Chapter 36 - Probability - Part III

In the previous section we completed a discussion on Polynomials. In this section, we will see Probability.

• We have already discussed the basic principles of probability.
    ♦ Probability part I consists of chapters 1.5, 1.6, . . . up to 1.9
    ♦ Probability part II consists of chapters 28, 28.1, . . . up to 28.4
• Our present discussion is a continuation from part II. 
• So the reader might want to revisit parts I and II thoroughly before taking up the present discussion.
• In part I, we saw theoretical probability
• In part II, we saw experimental probability
• We saw numerous examples in both the above two cases
• Let us recall some of them:
    ♦ While tossing a coin, the theoretical probability of getting head (or tail) is ¹⁄₂
    ♦ While rolling a die, the theoretical probability of getting 1 (or 2, 3, 4, 5, 6) is ¹⁄₆
■ But in real life, we never get such exact values.
• If we get exact values, the following situations will occur:
    ♦ If the coin is tossed ten times, heads will occur exactly 5 times and tails will occur exactly 5 times
    ♦ If a die is rolled 18 times, each number from 1 to 6 will occur exactly 3 times
• Such real life situations were discussed in part II.
    ♦ We saw that when a coin was tossed 30 times, no. of heads was not 15. It was 16
    ♦ So the experimental probability for heads is ¹⁶⁄₃₀ = 0.533 
    ♦ We saw that when a die was rolled 60 times, number '1' did not appear 10 times. It appeared 9 times
    ♦ So the experimental probability for is ⁹⁄₆₀ = 0.15

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■ In part II, we also saw that if the number of experiments increase, the experimental probability will get closer and closer to the theoretical probability
• For example, the eighteenth century French naturalist Comte de Buffon tossed a coin 4040 times and got 2048 heads. 
    ♦ So the experimental probability of getting a head = ²⁰⁴⁸⁄₄₀₄₀ = 0.507
• J.E. Kerrich, from Britain, tossed a coin 10000 times and got 5067 heads.
    ♦ So the experimental probability of getting a head = ⁵⁰⁶⁷⁄₁₀₀₀₀ = 0.5067
• Statistician Karl Pearson tossed a coin 24000 times and got 12012 heads.
    ♦ So the experimental probability of getting a head = ¹²⁰¹²⁄₂₄₀₀₀ = 0.5005
■ Note how the values are changing:
0.507 > 0.5067 > 0.5005
• The probability is getting closer and closer to the theoretical probability of 0.5000

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So it is obvious:
■ When the number of trials increases, the experimental probability approaches the theoretical probability
■ Let us see how we can put this into practical use:
Consider the following situation:
• A bag contains 3 red balls and 4 blue balls.
• For playing a game, a person would obviously choose the colour blue. Because, it has the upper hand
• We know how to write this 'upper hand' mathematically:
    ♦ The probability of getting a red ball is ³⁄₇
    ♦ The probability of getting a blue ball is ⁴⁄₇
• So blue has an upper hand
■ But what if we do not know the exact number of reds and blues in the bag?
• Then we will not be able to obtain ³⁄₇ or ⁴⁄₇

    ♦ That is., we will not be able to obtain theoretical probability values

• In such a situation, experiments (such as the ones we did in part II) will help us
• In our present case, the following 3 steps constitute one cycle (or one trial) of the experiment:
(i) Shuffling the balls well
(ii) Taking a ball with out looking
(iii) Recording the result in the note book.
• When the number of trials increases, the person doing the experiment will begin to feel the tendency:
    ♦ Blue will move towards ⁴⁄₇
    ♦ Red will move towards ³⁄₇
• So the person doing the experiment will begin to 'feel the upper hand possessed by the blue'
    ♦ Note that, he still does not know the number of each colour in the bag.
• The exact ⁴⁄₇ and ³⁄₇ will be obtained only if the number of cycles is infinity
• But as the number of cycles increase, accuracy will also increase

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Consider a similar situation:
■ There are two candidates in an election. Who will win?
• There is no value for theoretical probability.
• But interviewing the voters will give a tendency
• 'Interviewing one voter' is 'one trial'. 
• If a large number of voters are interviewed, the accuracy will increase.

