Chapter 28 - Probability- Increasing number of Trials

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

Number of times the coin is tossed	Number of times head comes up	Number of times tail comes up
10	6	4

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

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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

• 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:

Number of times the coin is tossed	Number of times head comes up	Number of times tail comes up
30	16	14

• 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

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So we did the same experiment 3 times. Before proceeding further, we will discuss the importance of the two ratios:

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 

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 

■ We are trying to prove this using our present experiments

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• 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

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• 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

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	_	_	_	_

■ 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 4^th 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 5^th 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 4^th 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 5^th 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 4^th 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 5^th 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

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■ 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 4^th 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 5^th column (which is written as B in the table) is:

^{Cumulative number of tails}⁄_{Total number of times the coin is tossed}

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Once the ratios are found out, we can analyse them:

■ Group 1:

• Number of tosses = 15
First ratio = 0.2
Second ratio = 0.8

■ Groups 1 and 2:

• Number of tosses = 30
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

• 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

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In the next section, we will see the above procedure applied to 'rolling of a die'.

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Chapter 1.8 - Solved examples on Probability

In the previous section we completed the discussion on the probability related to drawing a coloured ball, tossing a coin, rolling a die, drawing a card etc., We saw some solved examples also. In this section, we will see some solved examples with a higher level of difficulty.

Solved example 1.10
Tickets marked with numbers from 1 to 25 are put in a box and mixed well. Then one ticket is drawn from the box without looking. What is the probability that the drawn ticket is a multiple of 3 or 7?

Solution:
Step 1: Write the outcomes:
Outcome 1: The drawn ticket is the one which is marked with '1' 
Outcome 2: The drawn ticket is the one which is marked with '2'
Outcome 3: The drawn ticket is the one which is marked with '3'
- - -
- - -
Outcome 25: The drawn ticket is the one which is marked with '25'

So there are 25 outcomes. We need not write them all. In fact, once we understand the basics, we need to write only this:
Step 1: There are 25 possible outcomes.

Step 2: Analyse each outcome. According to the question, the card should be a multiple of 3 OR 7. So if the card drawn is either a multiple of 3 OR a multiple of 7, we can write 'favourable' towards that outcome. And we will have an event. Let us see on which all out comes, we can write 'favourable':

Ticket marked with '1' → 1 is not a multiple of 3 or 7 → Not favourable
Ticket marked with '2' → 2 is not a multiple of 3 or 7 → Not favourable
Ticket marked with '3' → 3 is a multiple 3 → Favourable
Ticket marked with '4' → 4 is not a multiple of 3 or 7 → Not favourable
Ticket marked with '5' → 5 is not a multiple of 3 or 7 → Not favourable
Ticket marked with '6' → 6 is a multiple 3 → Favourable
Ticket marked with '7' → 7 is a multiple 7 → Favourable
- - -
- - -
Ticket marked with '25' → 25 is not a multiple of 3 or 7 → Not favourable

In this step 2 also, it is not necessary to write the 25 lines. We need to write only this:
Step 2: Analysing each outcome, we find that, 3, 6, 7, 9, 12, 14, 15, 18, 21, and 24 satisfy the required conditions. So there are a total of 10 favourable outcomes.


Thus the required probability = ¹⁰⁄₂₅ = ²⁄₅ = 40%. So there is a 40% probability that we will get a ticket which is either a multiple of 3 or a multiple of 7.

Solved example 1.11
Two unbiased coins are tossed simultaneously. Find the probability of getting (i) Two heads (ii) One tail (iii) One head (iv) at most one head (v) at least one head (vi) no head

Solution:
Step 1: Write the outcomes:

When two coins are tossed simultaneously, the possible outcomes can be written in the form of a table:

Outcome	First coin	Second coin
Outcome 1	Heads	Heads
Outcome 2	Heads	Tails
Outcome 3	Tails	Tails
Outcome 4	Tails	Heads

The above 4 are the only possible outcomes. This table can be used for all 6 cases of the question.
(i) Two heads: Analysing each outcome, we find that only one outcome (Outcome 1) satisfies this condition. So the probability of getting 2 heads = ¹⁄₄ = 25%
(ii) One tail: Two outcomes (Outcome 2 and 4) satisfies this condition. So the probability = ²⁄₄ = ¹⁄₂ = 50%
(iii) One head: Two outcomes (Outcome 2 and 4) satisfies this condition. So the probability = ²⁄₄ = ¹⁄₂ = 50%
(iv) At most one head: This condition means that when two coins are tossed simultaneously, there must be only one head. There must not be more than one head.
Two outcomes (Outcome 2 and 4) satisfies this condition. So the probability = ²⁄₄ = ¹⁄₂ = 50%
(v) At least one head: This condition means that, when two coins are tossed simultaneously, there must be atleast one head. So even if there are two heads, the condition will be satisfied. Three outcomes (outcomes 1, 2 and 4) satisfies this condition. So the probability = ³⁄₄ = 75%
(vi) No head: This condition means that when two coins are tossed simultaneously, there must not be any head. Only one outcome (Outcome 3) satisfies this condition. So the probability = ¹⁄₄ = 25% 

