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PROBABILITY 15.
It is remarkable that a science, which began with the consideration of games of chance, should be elevated to the rank of the most important subject of human knowledge. —Pierre Simon Laplace
15.1 INTRODUCTION
In everyday life, we come across statements such as
It will probably rain today.
I doubt that he will pass the test.
Most probably, Kavita will stand first in the annual examination.
Chances are high that the prices of diesel will go up.
There is a 5050 chance of India winning a toss in today's match.
The words 'probably', 'doubt', 'most probably', 'chances', etc., used in the statements above involve an element of uncertainty. For example, in (1), 'probably rain' will mean it may rain or may not rain today. We are predicting rain today based on our past experience when it rained under similar conditions. Similar predictions are also made in other cases listed in (2) to (5).
The uncertainty of 'probably' etc can be measured numerically by means of 'probability' in many cases.
Though probability started with gambling, it has been used extensively in the fields of Physical Sciences, Commerce, Biological Sciences, Medical Sciences, Weather Forecasting, etc.
The concept of probability developed in a very strange manner. In 1654, a gambler Chevalier de Mere, approached the wellknown 17th century French philosopher and mathematician Blaise Pascal regarding certain dice problems. Pascal became interested in these problems, studied them and discussed them with another French mathematician, Pierre de Fermat. Both Pascal and Fermat solved the problems independently. This work was the beginning of Probability Theory.
The first book on the subject was written by the Italian mathematician, J.Cardan(1501–1576). The title of the book was 'Book on Games of Chance' (Liber de Ludo Aleae), published in 1663. Notable contributions were also made by mathematicians J. Bernoulli (1654–1705), P. Laplace (1749–1827), A.A. Markov (1856–1922) and A.N. Kolmogorov (born 1903).
Blaise Pascal
(16231662)
Fig. 15.1
Pierre de Fermat
(1601–1665)
Fig. 15.2
In earlier classes, you have had a glimpse of probability when you performed experiments like tossing of coins, throwing of dice, etc., and observed their outcomes. You will now learn to measure the chance of occurrence of a particular outcome in an experiment.
Activity 1
Take any coin, toss it ten times and note down the number of times a head and a tail come up. Record your observations in the form of the following table.
Table 15.1
Number of times the coin is tossed
Number of times head comes up
10

Write down the values of the following fractions:
Toss the coin twenty times and in the same way record your observations as above. Again find the values of the fractions given above for this collection of observations.
You will find that as the number of tosses gets larger, the values of the fractions come closer to 0.5. To record what happens in more and more tosses, the following group activity can also be performed:
Activity 2
Divide the class into groups of 2 or 3 students. Let a student in each group toss a coin 15 times. Another student in each group should record the observations regarding heads and tails. [Note that coins of the same denomination should be used in all the groups. It will be treated as if only one coin has been tossed by all the groups.]
Now, on the blackboard, make a table like Table 15.2. First, Group 1 can write down its observations and calculate the resulting fractions. Then Group 2 can write down its observations, but will calculate the fractions for the combined data of Groups 1 and 2, and so on. (We may call these fractions as cumulative fractions.) We have noted the first three rows based on the observations given by one class of students.
Group (1)
 Number of heads (2)
 Number of tails (3)
 Cumulative number of heads/_{Total number of times the coin is tossed} (4)
Cumulative number of heads/_{Total number of times the coin is tossed} (4)
1
3
12
3/_{15}
12/_{15}
2
 7
 8
7 + 3/_{15 + 15} = ^{10}/_{30}
8 + 12/_{15 + 15} = ^{20}/_{30}
7 + 10/_{15 + 30} = ^{17}/_{45}
 8 + 20/_{15 + 30} = ^{28}/_{45}
 4

What do you observe in the table? You will find that as the total number of tosses of the coin increases, the values of the fractions in Columns (4) and (5) come nearer and nearer to 0.5.
Activity 3

