In the previous section we completed a discussion on circle and it's chords. In this section we will see a discussion on Probability. We have seen the basic details of probability in an earlier chapter. We will continue our discussion from there.
Let us do an experiment. We will call it experiment IA:
1. Take a coin.
(i) Toss it once. The reading may be 'heads' or 'tails'.
(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 10 times.
3. In the note book, tabulate the readings as shown in table 28.1 below
Table 28.1
4. Determine the following ratios:
• Number of times head comes up⁄Total number of times the coin is tossed
• Number of times tail comes up⁄Total number of times the coin is tossed
• In our present case:
♦ the first ratio is 6⁄10 = 0.6
♦ the second ratio is 4⁄10 = 0.4
5. Once the ratios in (4) are determined, the experiment IA is complete
• For this experiment IB, the number of cycles in (2) must be 20
• Let the readings be as shown below:
• In this case:
♦ the first ratio is 9⁄20 = 0.45
♦ the second ratio is 11⁄20 = 0.55
• When the ratios in (4) are determined, the experiment IB is over
■ Once again repeat the experiment. We will call it experiment IC
• For this experiment IC, the number of cycles in (2) must be 30
• Let the readings be as shown below:
• In this case:
♦ the first ratio is 16⁄30 = 0.533
♦ the second ratio is 14⁄30 = 0.466
• When the ratios in (4) are determined, the experiment IC is over
So we did the same experiment 3 times. Before proceeding further, we will discuss the importance of the two ratios:
♦ Number of times head comes up⁄Total number of times the coin is tossed
♦ Number of times tail comes up⁄Total number of times the coin is tossed
■ Let us now analyse the ratios:
• First ratio when the number of trials is 10 = 0.6
• First ratio when the number of trials is 20 = 0.45
• First ratio when the number of trials is 30 = 0.533
■ As the number of trials increases, the first ratio gets closer and closer to 0.5
Consider the second ratio:
• Second ratio when the number of trials is 10 = 0.4
• Second ratio when the number of trials is 20 = 0.55
• Second ratio when the number of trials is 30 = 0.466
■ As the number of trials increases, the second ratio also gets closer and closer to 0.5
• What will happen if we go on increasing the 'total number of tosses'?
Ans: Each of the ratios will still become closer to 0.5
• Finally, when the 'total number of tosses' become a 'very large value', both the ratios will become equal to 0.5
• We can increase the 'number of tosses' to a 'considerably large value' in another way. We will do it as a new experiment. We will call it: Experiment ID
1. Divide the class into a number of groups. Each group must have three students
In our present case, let there be 12 groups, each with 3 students
2. Consider any one group
(i) A student in that group tosses the coin once. The reading may be 'heads' or 'tails'
(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 15 times. When the 15 cycles are completed, the task of this group of students is complete.
4. Now the coin can be given to another group. They can do the 15 cycles.
5. After the 15 cycles, the coin can be given to yet another group
6. So, when all the 12 groups have completed their tasks, the coin will be tossed 12×15 = 180 times.
• This 180 is a significantly large number. So we achieve a 'large task of tossing a coin 180 times'. We achieve it by dividing the task among 12 groups.
7. If there are 12 coins, all the groups can perform the task simultaneously. This will save time.
• There is only one condition: All the 12 coins must be of the same denomination and must be identical.
• If this condition is satisfied, it can be considered that a single student did all the 180 tosses
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 180 tosses.
9. For that, do the tabulation as shown in the table 28.2 below. The explanation about 'how this table is formed' is given below the table.
Table 28.2
■ 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 3⁄15.
♦ Obviously, it is the ratio: Number of times head comes up⁄Total number of times the coin is tossed
♦ In the 5th column, we have a ratio 12⁄15.
♦ Obviously, it is the ratio: Number of times tail comes up⁄Total number of times the coin is tossed
■ 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 10⁄30.
♦ The numerator is (7+3). So it is the total number of heads obtained by groups 1 and 2 together. It has the same effect of 'a single student doing 30 tosses, and obtaining 10 heads'.
♦ So in the denominator we have 30
♦ In the 5th column, we have a ratio 20⁄30.
♦ The numerator is (8+12). So it is the total number of tails obtained by groups 1 and 2 together. It has the same effect of 'a single student doing 30 tosses, and obtaining 20 tails'.
♦ So in the denominator we have 30
■ 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 16⁄45.
♦ The numerator is (6+10). So it is the total number of heads obtained by groups 1, 2 and 3 together. It has the same effect of 'a single student doing 45 tosses, and obtaining 16 heads'.
