Probability - Experimental Approach



In everyday life, we come across statements such as

(1) It will probably rain today.
(2) I doubt that he will pass the test.
(3) Most probably, Kavita will stand first in the annual examination.
(4) Chances are high that the prices of diesel will go up.
(5) There is a 50-50 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.


Probability – an Experimental Approach

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.



The concept of probability developed in a very
strange manner. In 1654, a gambler Chevalier
de Mere, approached the well-known 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).



Number of times the coin is tossed

Number of times head comes up Number of times tail comes up
                          10                            ---                  ---

Write down the values of the following fractions:

«math xmlns=¨¨»«mfrac»«mrow»«mi»N«/mi»«mi»u«/mi»«mi»m«/mi»«mi»b«/mi»«mi»e«/mi»«mi»r«/mi»«mo»§nbsp;«/mo»«mi»o«/mi»«mi»f«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»i«/mi»«mi»m«/mi»«mi»e«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»a«/mi»«mo»§nbsp;«/mo»«mi»h«/mi»«mi»e«/mi»«mi»a«/mi»«mi»d«/mi»«mo»§nbsp;«/mo»«mi»c«/mi»«mi»o«/mi»«mi»m«/mi»«mi»e«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»u«/mi»«mi»p«/mi»«/mrow»«mrow»«mi»T«/mi»«mi»o«/mi»«mi»t«/mi»«mi»a«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«mi»n«/mi»«mi»u«/mi»«mi»m«/mi»«mi»b«/mi»«mi»e«/mi»«mi»r«/mi»«mo»§nbsp;«/mo»«mi»o«/mi»«mi»f«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»i«/mi»«mi»m«/mi»«mi»e«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»h«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mi»c«/mi»«mi»o«/mi»«mi»i«/mi»«mi»n«/mi»«mo»§nbsp;«/mo»«mi»i«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»o«/mi»«mi»s«/mi»«mi»s«/mi»«mi»e«/mi»«mi»d«/mi»«/mrow»«/mfrac»«/math»
«math xmlns=¨¨»«mfrac»«mrow»«mi»N«/mi»«mi»u«/mi»«mi»m«/mi»«mi»b«/mi»«mi»e«/mi»«mi»r«/mi»«mo»§nbsp;«/mo»«mi»o«/mi»«mi»f«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»i«/mi»«mi»m«/mi»«mi»e«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»a«/mi»«mo»§nbsp;«/mo»«mi»h«/mi»«mi»e«/mi»«mi»a«/mi»«mi»d«/mi»«mo»§nbsp;«/mo»«mi»c«/mi»«mi»o«/mi»«mi»m«/mi»«mi»e«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»u«/mi»«mi»p«/mi»«mo»§nbsp;«/mo»«/mrow»«mrow»«mi»T«/mi»«mi»o«/mi»«mi»t«/mi»«mi»a«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«mi»n«/mi»«mi»u«/mi»«mi»m«/mi»«mi»b«/mi»«mi»e«/mi»«mi»r«/mi»«mo»§nbsp;«/mo»«mi»o«/mi»«mi»f«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»i«/mi»«mi»m«/mi»«mi»e«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»h«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mi»c«/mi»«mi»o«/mi»«mi»i«/mi»«mi»n«/mi»«mo»§nbsp;«/mo»«mi»i«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»o«/mi»«mi»s«/mi»«mi»s«/mi»«mi»e«/mi»«mi»d«/mi»«/mrow»«/mfrac»«/math»

(ii) 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.
(iii) Repeat the same experiment by increasing the number of tosses and record the number of heads and tails. Then find the values of the corresponding fractions.
      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:

Acitivity  : 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. 


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 : (i) Throw a die* 20 times and note down the number of times the numbers

*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.

1, 2, 3, 4, 5, 6 come up. Record your observations in the form of a table, as in Table 15.3:


Number of times a die is thrown
Number of times these scores turn up
1 2 3 4 5 6

(ii) 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 «math xmlns=¨¨»«mfrac»«mn»1«/mn»«mn»6«/mn»«/mfrac»«/math»
    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.




Total number of times a die is thrown in a group


«math xmlns=¨¨»«mfrac»«mrow»«mi»c«/mi»«mi»u«/mi»«mi»m«/mi»«mi»u«/mi»«mi»l«/mi»«mi»a«/mi»«mi»t«/mi»«mi»i«/mi»«mi»v«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mi»n«/mi»«mi»u«/mi»«mi»m«/mi»«mi»b«/mi»«mi»e«/mi»«mi»r«/mi»«mo»§nbsp;«/mo»«mi»o«/mi»«mi»f«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»i«/mi»«mi»m«/mi»«mi»e«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mn»1«/mn»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»u«/mi»«mi»r«/mi»«mi»n«/mi»«mi»e«/mi»«mi»d«/mi»«mo»§nbsp;«/mo»«mi»u«/mi»«mi»p«/mi»«/mrow»«mrow»«mi»t«/mi»«mi»o«/mi»«mi»t«/mi»«mi»a«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«mi»n«/mi»«mi»u«/mi»«mi»m«/mi»«mi»b«/mi»«mi»e«/mi»«mi»r«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»i«/mi»«mi»m«/mi»«mi»e«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»h«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mi»d«/mi»«mi»i«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mi»i«/mi»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»h«/mi»«mi»r«/mi»«mi»o«/mi»«mi»w«/mi»«mi»n«/mi»«/mrow»«/mfrac»«/math»




