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
■ 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
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
• 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
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 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.
• 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.
• So P(F) = Number of outcomes favourable to F⁄Number of all outcomes = 4⁄6 = 2⁄3.
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)] = [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
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 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)
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 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.
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.
• 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.
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.
• 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
■ 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
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
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 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.
• 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.
• 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.
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)] = [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
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 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)
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.♦ 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.
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.
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