Throw a die* 20 times and note down the number of times the numbers 1, 2, 3, 4, 5, 6 come up. Record your observations in the form of a table, as in Table 15.3:
*A die is a well balanced cube with its six faces marked with numbers from 1 to 6, one number on one face. Sometimes dots appear in place of numbers.
Table 15.3
Number of times a die is thrown
Number of times these scores turn up
5
2
620
Find the values of the following fractions:
Now throw the die 40 times, record the observations and calculate the fractions as done in (i).
As the number of throws of the die increases, you will find that the value of each fraction calculated in (i) and (ii) comes closer and closer to ⅙ .
To see this, you could perform a group activity, as done in Activity 2. Divide the students in your class, into small groups. One student in each group should throw a die ten times. Observations should be noted and cumulative fractions should be calculated.
The values of the fractions for the number 1 can be recorded in Table 15.4. This table can be extended to write down fractions for the other numbers also or other tables of the same kind can be created for the other numbers.
Table 15.4 Cumulative number of times 1 turned up/_{Total number of times the die is thrown} (4)

The dice used in all the groups should be almost the same in size and appearence. Then all the throws will be treated as throws of the same die.What do you observe in these tables? You will find that as the total number of throws gets larger, the fractions in Column (3) move closer and closer to ⅙ .
Activity 4
Toss two coins simultaneously ten times and record your observations in the form of a table as given below: Table 15.5
Number of times the two coins are tossed Number of times no head comes up Number of times one head comes up
Number of times two heads comes up
 Write down the fractions:
 Now increase the number of tosses (as in Activitiy 2). You will find that the more the number of tosses, the closer are the values of A, B and C to 0.25, 0.5 and 0.25, respectively.
In Activity 1, each toss of a coin is called a trial. Similarly in Activity 3, each throw of a die is a trial, and each simultaneous toss of two coins in Activity 4 is also a trial.
So, a trial is an action which results in one or several outcomes. The possible outcomes in Activity 1 were Head and Tail; whereas in Activity 3, the possible outcomes were 1, 2, 3, 4, 5 and 6.
In Activity 1, the getting of a head in a particular throw is an event with outcome 'head'. Similarly, getting a tail is an event with outcome 'tail'. In Activity 2, the getting of a particular number, say 1, is an event with outcome 1.
If our experiment was to throw the die for getting an even number, then the event would consist of three outcomes, namely, 2, 4 and 6.
So, an event for an experiment is the collection of some outcomes of the experiment. In Class X, you will study a more formal definition of an event.
So, can you now tell what the events are in Activity 4?
With this background, let us now see what probability is. Based on what we directly observe as the outcomes of our trials, we find the experimental or empirical probability. Let n be the total number of trials. The empirical probability P(E) of an event E happening, is given by
P(E) = ^{Number of trials in which the event happened}/_{The total numebr of trials}
In this chapter, we shall be finding the empirical probability, though we will write 'probability' for convenience.
Let us consider some examples.
To start with let us go back to Activity 2, and Table 15.2. In Column (4) of this table, what is the fraction that you calculated? Nothing, but it is the empirical probability of getting a head. Note that this probability kept changing depending on the number of trials and the number of heads obtained in these trials. Similarly, the empirical probability of getting a tail is obtained in Column (5) of Table 15.2. This is ^{12}/_{15} to start with, then it is ^{2}/_{3} , then ^{28}/_{45}, and so on.
Activity 5
 Before going further, look at the tables you drew up while doing Activity 3. Find the probabilities of getting a 3 when throwing a die a certain number of times. Also, show how it changes as the number of trials increases.
 Now let us consider some other examples.
Example 1