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 inColumn (3) move closer and closer to «math xmlns=¨¨»«mfrac»«mn»1«/mn»«mn»6«/mn»«/mfrac»«/math»

Example :

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 «math xmlns=¨¨»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mrow»«mi»t«/mi»«mi»o«/mi»«mi»t«/mi»«mi»a«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«mi»n«/mi»«mi»u«/mi»«mi»m«/mi»«mi»b«/mi»«mi»e«/mi»«mi»r«/mi»«mo»§nbsp;«/mo»«mi»h«/mi»«mi»e«/mi»«mi»a«/mi»«mi»d«/mi»«mi»s«/mi»«/mrow»«mrow»«mi»t«/mi»«mi»o«/mi»«mi»t«/mi»«mi»a«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«mi»n«/mi»«mi»u«/mi»«mi»m«/mi»«mi»b«/mi»«mi»e«/mi»«mi»r«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»a«/mi»«mi»i«/mi»«mi»l«/mi»«mi»s«/mi»«/mrow»«/mfrac»«/math»

i.e.,                        P (E)  = «math xmlns=¨¨»«mfrac»«mn»455«/mn»«mn»1000«/mn»«/mfrac»«/math» =0.455

Similarly, the probability of the event of getting a tail = «math xmlns=¨¨»«mfrac»«mrow»«mi»n«/mi»«mi»u«/mi»«mi»m«/mi»«mi»b«/mi»«mi»e«/mi»«mi»r«/mi»«mo»§nbsp;«/mo»«mi»o«/mi»«mi»f«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»a«/mi»«mi»i«/mi»«mi»l«/mi»«mi»s«/mi»«/mrow»«mrow»«mi»t«/mi»«mi»o«/mi»«mi»t«/mi»«mi»a«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«mi»n«/mi»«mi»u«/mi»«mi»m«/mi»«mi»b«/mi»«mi»e«/mi»«mi»r«/mi»«mo»§nbsp;«/mo»«mi»o«/mi»«mi»f«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»a«/mi»«mi»i«/mi»«mi»l«/mi»«mi»s«/mi»«/mrow»«/mfrac»«/math»

i.e.,                                                                              P(F) =«math xmlns=¨¨»«mfrac»«mn»545«/mn»«mn»1000«/mn»«/mfrac»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»545«/mn»«/math»

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 : .

Two coins are tossed simultaneously 500 times, and we get
                                    Two heads : 105 times
                                      One head : 275 times
                                        No head : 120 times
Find the probability of occurrence of each of these events.
Solution : Let us denote the events of getting two heads, one head and no head by E1,E2 and E3, respectively. So,

                                   P(E1)=«math xmlns=¨¨»«mfrac»«mn»105«/mn»«mn»500«/mn»«/mfrac»«/math»=0.021

                                   P(E2)=«math xmlns=¨¨»«mfrac»«mn»275«/mn»«mn»500«/mn»«/mfrac»«/math»=0.055

                                   P(E3)=«math xmlns=¨¨»«mfrac»«mn»120«/mn»«mn»500«/mn»«/mfrac»«/math»=0.024

Observe that P(E1) + P(E2) + P(E3) = 1. Also E1, E2 and E3 cover all the outcomes of a trial.


Example :

Consider the frequency distribution table (Table 14.3, Example 4, Chapter 14), which gives the weights of 38 students of a class.
(i) Find the probability that the weight of a student in the class lies in the interval 46-50 kg.
(ii) Give two events in this context, one having probability 0 and the other having probability 1.

Solution :

(i) The total number of students is 38, and the number of students with weight in the interval 46 - 50 kg is 3.
So, P(weight of a student is in the interval 46 - 50 kg) =«math xmlns=¨¨»«mfrac»«mn»3«/mn»«mn»38«/mn»«/mfrac»«/math»= 0.079
(ii) For instance, consider the event that a student weighs 30 kg. Since no student has this weight, the probability of occurrence of this event is 0. Similarly, the probability
of a student weighing more than 30 kg is «math xmlns=¨¨»«mfrac»«mn»38«/mn»«mn»38«/mn»«/mfrac»«/math»= 1.



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