:
■ Let E, F and G denote the events for HH, HT and TT respectively
1. The two coins are tossed simultaneously 500 times. So total number of trials = 500
2. In each trial there are 3 possible outcomes: HH, HT and TT
3. Consider the probability for HH.
• When we consider the probability for HH, obtaining HH in a trial is a favourable outcome.
• If a favourable outcome is obtained in a trial, that trial gives us an event.
• So when HH is considered, 105 trials give us an event.
• In other words, when HH is considered, we have 105 events
• Thus P(E) = Number of trials in which the event happened⁄Total number of trials = 105⁄500. = 0.21
4. Consider the probability for HT.
• When we consider the probability for HT, obtaining HT in a trial is a favourable outcome.
• If a favourable outcome is obtained in a trial, that trial gives us an event.
• So when HT is considered, 275 trials give us an event.
• In other words, when HT is considered, we have 275 events
• Thus P(F) = Number of trials in which the event happened⁄Total number of trials = 275⁄500. = 0.55
5. Consider the probability for TT.
• When we consider the probability for TT, obtaining TT in a trial is a favourable outcome.
• If a favourable outcome is obtained in a trial, that trial gives us an event.
• So when TT is considered, 120 trials give us an event.
• In other words, when TT is considered, we have 120 events
• Thus P(G) = Number of trials in which the event happened⁄Total number of trials = 120⁄500. = 0.24
6. Note that P(E) + P(F) + P(G) = 0.21 + 0.55 + 0.24 = 1
This is because HH, HT and TT are the only possible outcomes.
Solved example 28.3
A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as given in the table below:
Find the probability of getting each outcome
Solution:
Solved example 28.3
A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as given in the table below:
| Outcome | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Frequency | 179 | 150 | 157 | 149 | 175 | 190 |
Solution:
■ Let E1, E2, E3, E4, E5 and E6 denote the events for 1, 2, 3, 4, 5 and 6 respectively
1. The die is rolled 1000 times. So total number of trials = 1000
2. In each trial there are 6 possible outcomes: 1, 2, 3, 4, 5 or 6
3. Consider the probability for 1
• When we consider the probability for 1, obtaining 1 in a trial is a favourable outcome.
• If a favourable outcome is obtained in a trial, that trial gives us an event.
• So when 1 is considered, 179 trials give us an event.
• In other words, when 1 is considered, we have 179 events
• Thus P(E1) = Number of trials in which the event happened⁄Total number of trials = 179⁄1000 = 0.179
4. Consider the probability for 2
• When we consider the probability for 2, obtaining 2 in a trial is a favourable outcome.
• If a favourable outcome is obtained in a trial, that trial gives us an event.
• So when 2 is considered, 150 trials give us an event.
• In other words, when 2 is considered, we have 150 events
• Thus P(E2) = Number of trials in which the event happened⁄Total number of trials = 150⁄1000 = 0.150
• Thus P(E2) = Number of trials in which the event happened⁄Total number of trials = 150⁄1000 = 0.150
5. In a similar way, P(E3) = 157⁄1000 = 0.157
6. P(E4) = 149⁄1000 = 0.149
7. P(E5) = 175⁄1000 = 0.175
8. P(E6) = 190⁄1000 = 0.190
Note that P(E1) + P(E2) + P(E3) + P(E4) + P(E5) + P(E6) = 1
Solved example 28.4
On one page of a telephone directory, there are 200 telephone numbers. The unit place digit of those numbers were analysed. (For example, in the number 2578682, the unit place digit is 2). It was found that 0 occurred 22 times, 1 occurred 26 times, 2 occurred 22 times and so on.. The full frequency distribution is given in the table below:
With out looking at the page, the pencil is placed on a telephone number. What is the probability that, the digit in the unit place of that telephone number is 6?
Solution:
6. P(E4) = 149⁄1000 = 0.149
7. P(E5) = 175⁄1000 = 0.175
8. P(E6) = 190⁄1000 = 0.190
Note that P(E1) + P(E2) + P(E3) + P(E4) + P(E5) + P(E6) = 1
Solved example 28.4
On one page of a telephone directory, there are 200 telephone numbers. The unit place digit of those numbers were analysed. (For example, in the number 2578682, the unit place digit is 2). It was found that 0 occurred 22 times, 1 occurred 26 times, 2 occurred 22 times and so on.. The full frequency distribution is given in the table below:
| Digit | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
|---|---|---|---|---|---|---|---|---|---|---|
| Frequency | 22 | 26 | 22 | 22 | 20 | 10 | 14 | 28 | 16 | 20 |
Solution:
1. There are 200 telephone numbers. The pencil may be placed on any of those numbers. So there are 200 possible outcomes.
2. Out of the 200 possible outcomes, 14 are favourable
3. So probability = Number of favourable outcomes⁄Total number of possible outcomes = 14⁄200 = 0.07
Solved example 28.5
The record at a weather station shows that out of the past 250 consecutive days, the weather forecasts were correct on 175 days.
(i) What is the probability that on a given day it was correct
(ii) What is the probability that it was not correct on a given day?
Solution:
1. There are 250 possible outcomes
2. If the 'correct forecast' is considered, 175 outcomes are favourable
• So probability = Number of favourable outcomes⁄Total number of possible outcomes = 175⁄250 = 0.7
3. No. of incorrect forecasts = 250 - 175 = 75
• If the 'in correct forecast' is considered, 75 outcomes are favourable
• So probability = Number of favourable outcomes⁄Total number of possible outcomes = 75⁄250 = 0.3
4. Note that, the sum of two probabilities is 1
Solved example 28.6
A tyre manufacturing company kept a record of the distance covered before a tyre needed to be replaced. The table below shows the results of 1000 cases.