Solved example 1.12
Three unbiased coins are tossed simultaneously. Find the probability of getting (i) All heads (ii) two heads (iii) One head (iv) at least two heads

Solution:
Step 1: Write the outcomes:
When three coins are tossed simultaneously, the possible outcomes can be written in the form of a table:

The above 8 are the only possible outcomes. This table can be used for all 4 cases of the question.
(i) All heads: Analysing each outcome, we find that only one outcome (Outcome 1) satisfies this condition. So we can write: The probability of getting all heads = ¹⁄₈ = 12.5%
(ii) Two heads: Three outcomes (outcomes 2, 6 and 8) satisfies this condition. So the probability = ³⁄₈ = 37.5%
(iii) One head: Three outcomes (Outcome 3, 5 and 7) satisfies this condition. So the probability = ³⁄₈ = 37.5%
(iv) At least two heads: This condition means that when three coins are tossed simultaneously, there must be atleast two heads. So even if there are 3 heads, the condition will be satisfied.
Four outcomes (Outcome 1, 2, 6 and 8) satisfies this condition. So the probability = ⁴⁄₈ = ¹⁄₂ = 50%

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In the next section we will see some more solved examples

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Chapter 1.5 - Probability

In the previous section we completed the discussion on Pie charts. In this section we will discuss about Probability.
In our day to day life, we use words related to 'probability' many times. Some examples are:
• There is a high probability that a question may be asked from this particular topic in the examination.
• Probably it will rain today. The sky is a bit cloudy
• It is less probable that we will get the car serviced by today evening as they had promised. Some workers are on leave.

In the above statements, we notice that there is an element of chance. An event may or may not occur. In those situations, one can not say for certain that an event ‘will’ occur. Neither can he say that an event ‘will not’ occur.
• A topic may be important. But the question sheet may not contain any questions from it.
• It may appear that the ‘rain will soon start’ on a cloudy day. But a wind may blow away the clouds, and it will become a sunny day. On the other hand, the wind can bring in more clouds and it may soon begin to rain.
• The car may not be ready, as some workers are on leave. But other workers might do ‘over time jobs’, and the car might be ready in time.

So we see that there is probability for an occurrence and non-occurrence. In the topic of ‘probability’, we try to predict an occurrence or non-occurrence by making simple mathematical models. First we will see the ‘degree of probability’.

On a very cloudy day there is a high probability that it will rain. The chance of a wind blowing away the clouds is low. So
• The degree of probability is high for one event. → The occurrence of rain

Here is another example:
A bag has 2 red balls and 5 blue balls. All the balls are identical except the colours. The balls are well shuffled and one ball is drawn with out looking. There is a high probability that, the drawn ball will be blue because the number of blue balls is greater. The chance for the drawn ball to be red is low. Then
• The degree of probability is high for one event. → The drawn ball to be blue.

In the two examples that we saw above, one event has a high degree of probability. So that event has an upper hand. For us, the other event, which is less probable, is inferior (even though, breaking all expectations, the inferior event ‘can’ indeed occur). Our aim is to show or present this high degree in a mathematical way.

But there are some situations where no event is inferior. All the possible events have equal chances of occurrence. Some examples are:
Tossing of a coin. Here, the two possible events are:
• The coin lands with Heads on the upper face
• The coin lands with Tails on the upper face
Both these two events have equal probability. None of them have an upper hand.

Rolling an unbiased cubical die. It has 6 numbers, from 1 to 6, with one number marked on each side. There are six possible events. They are: The die lands with
• 1 on the upper face
• 2 on the upper face
• 3 on the upper face
• 4 on the upper face
• 5 on the upper face
• 6 on the upper face

Each of the above six events have equal probability. None of them have an upper hand on the others because, it is an unbiased die. We must be able to present such equal probability also in a mathematical way.