♦ So in the denominator we have 45
♦ In the 5th column, we have a ratio 29⁄45.
♦ The numerator is (9+20). So it is the total number of tails obtained by groups 1, 2 and 3 together. It has the same effect of 'a single student doing 45 tosses, and obtaining 29 tails'.
♦ So in the denominator we have 45
Let us do an experiment. We will call it experiment IA:
1. Take a coin.
(i) Toss it once. The reading may be 'heads' or 'tails'.
(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 10 times.
3. In the note book, tabulate the readings as shown in table 28.1 below
Table 28.1
| Number of times the coin is tossed | Number of times head comes up | Number of times tail comes up |
|---|---|---|
| 10 | 6 | 4 |
• Number of times head comes up⁄Total number of times the coin is tossed
• Number of times tail comes up⁄Total number of times the coin is tossed
• In our present case:
♦ the first ratio is 6⁄10 = 0.6
♦ the second ratio is 4⁄10 = 0.4
5. Once the ratios in (4) are determined, the experiment IA is complete
But our work is not over
■ Repeat the above experiment. We will call it experiment IB.• For this experiment IB, the number of cycles in (2) must be 20
• Let the readings be as shown below:
| Number of times the coin is tossed | Number of times head comes up | Number of times tail comes up |
|---|---|---|
| 20 | 9 | 11 |
♦ the first ratio is 9⁄20 = 0.45
♦ the second ratio is 11⁄20 = 0.55
• When the ratios in (4) are determined, the experiment IB is over
■ Once again repeat the experiment. We will call it experiment IC
• For this experiment IC, the number of cycles in (2) must be 30
• Let the readings be as shown below:
| Number of times the coin is tossed | Number of times head comes up | Number of times tail comes up |
|---|---|---|
| 30 | 16 | 14 |
♦ the first ratio is 16⁄30 = 0.533
♦ the second ratio is 14⁄30 = 0.466
• When the ratios in (4) are determined, the experiment IC is over
1. We know that, the probability of obtaining heads or tails when tossing a coin once is 0.5
• It means that, there is a 50% chance of getting heads. Also there is a 50% chance of getting tails.
2. But this '0.5' is a theoretical value. If 0.5 is always obtained in the real life also, we will get results such as these:
• Toss the coin 10 times
♦ Heads will be obtained 5 times and tails will be obtained 5 times
♦ That is., 50% heads (5⁄10) and 50% tails (5⁄10)
• Toss the coin 50 times
♦ Heads will be obtained 25 times and tails will be obtained 25 times
♦ That is., 50% heads (25⁄50) and 50% tails (25⁄50)
3. But we never get such exact values.
4. However, as the number of tosses increase, both the ratios become closer and closer to 0.5
4. However, as the number of tosses increase, both the ratios become closer and closer to 0.5
5. If, instead of 10 times or 50 times, if we toss it for a 'very large number of times', then:
• No. of times head is obtained = No. of times tail is obtained = k
• That is., 'the equality of probability while tossing a coin' will become clear
• Let the 'very large number of times' for which the coin is tossed be 'n'
• Then k⁄n will be equal to 0.5
• No. of times head is obtained = No. of times tail is obtained = k
• That is., 'the equality of probability while tossing a coin' will become clear
• Let the 'very large number of times' for which the coin is tossed be 'n'
• Then k⁄n will be equal to 0.5
■ 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 two ratios:♦ Number of times head comes up⁄Total number of times the coin is tossed
♦ Number of times tail comes up⁄Total number of times the coin is tossed
■ Let us now analyse the ratios:
• First ratio when the number of trials is 10 = 0.6
• First ratio when the number of trials is 20 = 0.45
• First ratio when the number of trials is 30 = 0.533
■ As the number of trials increases, the first ratio gets closer and closer to 0.5
Consider the second ratio:
• Second ratio when the number of trials is 10 = 0.4
• Second ratio when the number of trials is 20 = 0.55
• Second ratio when the number of trials is 30 = 0.466
■ As the number of trials increases, the second ratio also gets closer and closer to 0.5
• What will happen if we go on increasing the 'total number of tosses'?
Ans: Each of the ratios will still become closer to 0.5
• Finally, when the 'total number of tosses' become a 'very large value', both the ratios will become equal to 0.5
1. Divide the class into a number of groups. Each group must have three students
In our present case, let there be 12 groups, each with 3 students
2. Consider any one group
(i) A student in that group tosses the coin once. The reading may be 'heads' or 'tails'
(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 15 times. When the 15 cycles are completed, the task of this group of students is complete.