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• So the relation between theoretical probability and experimental probability is obvious. 
• Today, many fields like Science, Engineering, Sociology etc., uses the principles of Probability to solve many problems. 
• But to apply them in such real life situations, we need to have a deep understanding on the topics of probability and statistics as a whole. 
• In this chapter we will see some more basic principles of probability. We will see more in higher classes.

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• In this chapter, we will be discussing about Theoretical probability
• We will first do some problems (some of which we have already done in parts I and II) and will write the results using 'more official looking' notations.
• We have seen the definition of Theoretical probability in part I
Let P(E) be the theoretical probability for an event to occur. From what we saw in part I, we can write:
P(E) = ^{Number of outcomes favourable to E}⁄_{Number of all outcomes}

Example 1
Find the probability of getting a head when a coin is tossed once. Also find the probability of getting a tail.
Solution:
• We want the probability for Heads. Let us denote it as P(E)
    ♦ The experiment is 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.
• We say that: The number of all possible outcomes is equal to 2
    ♦ So in the denominator, we will have '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.
• Thus, out of the 2 possible outcomes, 1 is favourable, and gives us an event.
    ♦ So in the numerator, we will have '1'
• So P(E) = ^{Number of outcomes favourable to E}⁄_{Number of all outcomes} = ¹⁄₂
■ We want to know the probability for the tails also. Then, getting tails is the event. Let us denote it as P(F). The steps can be written as:
• 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.
• We say that: The number of all possible outcomes is equal to 2
    ♦ So in the denominator, we will have '2'
• We want to present the probability for ‘getting tails’.
■ If we get tails, we will call it an event. Let us examine each outcome:
• 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.
• Thus, out of the 2 possible outcomes, 1 is favourable, and gives us an event.
    ♦ So in the numerator, we will have '1'
• So P(F) = ^{Number of outcomes favourable to E}⁄_{Number of all outcomes} = ¹⁄₂

