In the previous section we saw the results when the number of times a coin is tossed is increased. In this section we will see such experiments related to 'rolling a die'.
Let us do an experiment. We will call it experiment IIA:
1. Take a die.
(i) Roll it once. The reading may be any one of the following:
1, 2, 3, 4, 5 or 6
(ii) What ever be the reading, note that reading on a piece of paper.
• (i) and (ii) constitutes one cycle of our experiment.
2. Repeat the cycle 20 times.
3. In the note book, tabulate the readings as shown in table 28.3 below
Table 28.3
4. Determine the following ratios:
• Number of times 1 turned up⁄Total number of times the die is rolled
• Number of times 2 turned up⁄Total number of times the die is rolled
• Number of times 3 turned up⁄Total number of times the die is rolled
• - - -
• - - -
• Number of times 6 turned up⁄Total number of times the die is rolled
• In our present case:
♦ the 1st ratio is 7⁄20 = 0.35
♦ the 2nd ratio is 1⁄20 = 0.05
♦ the 3rd ratio is 1⁄20 = 0.05
♦ the 4th ratio is 3⁄20 = 0.15
♦ the 5th ratio is 5⁄20 = 0.25
♦ the 6th ratio is 3⁄20 = 0.15
5. Once the ratios in (4) are determined, the experiment IIA is complete
• For this experiment IIB, the number of cycles in (2) must be 40
• Let the readings be as shown below:
• In this case:
♦ the 1st ratio is 4⁄40 = 0.1
♦ the 2nd ratio is 8⁄40 = 0.2
♦ the 3rd ratio is 9⁄40 = 0.225
♦ the 4th ratio is 4⁄40 = 0.1
♦ the 5th ratio is 9⁄40 = 0.225
♦ the 6th ratio is 6⁄40 = 0.15
• When the ratios in (4) are determined, the experiment IIB is over
■ Once again repeat the experiment. We will call it experiment IIC
• For this experiment IIC, the number of cycles in (2) must be 60
• Let the readings be as shown below:
• In this case:
♦ the 1st ratio is 9⁄60 = 0.15
♦ the 2nd ratio is 10⁄60 = 0.167
♦ the 3rd ratio is 8⁄60 = 0.133
♦ the 4th ratio is 10⁄60 = 0.167
♦ the 5th ratio is 12⁄60 = 0.2
♦ the 6th ratio is 11⁄60 = 0.183
• When the ratios in (4) are determined, the experiment IIC is over
So we did the same experiment 3 times. Before proceeding further, we will discuss the importance of the six ratios:
♦ Number of times 1 turned up⁄Total number of times the die is rolled
♦ Number of times 2 turned up⁄Total number of times the die is rolled
♦ Number of times 3 turned up⁄Total number of times the die is rolled
♦ - - -
♦ - - -
♦ Number of times 6 turned up⁄Total number of times the die is rolled
■ Let us now analyse the ratios:
1st ratio when number of trials is 20 = 0.35
1st ratio when number of trials is 40 = 0.1
1st ratio when number of trials is 60 = 0.15
• We have, 1⁄6 = 0.1667
• As the number of trial increases, the 1st ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 1st ratio gets closer and closer to 1⁄6.
Consider the 2nd ratio:
2nd ratio when number of trials is 20 = 0.05
2nd ratio when number of trials is 40 = 0.2
2nd ratio when number of trials is 60 = 0.167
• We have, 1⁄6 = 0.1667
• As the number of trial increases, the 2nd ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 2nd ratio gets closer and closer to 1⁄6.
Consider the 3rd ratio:
3rd ratio when number of trials is 20 = 0.05
3rd ratio when number of trials is 40 = 0.225
3rd ratio when number of trials is 60 = 0.133
• We have, 1⁄6 = 0.1667
• As the number of trial increases, the 3rd ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 3rd ratio gets closer and closer to 1⁄6.
Consider the 4th ratio:
4th ratio when number of trials is 20 = 0.15
4th ratio when number of trials is 40 = 0.1
4th ratio when number of trials is 60 = 0.167
• We have, 1⁄6 = 0.1667
• As the number of trial increases, the 4th ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 4th ratio gets closer and closer to 1⁄6.
Consider the 5th ratio:
5th ratio when number of trials is 20 = 0.25
5th ratio when number of trials is 40 = 0.225
5th ratio when number of trials is 60 = 0.2
• We have, 1⁄6 = 0.1667
• As the number of trial increases, the 5th ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 5th ratio gets closer and closer to 1⁄6.
