Tuesday, February 27, 2018

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 12
    ♦ While rolling a die, the theoretical probability of getting 1 (or 2, 3, 4, 5, 6) is 16
■ 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 1630 = 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 960 = 0.15

■ 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 = 20484040 = 0.507
• J.E. Kerrich, from Britain, tossed a coin 10000 times and got 5067 heads.
    ♦ So the experimental probability of getting a head = 506710000 = 0.5067
• Statistician Karl Pearson tossed a coin 24000 times and got 12012 heads.
    ♦ So the experimental probability of getting a head = 1201224000 = 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

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 37
    ♦ The probability of getting a blue ball is 47
• 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 3or 47
    ♦ 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 47
    ♦ Red will move towards 37
• 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 4and 3will be obtained only if the number of cycles is infinity
• But as the number of cycles increase, accuracy will also increase

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.

• 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.

• 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 ENumber 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 ENumber of all outcomes = 12
■ 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 ENumber of all outcomes = 12

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 YNumber of all outcomes = 13.
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 RNumber of all outcomes = 13.
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 BNumber of all outcomes = 13.

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 ENumber of all outcomes = 213.
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 FNumber of all outcomes = 423.

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

We can see some interesting results:
• In example 1, [P(E) + P(F)] = [112] = 1
• In example 2, [P(Y) + P(R) + P(B)] = [1113] = 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

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'

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 13
• But we did not obtain P(E)
• However, we saw in step (3) that, P(F) = P(E)
• So P(E) = P(F) = 23.
• Thus the sum P(E) + P(E) = (13 + 23) = 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 23
• But we did not obtain P(F)
• However, we saw in step (2) that, P(E) = P(F)
• So P(F) = P(E) = 13.
• Thus the sum P(F) + P(F) = (23 + 13) = 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)

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 ENumber of all outcomes = 0= 0
■ An event whose probability is zero is called an impossible event.

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 ENumber of all outcomes = 6= 1
■ An event whose probability is 1 is called a sure event or a certain event.

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



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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