Example 2
A bag contains a red ball, a blue ball and a yellow ball, all the balls being of the same size. Kritika takes out a ball from the bag without looking into it. What is the probability that she takes out the
(i) yellow ball? (ii) red ball? (iii) blue ball?
Solution:
Part (i):
• We want the probability for yellow ball. Let us denote it as P(Y)
    ♦ The experiment to find out P(Y) is done only once. The following three steps constitute one experiment:
(i) Shuffling the balls well
(ii) Taking a ball with out looking
(iii) Recording the result in the note book.
• The possible outcomes are:
    ♦ Outcome 1: The ball taken is yellow
    ♦ Outcome 2: The ball taken is red
    ♦ Outcome 3: The ball taken is blue
• No outcomes other than the above 3 can possibly occur.
• We say that: The number of all possible outcomes is equal to 3
    ♦ So in the denominator, we will have '3'
• We want to present the probability for ‘getting yellow’.
■ If we get yellow, we will call it an event. Let us examine each outcome:
• Outcome 1: The ball taken is yellow. → A favourable outcome → We have an event.
• Outcome 2: The ball taken is red. → Not a favourable outcome → We don’t have an event.
• Outcome 3: The ball taken is blue. → Not a favourable outcome → We don’t have an event.
• Thus, out of the 3 possible outcomes, 1 is favourable, and gives us an event.
    ♦ So in the numerator, we will have '1'
• So P(Y) = ^{Number of outcomes favourable to Y}⁄_{Number of all outcomes} = ¹⁄₃.
Part (ii):
• We want the probability for red ball. Let us denote it as P(R)
    ♦ The experiment to find out P(R) is done only once. The following three steps constitute one experiment:
(i) Shuffling the balls well
(ii) Taking a ball with out looking
(iii) Recording the result in the note book.
• The possible outcomes are:
    ♦ Outcome 1: The ball taken is yellow
    ♦ Outcome 2: The ball taken is red
    ♦ Outcome 3: The ball taken is blue
• No outcomes other than the above 3 can possibly occur.
• We say that: The number of all possible outcomes is equal to 3
    ♦ So in the denominator, we will have '3'
• We want to present the probability for ‘getting red’.
■ If we get red, we will call it an event. Let us examine each outcome:
• Outcome 1: The ball taken is yellow. → Not a favourable outcome → We don’t have an event.
• Outcome 2: The ball taken is red. → A favourable outcome → We have an event.
• Outcome 3: The ball taken is blue. → Not a favourable outcome → We don’t have an event.
• Thus, out of the 3 possible outcomes, 1 is favourable, and gives us an event.
    ♦ So in the numerator, we will have '1'
• So P(R) = ^{Number of outcomes favourable to R}⁄_{Number of all outcomes} = ¹⁄₃.
Part (iii):
• We want the probability for blue ball. Let us denote it as P(B)
    ♦ The experiment to find out P(B) is done only once. The following three steps constitute one experiment:
(i) Shuffling the balls well
(ii) Taking a ball with out looking
(iii) Recording the result in the note book.
• The possible outcomes are:
    ♦ Outcome 1: The ball taken is yellow
    ♦ Outcome 2: The ball taken is red
    ♦ Outcome 3: The ball taken is blue
• No outcomes other than the above 3 can possibly occur.
• We say that: The number of all possible outcomes is equal to 3
    ♦ So in the denominator, we will have '3'
• We want to present the probability for ‘getting blue’.
■ If we get blue, we will call it an event. Let us examine each outcome:
• Outcome 1: The ball taken is yellow. → Not a favourable outcome → We don’t have an event.
• Outcome 2: The ball taken is red. → Not a favourable outcome → We don’t have an event.
• Outcome 3: The ball taken is blue. → A favourable outcome → We have an event.
• Thus, out of the 3 possible outcomes, 1 is favourable, and gives us an event.
    ♦ So in the numerator, we will have '1'
• So P(B) = ^{Number of outcomes favourable to B}⁄_{Number of all outcomes} = ¹⁄₃.

Example 3
Suppose we throw a die once. (i) What is the probability of getting a number greater than 4? (ii) What is the probability of getting a number less than or equal to 4?
Solution:
Part (i):
• We want the probability for getting a number greater than 4. Let us denote it as P(E)
    ♦ The experiment to find out P(E) is done only once. That is., rolling the die is done only once.
• The possible outcomes are:
    ♦ Outcome 1: The die lands with 1 on upper face
    ♦ Outcome 2: The die lands with 2 on upper face
    ♦ Outcome 3: The die lands with 3 on upper face
    ♦ Outcome 4: The die lands with 4 on upper face
    ♦ Outcome 5: The die lands with 5 on upper face
    ♦ Outcome 6: The die lands with 6 on upper face
• No outcomes other than the above 6 can possibly occur.
• We say that: The number of all possible outcomes is equal to 6
    ♦ So in the denominator, we will have '6'
• We want to present the probability for ‘getting  a number greater than 4’.
■ If we get  a number greater than 4, we will call it an event. Let us examine each outcome:
• Outcome 1: The die lands with 1 on upper face. → Not a favourable outcome → We don’t have an event.
• Outcome 2: The die lands with 2 on upper face. → Not a favourable outcome → We don’t have an event.
• Outcome 3: The die lands with 3 on upper face. → Not a favourable outcome → We don’t have an event.

• Outcome 4: The die lands with 4 on upper face. → Not a favourable outcome → We don’t have an event.

• Outcome 5: The die lands with 5 on upper face. → A favourable outcome → We have an event.
• Outcome 6: The die lands with 6 on upper face. → A favourable outcome → We have an event.

• Thus, out of the 6 possible outcomes, 2 are favourable, and gives us an event. 