A coin is tossed 1000 times with the following frequencies:
Head : 455, Tail : 545
Compute the probability for each event.
Solution
Since the coin is tossed 1000 times, the total number of trials is 1000. Let us call the events of getting a head and of getting a tail as E and F, respectively. Then, the number of times E happens, i.e., the number of times a head come up, is 455.
So, the probability of E = ^{Number of heads}/_{Total number of trials}
i.e. P (E) = ^{455}/_{1000} = 0.455
Similarly, the probability of the event of getting a tail = ^{Number of tails}/_{Total number of trials}
i.e. P(F) = ^{545}/_{1000} = 0.545
Note that in the example above, P(E) + P(F) = 0.455 + 0.545 = 1, and E and F are the only two possible outcomes of each trial. Example 2 Two coins are tossed simultaneously 500 times, and we getOne head : 275 times
No head : 120 times
Find the probability of occurrence of each of these events.
Let us denote the events of getting two heads, one head and no head by E1, E2 and E3, respectively. So,
P(E_{1}) = ^{105}/_{500} = 0.21
P(E_{3}) = ^{120}/_{500} = 0.24
 Observe that P(E_{1}) + P(E_{2}) + P(E_{3}) = 1. Also E_{1}, E_{2} and E_{3} cover all the outcomes of a trial.
 Example 3
A die is thrown 1000 times with the frequencies for the outcomes 1, 2, 3, 4, 5 and 6 as given in the following table :
Table 15.6
 Outcome
Frequency 
179
150
190
 Find the probability of getting each outcome.
 Let E_{i} denote the event of getting the outcome i, where i = 1, 2, 3, 4, 5, 6.
Then
 Probability of the outcome 1 = P(E_{1}) = ^{Frequency of 1}/_{Total number of times the die is thrown } = ^{179}/_{1000} = 0.179
Similarly, P(E_{2}) = ^{150}/_{1000} = 0.15, P(E_{3}) = ^{157}/_{1000} = 0.157,
P(E_{4}) = ^{149}/_{1000} = 0.149, P(E_{5}) = ^{175 }/_{1000} = 0.175 
and P(E_{6}) = ^{190}/_{1000 } = 0.19.
Note that P(E_{1}) + P(E_{2}) + P(E_{3}) + P(E_{4}) + P(E_{5}) + P(E_{6}) = 1
Also note that:
The probability of each event lies between 0 and 1.
E_{1}, E_{2}, . . ., E_{6} cover all the possible outcomes of a trial.
 Example 4
 On one page of a telephone directory, there were 200 telephone numbers. The frequency distribution of their unit place digit (for example, in the number 25828573, the unit place digit is 3) is given in Table 15.7 :
 Table 15.7
Digit
0
9

22
26
14
28
Without looking at the page, the pencil is placed on one of these numbers, i.e., the number is chosen at random. What is the probability that the digit in its unit place is 6?
The probability of digit 6 being in the unit place = ^{Frequency of 6}/_{Total number of selected phone numbers} = ^{14}/_{200} = 0.07
You can similarly obtain the empirical probabilities of the occurrence of the numbers having the other digits in the unit place.
Example 5
The record of a weather station shows that out of the past 250 consecutive days, its weather forecasts were correct 175 times.
What is the probability that on a given day it was correct?
What is the probability that it was not correct on a given day?
The total number of days for which the record is available = 250P(the forecast was correct on a given day) = ^{Number of days when forecast was correct}/_{total number of days which the record is available} = ^{175}/_{250} = 0.7The number of days when the forecast was not correct = 250 – 175 = 75
So, P(the forecast was not correct on a given day) = ^{75}/_{250} = 0.3
Notice that:
Example 6
A tyre manufacturing company kept a record of the distance covered before a tyre needed to be replaced. The table shows the results of 1000 cases.
Table 15.8
Distance (in km)
less than 4000
40009000
900114000
more than 14000
210
445
If you buy a tyre of this company, what is the probability that :
it will need to be replaced before it has covered 4000 km?
it will last more than 9000 km?
it will need to be replaced after it has covered somewhere between 4000 km and 14000 km?
The total number of trials = 1000.
The frequency of a tyre that needs to be replaced before it covers 4000 km is 20.
The frequency of a tyre that will last more than 9000 km is 325 + 445 = 770
 So, P(tyre will last more than 9000 km) = ^{770}/_{1000} = 0.77
 The frequency of a tyre that requires replacement between 4000 km and 14000 km is 210 + 325 = 535.
So, P(tyre requiring replacement between 4000 km and 14000 km)= ^{535}/_{1000} = 0.535
 Example 7
The percentage of marks obtained by a student in the monthly unit tests are given below:  Table 15.9
Unit test
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.  The total number of unit tests held is 5.
The number of unit tests in which the student obtained more than 70% marks is 3.  So, P(scoring more than 70% marks) = ^{3}/_{5} = 0.6
Example 8
To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table.
 Opinion
 Number of students
Like
135  Dislike
 65
 Find the probability that a student chosen at random
likes statistics,  does not like it.
 Refer to Q.2, Exercise 14.2. What is the empirical probability that an engineer lives: less than 7 km from her place of work? more than or equal to 7 km from her place of work? within km from her place of work?