Consider the 6th ratio:
6th ratio when number of trials is 20 = 0.15
6th ratio when number of trials is 40 = 0.15
6th ratio when number of trials is 60 = 0.183
• We have, 1⁄6 = 0.1667
• As the number of trial increases, the 6th ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 6th ratio gets closer and closer to 1⁄6.
• We can increase the 'number of trials' to a 'considerably large value' in another way. We will do it as a new experiment. We will call it: Experiment IIC
1. Divide the class into a number of groups. Each group must have three students
In our present case, let there be 15 groups, each with 3 students
2. Consider any one group
(i) A student in that group rolls the die once. The reading may be 1, 2, 3, 4, 5 or 6
(ii) Whatever be the reading, the other two students must check it and record it
• (i) and (ii) constitutes one cycle of our experiment.
3. Repeat the cycle 20 times. When the 20 cycles are completed, the task of this group of students is complete.
4. Now the die can be given to another group. They can do the 20 cycles.
5. After the 20 cycles, the die can be given to yet another group
6. So, when all the 15 groups have completed their tasks, the die will be rolled 15×20 = 300 times.
• This 300 is a significantly large number. So we achieve a 'large task of rolling a die 300 times'. We achieve it by dividing the task among 15 groups.
7. If there are 15 dice, all the groups can perform the task simultaneously. This will save time.
• There is only one condition: All the 15 dice must be identical.
If this condition is satisfied, it can be considered that a single student did all the 300 rolls.
8. When all the groups have completed their tasks, we can proceed to do the calculations. We have to combine the results in such a way that, one student did all the 300 rolls.
9. For that, do the tabulation as shown in the table 28.4 below.
• In the table, only 1 and 4 is given. This is to keep the size of the table small. Students may use a large sheet of paper and include all the numbers from 1 to 6.
• The explanation about 'how this table is formed' is given below the table.
Table 28.4
■ Consider the row corresponding to group 1.
• The first three columns in this row does not need any explanation because, they are the readings obtained during the experiment.
♦ In the 4th column, we have a ratio 3⁄20.
♦ Obviously, it is the ratio: Number of times 1 comes up⁄Total number of times the die is rolled
♦ In the 5th column, we have a ratio 4⁄20.
♦ Obviously, it is the ratio: Number of times 4 comes up⁄Total number of times the die is rolled
■ Consider the row corresponding to group 2.
• The first three columns in this row does not need any explanation because, they are the readings obtained during the experiment.
♦ In the 4th column, we have a ratio 5⁄40.
♦ The numerator is (2+3). So it is the total number of 1 obtained by groups 1 and 2 together. It has the same effect of 'a single student doing 40 rolls, and obtaining 1 five times'.
♦ So in the denominator we have 40
♦ In the 5th column, we have a ratio 5⁄40.
♦ The numerator is (1+4). So it is the total number of 4 obtained by groups 1 and 2 together. It has the same effect of 'a single student doing 40 rolls, and obtaining 4 five times'.
♦ So in the denominator we have 40
■ Consider the row corresponding to group 3.
• The first three columns in this row does not need any explanation because, they are the readings obtained during the experiment.
♦ In the 4th column, we have a ratio 7⁄60.
♦ The numerator is (2+5). So it is the total number of 1 obtained by groups 1, 2 and 3 together. It has the same effect of 'a single student doing 60 rolls, and obtaining 1 seven times'.
♦ So in the denominator we have 60
♦ In the 5th column, we have a ratio 10⁄60.
♦ The numerator is (5+5). So it is the total number of 4 obtained by groups 1, 2 and 3 together. It has the same effect of 'a single student doing 60 rolls, and obtaining 4 ten times'.
♦ So in the denominator we have 60
Let us do an experiment. We will call it experiment IIA:
1. Take a die.
(i) Roll it once. The reading may be any one of the following:
1, 2, 3, 4, 5 or 6
(ii) What ever be the reading, note that reading on a piece of paper.
• (i) and (ii) constitutes one cycle of our experiment.