    ♦ So in the numerator, we will have '2'
• So P(E) = ^{Number of outcomes favourable to E}⁄_{Number of all outcomes} = ²⁄₆ = ¹⁄₃.
Part (ii):
• We want the probability for getting a number less than or equal to 4. Let us denote it as P(F)
    ♦ The experiment to find out P(F) is done only once. That is., rolling the die is done only once.
• The possible outcomes are:
    ♦ Outcome 1: The die lands with 1 on upper face
    ♦ Outcome 2: The die lands with 2 on upper face
    ♦ Outcome 3: The die lands with 3 on upper face
    ♦ Outcome 4: The die lands with 4 on upper face
    ♦ Outcome 5: The die lands with 5 on upper face
    ♦ Outcome 6: The die lands with 6 on upper face
• No outcomes other than the above 6 can possibly occur.
• We say that: The number of all possible outcomes is equal to 6
    ♦ So in the denominator, we will have '6'
• We want to present the probability for ‘getting a number less than or equal to 4’.
■ If we get  a number less than or equal to 4, we will call it an event. Let us examine each outcome:
• Outcome 1: The die lands with 1 on upper face. → A favourable outcome → We have an event.
• Outcome 2: The die lands with 2 on upper face. → A favourable outcome → We have an event.
• Outcome 3: The die lands with 3 on upper face. → A favourable outcome → We have an event.
• Outcome 4: The die lands with 4 on upper face. → A favourable outcome → We have an event.
• Outcome 5: The die lands with 5 on upper face. → Not a favourable outcome → We don’t have an event.

• Outcome 6: The die lands with 6 on upper face. → Not a favourable outcome → We don’t have an event.

• Thus, out of the 6 possible outcomes, 4 are favourable, and gives us an event. 

    ♦ So in the numerator, we will have '4'
• So P(F) = ^{Number of outcomes favourable to F}⁄_{Number of all outcomes} = ⁴⁄₆ = ²⁄₃.

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We have seen three examples. This is a good time to learn some new terms
■ An event having only one favourable outcome in an experiment is called an elementary event
1. In example 1, we defined an event 'E'
• In that experiment, there are 2 possible outcomes
    ♦ But only one outcome is favourable for 'E'
• So 'E' is an elementary event
• We also defined an event 'F'
• In that experiment, there are 2 possible outcomes
    ♦ But only one outcome is favourable for 'F'
• So 'F' is an elementary event
2. In example 2, we defined an event 'Y'
• In that experiment, there are 3 possible outcomes
    ♦ But only one outcome is favourable for 'Y'
• So 'Y' is an elementary event
• We also defined an event 'R'
• In that experiment, there are 3 possible outcomes
    ♦ But only one outcome is favourable for 'R'
• So 'R' is an elementary event
• We also defined an event 'B'
• In that experiment, there are 3 possible outcomes
    ♦ But only one outcome is favourable for 'B'
• So 'B' is an elementary event

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We can see some interesting results:
• In example 1, [P(E) + P(F)] = [¹⁄₂ + ¹⁄₂] = 1
• In example 2, [P(Y) + P(R) + P(B)] = [¹⁄₃ + ¹⁄₃ + ¹⁄₃] = 1
In an experiment, the sum of the probabilities of all elementary events is equal to 1
We will write it as a theorem:
Theorem 36.1
• Let E, F, G, . . . be the elementary events in an experiment
• Let P(E), P(E), P(E), . . .  be the probabilities of those elementary events
■ Then we will get:
P(E) + P(F) + P(G) + . . . = 1

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3. In example 3, we defined an event 'E'
• In that experiment, there are 6 possible outcomes
    ♦ Two outcomes are favourable for 'E'
• So 'E' is not an elementary event
• We also defined an event 'F'
• In that experiment, there are 6 possible outcomes
    ♦ Two outcomes are favourable for 'F
• So 'F' is not an elementary event'