4. Now the coin can be given to another group. They can do the 15 cycles.
5. After the 15 cycles, the coin can be given to yet another group
6. So, when all the 12 groups have completed their tasks, the coin will be tossed 12×15 = 180 times.
• This 180 is a significantly large number. So we achieve a 'large task of tossing a coin 180 times'. We achieve it by dividing the task among 12 groups.
7. If there are 12 coins, all the groups can perform the task simultaneously. This will save time.
• There is only one condition: All the 12 coins must be of the same denomination and must be identical.
• If this condition is satisfied, it can be considered that a single student did all the 180 tosses
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 180 tosses.
9. For that, do the tabulation as shown in the table 28.2 below. The explanation about 'how this table is formed' is given below the table.
Table 28.2
| Group | Number of Heads | Number of Tails | A | B |
|---|---|---|---|---|
| (1) | (2) | (3) | (4) | (5) |
| 1 | 3 | 12 | 3⁄15 = 0.2 | 12⁄15 = 0.8 |
| 2 | 7 | 8 | (7+3)⁄(15+15) = 10⁄30 = 0.333 | (8+12)⁄(15+15) = 20⁄30 = 0.666 |
| 3 | 5 | 10 | (6+10)⁄(15+30) = 16⁄45 = 0.356 | (9+20)⁄(15+30) = 29⁄45 = 0.644 |
| _ | _ | _ | _ | _ |
| _ | _ | _ | _ | _ |
| 12 | _ | _ | _ | _ |
• 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 3⁄15.
♦ Obviously, it is the ratio: Number of times head comes up⁄Total number of times the coin is tossed
♦ In the 5th column, we have a ratio 12⁄15.
♦ Obviously, it is the ratio: Number of times tail comes up⁄Total number of times the coin is tossed
■ 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 10⁄30.
♦ The numerator is (7+3). So it is the total number of heads obtained by groups 1 and 2 together. It has the same effect of 'a single student doing 30 tosses, and obtaining 10 heads'.
♦ So in the denominator we have 30
♦ In the 5th column, we have a ratio 20⁄30.
♦ The numerator is (8+12). So it is the total number of tails obtained by groups 1 and 2 together. It has the same effect of 'a single student doing 30 tosses, and obtaining 20 tails'.
♦ So in the denominator we have 30
■ 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 16⁄45.
♦ The numerator is (6+10). So it is the total number of heads obtained by groups 1, 2 and 3 together. It has the same effect of 'a single student doing 45 tosses, and obtaining 16 heads'.
♦ So in the denominator we have 45
♦ In the 5th column, we have a ratio 29⁄45.
♦ The numerator is (9+20). So it is the total number of tails obtained by groups 1, 2 and 3 together. It has the same effect of 'a single student doing 45 tosses, and obtaining 29 tails'.
♦ So in the denominator we have 45
■ In this way, we can fill up the rows of all the 12 groups, and find all the required ratios.
■ Note: As the number of heads and tails in each row is added to the 'total values up to the previous row', the word cumulative is used.
• So the heading of the 4th column (which is written as A in the table) is:
Cumulative number of heads⁄Total number of times the coin is tossed
• Similarly, the heading of the 5th column (which is written as B in the table) is:
Cumulative number of tails⁄Total number of times the coin is tossed
Once the ratios are found out, we can analyse them:
■ Group 1:
• Number of tosses = 15
First ratio = 0.2
Second ratio = 0.8
First ratio = 0.2
Second ratio = 0.8
■ Groups 1 and 2:
• Number of tosses = 30
First ratio = 0.333
Second ratio = 0.666
First ratio = 0.333
Second ratio = 0.666
■ Groups 1, 2 and 3:
• Number of tosses = 45
First ratio = 0.356
Second ratio = 0.644
■ So the first ratio takes the values 0.2, 0.333, 0.356, . . .
It is getting closer and closer to 0.5
■ The second ratio takes the values 0.8, 0.666, 0.644, . . .
It is getting closer and closer to 0.5
First ratio = 0.356
Second ratio = 0.644
■ So the first ratio takes the values 0.2, 0.333, 0.356, . . .
It is getting closer and closer to 0.5
■ The second ratio takes the values 0.8, 0.666, 0.644, . . .
It is getting closer and closer to 0.5
• In this experiment, we made 180 trials. This '180' 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', both the ratios will become equal to 0.5
• A 'very large number' is like 'infinity'
• At such a 'very large number', both the ratios will become equal to 0.5
In the next section, we will see the above procedure applied to 'rolling of a die'.
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