If you buy a tyre from this company, what is the probability that:
(i) It will need to be replaced before it has covered 4000 km?
(ii) It will last more than 9000 km?
(iii) It will need to be replaced after it has covered 4000 km, but before reaching 14000 km?
Solution:
1. The company did an experiment 1000 times. The procedure for each experiment was the same. Let us write that procedure:
Step 1: Make four segments:
• Less than 4000
• 4001 to 9000
• 9001 to 14000
• More than 14000
Step 2: This step is carried out when a vehicle comes for tyre replacement.
(a) Verify the records and find the distance travelled by the tyre.
(b) Based on the distance, a tally mark can be placed in the appropriate segment
For example, if the distance is 5200 km, the tally mark is placed in the second segment.
If the distance is 15400 km, the tally mark is placed in the fourth segment
2. When each vehicle comes for tyre replacement, a tally mark is placed in the appropriate segment.
When a tally mark is placed, one experiment is complete.
So there will be 1000 tally marks when 1000 vehicles come for tyre replacement
3. When the 1000 tally marks are complete, they can be counted and the given table can be obtained.
4. If the company publish the table, future buyers can calculate different probabilities
5. If various manufacturing companies publish such a table of their own, and if the authenticity of such tables can be verified, buyers can make a comparison between companies.
6. Let us now calculate the different probabilities of the given company:
7. Replacement before 4000 km
• Probability = 20⁄1000 = 0.02
8. Replacement after 9000 km
• Probability = (325+445)⁄1000 = 770⁄1000 = 0.77
9. Replacement between 4000 and 14000 km
• Probability = (210+325)⁄1000 = 535⁄1000 = 0.535
Solved example 28.7
The percentage of marks obtained by a student in the monthly unit tests are given below:
Based on this data, find the probability that the student gets more than 70% marks in a unit test
Solution:
Each unit test can be considered as a trial. So there are 5 trials.
When we consider 'more than 70%', there are 3 events
So probability = Number of trials in which the event happened⁄Total number of trials = 3⁄5 = 0.6
2. Out of the 200 possible outcomes, 14 are favourable
3. So probability = Number of favourable outcomes⁄Total number of possible outcomes = 14⁄200 = 0.07
Solved example 28.5
The record at a weather station shows that out of the past 250 consecutive days, the weather forecasts were correct on 175 days.
(i) What is the probability that on a given day it was correct
(ii) What is the probability that it was not correct on a given day?
Solution:
1. There are 250 possible outcomes
2. If the 'correct forecast' is considered, 175 outcomes are favourable
• So probability = Number of favourable outcomes⁄Total number of possible outcomes = 175⁄250 = 0.7
3. No. of incorrect forecasts = 250 - 175 = 75
• If the 'in correct forecast' is considered, 75 outcomes are favourable
• So probability = Number of favourable outcomes⁄Total number of possible outcomes = 75⁄250 = 0.3
4. Note that, the sum of two probabilities is 1
Solved example 28.6
A tyre manufacturing company kept a record of the distance covered before a tyre needed to be replaced. The table below shows the results of 1000 cases.
| Distance (km) | less than 4000 | 4000 to 9000 | 9001 to 14000 | more than 14000 |
|---|---|---|---|---|
| Frequency | 20 | 210 | 325 | 445 |
(i) It will need to be replaced before it has covered 4000 km?
(ii) It will last more than 9000 km?
(iii) It will need to be replaced after it has covered 4000 km, but before reaching 14000 km?
Solution:
1. The company did an experiment 1000 times. The procedure for each experiment was the same. Let us write that procedure:
Step 1: Make four segments:
• Less than 4000
• 4001 to 9000
• 9001 to 14000
• More than 14000
Step 2: This step is carried out when a vehicle comes for tyre replacement.
(a) Verify the records and find the distance travelled by the tyre.
(b) Based on the distance, a tally mark can be placed in the appropriate segment
For example, if the distance is 5200 km, the tally mark is placed in the second segment.
If the distance is 15400 km, the tally mark is placed in the fourth segment
2. When each vehicle comes for tyre replacement, a tally mark is placed in the appropriate segment.
When a tally mark is placed, one experiment is complete.
So there will be 1000 tally marks when 1000 vehicles come for tyre replacement
3. When the 1000 tally marks are complete, they can be counted and the given table can be obtained.
4. If the company publish the table, future buyers can calculate different probabilities
5. If various manufacturing companies publish such a table of their own, and if the authenticity of such tables can be verified, buyers can make a comparison between companies.
6. Let us now calculate the different probabilities of the given company:
7. Replacement before 4000 km
• Probability = 20⁄1000 = 0.02
8. Replacement after 9000 km
• Probability = (325+445)⁄1000 = 770⁄1000 = 0.77
9. Replacement between 4000 and 14000 km
• Probability = (210+325)⁄1000 = 535⁄1000 = 0.535
Solved example 28.7
The percentage of marks obtained by a student in the monthly unit tests are given below:
| Unit test | I | II | III | IV | V |
|---|---|---|---|---|---|
| Percentage of marks obtained |
69 | 71 | 73 | 68 | 74 |
Solution:
Each unit test can be considered as a trial. So there are 5 trials.
When we consider 'more than 70%', there are 3 events
So probability = Number of trials in which the event happened⁄Total number of trials = 3⁄5 = 0.6
In the next section, we will see a few more solved examples.
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