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Let us see some more interesting results. We will write it in steps:
1. Consider E. It is the event of getting a number greater than 4
• Consider F. It is the event of getting a number less than or equal to 4
2. But 'getting a number greater than 4' is same as 'not getting a number less than or equal to 4'
• So if event E occurs, F will not occur
• So 'probability of the occurrence of E' is same as the 'probability of 'non-occurrence of F'
• That is., P(E) = P(not F)
    ♦ 'not F' is denoted as '(F)'
• So we can write: P(E) = P(F)
3. Similarly, 'getting a number less than or equal to 4' is same as 'not getting a number greater than 4'
• So if event F occurs, E will not occur
• So 'probability of the occurrence of F' is same as the 'probability of 'non-occurrence of E'
• That is., P(F) = P(not E)
    ♦ 'not E' is denoted as '(E)'
• So we can write: P(F) = P(E)
4. The event E, representing ‘not E’, is called the complement of the event E.
    ♦ We also say that E and E are complementary events.
• Similarly, the event F, representing ‘not F’, is called the complement of the event F.
    ♦ We also say that F and F are complementary events.
5. Now we will try to find the relation between the following two:
• Probability of an event
• Probability of it's complementary event
6. Let us add the two and find the sum. That is., we want the following sum:
P(E) + P(E)
• In the example 3, we obtained P(E) as ¹⁄₃
• But we did not obtain P(E)
• However, we saw in step (3) that, P(F) = P(E)
• So P(E) = P(F) = ²⁄₃.
• Thus the sum P(E) + P(E) = (¹⁄₃ + ²⁄₃) = 1
7. Now let us add P(F) and P(F). That is., we want the following sum:
P(F) + P(F)
• In the example 3, we obtained P(F) as ²⁄₃
• But we did not obtain P(F)
• However, we saw in step (2) that, P(E) = P(F)
• So P(F) = P(E) = ¹⁄₃.
• Thus the sum P(F) + P(F) = (²⁄₃ + ¹⁄₃) = 1
8. From steps (6) and (7) we see that the sum of the following two quantities is 1:
• Probability of an event
• Probability of it's complementary event
Example: P(E) + P(E) = 1 
9. So if we know the probability of an event, we can readily calculate the probability of 'that event not occurring'.
• All we need to do is: Subtract the probability from 1
That is.,  P(E) = 1 - P(E)

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Some more results:
In the example 3 above, we saw the probabilities related to the rolling of a die.
What is the probability of getting '8' in a single roll of the die?
Solution:
• It is specially mentioned that, the '8' should be obtained in a 'single roll of a die'. 
• Why is it specially mentioned? 
• Let us analyse:  
• Many often a player will want to get a number '8'.
    ♦ He may be able to obtain it if two dice are rolled together
    ♦ He may be able to obtain it if he is allowed to roll a die more than once
■ But in our problems these are not allowed. That is why it is specially mentioned 'single roll of a die'.
Let us write the steps:
• We want the probability for getting number 8. Let us denote it as P(E)
    ♦ The experiment to find out P(E) is done only once. That is., rolling the die is done only once.
• The possible outcomes are:
    ♦ Outcome 1: The die lands with 1 on upper face
    ♦ Outcome 2: The die lands with 2 on upper face
    ♦ Outcome 3: The die lands with 3 on upper face
    ♦ Outcome 4: The die lands with 4 on upper face
    ♦ Outcome 5: The die lands with 5 on upper face
    ♦ Outcome 6: The die lands with 6 on upper face
• No outcomes other than the above 6 can possibly occur.
• We say that: The number of all possible outcomes is equal to 6
    ♦ So in the denominator, we will have '6'
• We want to present the probability for ‘getting number 8’.
■ If we get  number 8, we will call it an event. Let us examine each outcome:
• Outcome 1: The die lands with 1 on upper face. → Not a favourable outcome → We don’t have an event.
• Outcome 2: The die lands with 2 on upper face. → Not a favourable outcome → We don’t have an event.
• Outcome 3: The die lands with 3 on upper face. → Not a favourable outcome → We don’t have an event.