2. Repeat the cycle 20 times.
3. In the note book, tabulate the readings as shown in table 28.3 below
Table 28.3
| Number of times a die is thrown | Number of times these scores turn up | |||||
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | |
| 20 | 7 | 1 | 1 | 3 | 5 | 3 |
• Number of times 1 turned up⁄Total number of times the die is rolled
• Number of times 2 turned up⁄Total number of times the die is rolled
• Number of times 3 turned up⁄Total number of times the die is rolled
• - - -
• - - -
• Number of times 6 turned up⁄Total number of times the die is rolled
• In our present case:
♦ the 1st ratio is 7⁄20 = 0.35
♦ the 2nd ratio is 1⁄20 = 0.05
♦ the 3rd ratio is 1⁄20 = 0.05
♦ the 4th ratio is 3⁄20 = 0.15
♦ the 5th ratio is 5⁄20 = 0.25
♦ the 6th ratio is 3⁄20 = 0.15
5. Once the ratios in (4) are determined, the experiment IIA is complete
But our work is not over
■ Repeat the above experiment. We will call it experiment IIB.• For this experiment IIB, the number of cycles in (2) must be 40
• Let the readings be as shown below:
| Number of times a die is thrown | Number of times these scores turn up | |||||
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | |
| 40 | 4 | 8 | 9 | 4 | 9 | 6 |
♦ the 1st ratio is 4⁄40 = 0.1
♦ the 2nd ratio is 8⁄40 = 0.2
♦ the 3rd ratio is 9⁄40 = 0.225
♦ the 4th ratio is 4⁄40 = 0.1
♦ the 5th ratio is 9⁄40 = 0.225
♦ the 6th ratio is 6⁄40 = 0.15
• When the ratios in (4) are determined, the experiment IIB is over
■ Once again repeat the experiment. We will call it experiment IIC
• For this experiment IIC, the number of cycles in (2) must be 60
• Let the readings be as shown below:
| Number of times a die is thrown | Number of times these scores turn up | |||||
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | |
| 60 | 9 | 10 | 8 | 10 | 12 | 11 |
♦ the 1st ratio is 9⁄60 = 0.15
♦ the 2nd ratio is 10⁄60 = 0.167
♦ the 3rd ratio is 8⁄60 = 0.133
♦ the 4th ratio is 10⁄60 = 0.167
♦ the 5th ratio is 12⁄60 = 0.2
♦ the 6th ratio is 11⁄60 = 0.183
• When the ratios in (4) are determined, the experiment IIC is over
1. We know that, the probability of obtaining 1, 2, 3, 4, 5 or 6 when rolling a die once is 1⁄6
• Each of the numbers 1, 2, 3, 4, 5 and 6 has the same probability 1⁄6.
2. But this '1⁄6' is a theoretical value. If 1⁄6 is always obtained in the real life also, we will get results such as these:
• Roll the die 6 times
♦ 1 will be obtained once
♦ 2 will be obtained once
♦ 3 will be obtained once
♦ - - -
♦ - - -
♦ 6 will be obtained once
• Roll the die 18 times
♦ 1 will be obtained thrice
♦ 2 will be obtained thrice
♦ 3 will be obtained thrice
♦ - - -
♦ - - -
♦ 6 will be obtained thrice
♦ 1 will be obtained once
♦ 2 will be obtained once
♦ 3 will be obtained once
♦ - - -
♦ - - -
♦ 6 will be obtained once
• Roll the die 18 times
♦ 1 will be obtained thrice
♦ 2 will be obtained thrice
♦ 3 will be obtained thrice
♦ - - -
♦ - - -
♦ 6 will be obtained thrice
3. But we never get such exact values.
4. However, as the number of trials increase, each of the 6 ratios become closer and closer to 1⁄6
4. However, as the number of trials increase, each of the 6 ratios become closer and closer to 1⁄6
5. If, instead of 20,40 or 60 times, if we toss it for a 'very large number of times', then:
• No. of times 1 is obtained = No. of times 2 is obtained = No. of times 3 is obtained = No. of times 4 is obtained = No. of times 5 is obtained = No. of times 6 is obtained. = k
• That is., 'the equality of probability while rolling a die' will become clear.
• Let the 'very large number of times' for which the die is rolled be 'n'
• Then k⁄n will be equal to 1⁄6.
• Let the 'very large number of times' for which the die is rolled be 'n'
• Then k⁄n will be equal to 1⁄6.