Let us see how the mathematical formulation can be done. We will consider an experiment with the bag of balls that was mentioned earlier. The bag contains 2 red balls and 5 blue balls. All the balls are identical except the colours. Let us mark the two red balls as R1 and R2. And the 5 blue balls as B1, B2, B3, B4 and B5. When a person draws a ball with out looking, any one of the results (or outcomes) given below will occur:
• Outcome 1: The drawn ball is R1
• Outcome 2:The drawn ball is R2
• Outcome 3:The drawn ball is B1
• Outcome 4:The drawn ball is B2
• Outcome 5:The drawn ball is B3
• Outcome 6:The drawn ball is B4
• Outcome 7:The drawn ball is B5

● No outcomes other than the above 7 can possibly occur. So we say that the maximum number of outcomes possible is equal to 7.
● We want to present the high probability for the ‘occurrence of drawing a blue ball’. If a blue ball is drawn, we will call it an event. Let us examine each outcome:

Outcome 1: The drawn ball is R1 → Not a favourable outcome → We don’t have an event
Outcome 2: The drawn ball is R2 → Not a favourable outcome → We don’t have an event
Outcome 3: The drawn ball is B1 → A favourable outcome → We have an event
Outcome 4: The drawn ball is B2 → A favourable outcome → We have an event
Outcome 5: The drawn ball is B3 → A favourable outcome → We have an event
Outcome 6: The drawn ball is B4 → A favourable outcome → We have an event
Outcome 7: The drawn ball is B5 → A favourable outcome → We have an event

So, out of the 7 possible outcomes, 5 are favourable, and gives us an event. So,the probability for the occurance of the event (which is the ‘drawing of a blue ball’) is 5 in 7. In mathematical form, we write: The probability for the event of drawing a blue ball = ⁵⁄₇

We want to know the probability for the red ball also. Then, drawing of a red ball is the event. The steps can be written as:

Outcome 1: The drawn ball is R1 → A favourable outcome → We have an event
Outcome 2: The drawn ball is R2 → A favourable outcome → We have an event
Outcome 3: The drawn ball is B1 → Not a favourable outcome → We don’t have an event
Outcome 4: The drawn ball is B2 → Not a favourable outcome → We don’t have an event
Outcome 5: The drawn ball is B3 → Not a favourable outcome → We don’t have an event
Outcome 6: The drawn ball is B4 → Not a favourable outcome → We don’t have an event
Outcome 7: The drawn ball is B5 → Not a favourable outcome → We don’t have an event

So, out of the 7 possible outcomes, 2 are favourable, and gives us an event. So,the probability for the occurance of the event (which is the ‘drawing of a red ball’) is 2 in 7. In mathematical form, we write: The probability for the event of drawing a red ball = ²⁄₇

Now, we compare the quantities that we derived:
⁵⁄₇ is greater than ²⁄₇ . That is., ⁵⁄₇ > ²⁄₇.
The comparison can be presented in the form of a pie chart also. We learned how to draw pie chart in the previous section. For our present case, it will be as shown in the fig.1.38 below:

Fig.1.38 Probability for Red and blue ball

Thus, the high probability of the blue ball is proved.

[In the above experiment, it is important to note that the experiment is done only once. That is., the ball is drawn only once. We wrote a large number of sequential steps. Each of those steps is a ‘possible’ outcome in that single experiment. It should not be misunderstood that each outcome is an experiment.]

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Now let us consider the experiment of tossing a coin. We want the probability for Heads. [This experiment is also done only once. That is., the coin is tossed only once] The possible outcomes are:

Outcome 1: The coin lands with Heads on the upper face.
Outcome 2: The coin lands with Tails on the upper face.

● No outcomes other than the above 2 can possibly occur. So we say that the maximum number of outcomes possible is equal to 2.
● We want to present the probability for ‘getting heads’. If we get heads, we will call it an event. Let us examine each outcome:

Outcome 1: The coin lands with Heads on the upper face. → A favourable outcome → We have an event.
Outcome 2: The coin lands with Tails on the upper face. → Not a favourable outcome → We don’t have an event.

So, out of the 2 possible outcomes, 1 is favourable, and gives us an event. So,the probability for the occurance of the event (which is ‘getting heads’) is 1 in 2. In mathematical form, we write: The probability for the event of getting heads = ¹⁄₂

We want to know the probability for the tails also. Then, getting tails is the event. The steps can be written as:

Outcome 1: The coin lands with Heads on the upper face. Not a favourable outcome. We don’t have an event.
Outcome 2: The coin lands with Tails on the upper face. A favourable outcome. We have an event.

So, out of the 2 possible outcomes, 1 is favourable, and gives us an event. So,the probability for the occurance of the event (which is ‘getting tails’) is 1 in 2. In mathematical form, we write: The probability for the event of getting tails = ¹⁄₂

Now, we compare the quantities that we derived:
Both are equal to ¹⁄₂. So the equal probability for heads and tails is proved. The presentation in the form of a pie chart is given below:

Fig.1.39 Probability for Heads or Tails

In the next section, we will consider the rolling of a die.

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