• Outcome 4: The die lands with 4 on upper face. → Not a favourable outcome → We don’t have an event.

• Outcome 5: The die lands with 5 on upper face. → Not a favourable outcome → We don’t have an event.

• Outcome 6: The die lands with 6 on upper face. → Not a favourable outcome → We don’t have an event.

• Thus, out of the 6 possible outcomes, 0 are favourable. 
    ♦ So in the numerator, we will have '0'
• So P(E) = ^{Number of outcomes favourable to E}⁄_{Number of all outcomes} = ⁰⁄₆ = 0
■ An event whose probability is zero is called an impossible event.

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We are discussing the probabilities related to the rolling of a die.
What is the probability of getting a number less than 7?
Solution:
Let us write the steps:
• We want the probability for getting number less than 7. Let us denote it as P(F)
    ♦ The experiment to find out P(F) is done only once. That is., rolling the die is done only once.
• The possible outcomes are:
    ♦ Outcome 1: The die lands with 1 on upper face
    ♦ Outcome 2: The die lands with 2 on upper face
    ♦ Outcome 3: The die lands with 3 on upper face
    ♦ Outcome 4: The die lands with 4 on upper face
    ♦ Outcome 5: The die lands with 5 on upper face
    ♦ Outcome 6: The die lands with 6 on upper face
• No outcomes other than the above 6 can possibly occur.
• We say that: The number of all possible outcomes is equal to 6
    ♦ So in the denominator, we will have '6'
• We want to present the probability for ‘getting number less than 7’.
■ If we get  a number less than 7, we will call it an event. Let us examine each outcome:
• Outcome 1: The die lands with 1 on upper face. → A favourable outcome → We have an event.
• Outcome 2: The die lands with 2 on upper face. → A favourable outcome → We have an event.
• Outcome 3: The die lands with 3 on upper face. → A favourable outcome → We have an event.
• Outcome 4: The die lands with 4 on upper face. → A favourable outcome → We have an event.
• Outcome 5: The die lands with 5 on upper face. → A favourable outcome → We have an event.
• Outcome 6: The die lands with 6 on upper face. → A favourable outcome → We have an event.

• Thus, out of the 6 possible outcomes, 6 are favourable. 

    ♦ So in the numerator, we will have '6'
• So P(F) = ^{Number of outcomes favourable to E}⁄_{Number of all outcomes} = ⁶⁄₆ = 1
■ An event whose probability is 1 is called a sure event or a certain event.

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From the above discussion, we get a range for P(E). It can be explained as follows: 
• P(E) can be greater than or equal to zero
    ♦ Zero if E is an impossible event
    ♦ Greater than zero if E is not an impossible event
• P(E) can be less than or equal to one
    ♦ One if E is a certain event
    ♦ Less than one if E is not a certain event
■ Note that, P(E) can never be greater than one
• Because, the numerator will always be less than or equal to the denominator. This is because, 'number of favourable outcomes' cannot be greater than the 'total number of possible outcomes'  
• So we can write: 0 ≤ P(E) ≤ 1

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In this section, we saw 3 examples and learned some new terms related to Probability. In the next section, we will see more examples.

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chapter 28.3 - Solved examples on Experimental probability

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

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• 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

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■ 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

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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 

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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:

Outcome	1	2	3	4	5	6
Frequency	179	150	157	149	175	190

Find the probability of getting each outcome
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

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:

Digit	0	1	2	3	4	5	6	7	8	9
Frequency	22	26	22	22	20	10	14	28	16	20

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:

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.

Distance (km)	less than 4000	4000 to 9000	9001 to 14000	more than 14000
Frequency	20	210	325	445

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:

Unit test	I	II	III	IV	V
Percentage of marks obtained	69	71	73	68	74

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

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In the next section, we will see a few more solved examples.

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