5. We are trying to prove this using our present experiments
• From the three experiments, we have three sets of ratios. One set from each experiment
• Each set has six ratios:♦ Number of times 1 turned up⁄Total number of times the die is rolled
♦ Number of times 2 turned up⁄Total number of times the die is rolled
♦ Number of times 3 turned up⁄Total number of times the die is rolled
♦ - - -
♦ - - -
♦ Number of times 6 turned up⁄Total number of times the die is rolled
■ Let us now analyse the ratios:
1st ratio when number of trials is 20 = 0.35
1st ratio when number of trials is 40 = 0.1
1st ratio when number of trials is 60 = 0.15
• We have, 1⁄6 = 0.1667
• As the number of trial increases, the 1st ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 1st ratio gets closer and closer to 1⁄6.
Consider the 2nd ratio:
2nd ratio when number of trials is 20 = 0.05
2nd ratio when number of trials is 40 = 0.2
2nd ratio when number of trials is 60 = 0.167
• We have, 1⁄6 = 0.1667
• As the number of trial increases, the 2nd ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 2nd ratio gets closer and closer to 1⁄6.
Consider the 3rd ratio:
3rd ratio when number of trials is 20 = 0.05
3rd ratio when number of trials is 40 = 0.225
3rd ratio when number of trials is 60 = 0.133
• We have, 1⁄6 = 0.1667
• As the number of trial increases, the 3rd ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 3rd ratio gets closer and closer to 1⁄6.
Consider the 4th ratio:
4th ratio when number of trials is 20 = 0.15
4th ratio when number of trials is 40 = 0.1
4th ratio when number of trials is 60 = 0.167
• We have, 1⁄6 = 0.1667
• As the number of trial increases, the 4th ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 4th ratio gets closer and closer to 1⁄6.
Consider the 5th ratio:
5th ratio when number of trials is 20 = 0.25
5th ratio when number of trials is 40 = 0.225
5th ratio when number of trials is 60 = 0.2
• We have, 1⁄6 = 0.1667
• As the number of trial increases, the 5th ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 5th ratio gets closer and closer to 1⁄6.
Consider the 6th ratio:
6th ratio when number of trials is 20 = 0.15
6th ratio when number of trials is 40 = 0.15
6th ratio when number of trials is 60 = 0.183
• We have, 1⁄6 = 0.1667
• As the number of trial increases, the 6th ratio gets closer and closer to 0.1667
■ That is., as the number of trial increases, the 6th ratio gets closer and closer to 1⁄6.
So all the ratios are getting closer and closer to 1⁄6. When the number of trials become very large, each of the ratios will become equal to 1⁄6.
1. Divide the class into a number of groups. Each group must have three students
In our present case, let there be 15 groups, each with 3 students
2. Consider any one group
(i) A student in that group rolls the die once. The reading may be 1, 2, 3, 4, 5 or 6
(ii) Whatever be the reading, the other two students must check it and record it
• (i) and (ii) constitutes one cycle of our experiment.
3. Repeat the cycle 20 times. When the 20 cycles are completed, the task of this group of students is complete.
4. Now the die can be given to another group. They can do the 20 cycles.
5. After the 20 cycles, the die can be given to yet another group
6. So, when all the 15 groups have completed their tasks, the die will be rolled 15×20 = 300 times.
• This 300 is a significantly large number. So we achieve a 'large task of rolling a die 300 times'. We achieve it by dividing the task among 15 groups.
7. If there are 15 dice, all the groups can perform the task simultaneously. This will save time.
• There is only one condition: All the 15 dice must be identical.
If this condition is satisfied, it can be considered that a single student did all the 300 rolls.
8. When all the groups have completed their tasks, we can proceed to do the calculations. We have to combine the results in such a way that, one student did all the 300 rolls.
9. For that, do the tabulation as shown in the table 28.4 below.
• In the table, only 1 and 4 is given. This is to keep the size of the table small. Students may use a large sheet of paper and include all the numbers from 1 to 6.
• The explanation about 'how this table is formed' is given below the table.
Table 28.4
| Group | Number of 1 | Number of 4 | A | B |
|---|---|---|---|---|
| (1) | (2) | (3) | (4) | (5) |
| 1 | 3 | 4 | 3⁄20 = 0.15 | 4⁄20 = 0.2 |
| 2 | 2 | 1 | (2+3)⁄(20+20) = 5⁄40 = 0.125 | (1+4)⁄(20+20) = 5⁄40 = 0.125 |
| 3 | 2 | 5 | (2+5)⁄(20+40) = 7⁄60 = 0.116 | (5+5)⁄(20+40) = 10⁄60 = 0.16 |
| _ | _ | _ | _ | _ |
| _ | _ | _ | _ | _ |
| 15 | _ | _ | _ | _ |
• The first three columns in this row does not need any explanation because, they are the readings obtained during the experiment.
♦ In the 4th column, we have a ratio 3⁄20.
♦ Obviously, it is the ratio: Number of times 1 comes up⁄Total number of times the die is rolled
♦ In the 5th column, we have a ratio 4⁄20.
♦ Obviously, it is the ratio: Number of times 4 comes up⁄Total number of times the die is rolled
■ Consider the row corresponding to group 2.
• The first three columns in this row does not need any explanation because, they are the readings obtained during the experiment.
♦ In the 4th column, we have a ratio 5⁄40.
♦ The numerator is (2+3). So it is the total number of 1 obtained by groups 1 and 2 together. It has the same effect of 'a single student doing 40 rolls, and obtaining 1 five times'.
♦ So in the denominator we have 40
♦ In the 5th column, we have a ratio 5⁄40.
♦ The numerator is (1+4). So it is the total number of 4 obtained by groups 1 and 2 together. It has the same effect of 'a single student doing 40 rolls, and obtaining 4 five times'.
♦ So in the denominator we have 40
■ Consider the row corresponding to group 3.
• The first three columns in this row does not need any explanation because, they are the readings obtained during the experiment.
♦ In the 4th column, we have a ratio 7⁄60.
♦ The numerator is (2+5). So it is the total number of 1 obtained by groups 1, 2 and 3 together. It has the same effect of 'a single student doing 60 rolls, and obtaining 1 seven times'.
♦ So in the denominator we have 60
♦ In the 5th column, we have a ratio 10⁄60.
♦ The numerator is (5+5). So it is the total number of 4 obtained by groups 1, 2 and 3 together. It has the same effect of 'a single student doing 60 rolls, and obtaining 4 ten times'.
♦ So in the denominator we have 60
■ In this way, we can fill up the rows of all the 15 groups, and find all the required ratios.
■ Note: As the number of '1' and '4' in each row is added to the 'total values up to the previous row', the word cumulative can be used.
• So the heading of the 4th column (which is written as A in the table) is:
Cumulative number of 1⁄Total number of times the die is rolled
• Similarly, the heading of the 5th column (which is written as B in the table) is:
Cumulative number of 4⁄Total number of times the die is rolled
Once the ratios are found out, we can analyse them:
■ Group 1:
• Number of rolls = 20
Ratio for '1' = 0.15
Ratio for '4' = 0.2
Ratio for '1' = 0.15
Ratio for '4' = 0.2
■ Groups 1 and 2:
• Number of rolls = 40
Ratio for '1' = 0.125
Ratio for '4' = 0.125
• Number of rolls = 40
Ratio for '1' = 0.125
Ratio for '4' = 0.125
■ Groups 1, 2 and 3:
• Number of rolls = 60
Ratio for '1' = 0.116
Ratio for '4' = 0.16
■ So the ratio for '1' takes the values 0.15, 0.125, 0.116, . . .
It is getting closer and closer to 1⁄6.
■ The ratio for '4' takes the values 0.2, 0.125, 0.16
It is getting closer and closer to 1⁄6.
• If we had made the above table 28.4 on a large sheet of paper, that included numbers 2, 3, 5 and 6 also, we would get the same result for them also.
Ratio for '1' = 0.116
Ratio for '4' = 0.16
■ So the ratio for '1' takes the values 0.15, 0.125, 0.116, . . .
It is getting closer and closer to 1⁄6.
■ The ratio for '4' takes the values 0.2, 0.125, 0.16
It is getting closer and closer to 1⁄6.
• If we had made the above table 28.4 on a large sheet of paper, that included numbers 2, 3, 5 and 6 also, we would get the same result for them also.
■ In this experiment we made 300 trials. This '300' is significantly large but not even close to 'a very large number n'.
• A very large number is like 'infinity'
• At such a very large number, all the ratios will become equal to 1⁄6.
• A very large number is like 'infinity'
• At such a very large number, all the ratios will become equal to 1⁄6.
In the next section, we will see the above procedure when 'two coins are tossed simultaneously'.
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