Random Experiment

An experiment, whose all possible outcomes are known in advance but the outcome of any specific performance can not predicted before the completion of the experiment, is known as random experiment. An example of random experiment might be tossing of a coin.


A set of all possible outcomes associated with same random experiment is called sample-space and is usually denoted by ‘S’. Consider the experiment of tossing a die. If we are interested in the number that shows on the top face, then sample space would be

S1 = {1, 2, 3, 4, 5, 6}


An event is a subset of sample – space.
In any sample space we may be interested in the occurrence of certain events rather than in the occurrence of a specific element in the sample space.

Simple Event

If an event is a set containing only one element of the sample-space, then it is called a simple event.

Compound Event

A compound event is one that can be represented as a union of sample points.

For instance, the event of drawing a heart from a deck of cards is the subset A ={heart} of the sample space S = {heart, spade, club, diamond}. Therefore A is a simple event. None the event B of drawing a red card is a compound event since B = {heart U diamond} = {heart, diamond}.


If a random experiment can result in any one of N different equally likely outcomes, and if exactly n of these outcomes favors to A, then the probability of event A, P (A) = n/N
i.e. favourbale cases/total no. of cases.


  1. If the probability of certain event is one, it doesn’t mean that event is going to happen with certainty! Infact we would be just predicting that, the event is most likely to occur in comparison to other events. Predictions depend upon the past information and of course also on the way of analysing the information at hand!!
  2. Similarly if the probability of certain event is zero, it doesn’t mean that, the event can never occur!


Example -:  Two fair coins are tossed. What is the probability that atleast one head occurs?

Sol:                The sample space ‘S’ for this experiment is, S = {HH, HT, TH, TT}

As the coin is fair all these outcomes are equally likely. Let ‘w’ be the weight assigned to any sample point then w + w + w + w = 1 Þ w = ¼

If ‘A’ is the event representing the event of atleast one head occuring,

Then   A = {HT, TH, HH}  Þ P(A) = 3/4

Example -: In a single cast with two fair dice, what is the chance of throwing

(i) two 4¢s                                         (ii) a doublet

(iii) five – six                                                 (iv) a sum of 7

Sol:                  (i) There are 6 ´ 6 equally likely cases ( as any face of any die may turn up)

=> 36 possible outcomes. For this event, only one outcome (4-4) is favourable \ probability = 1/36).

(ii)A doublet can occur in six ways {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}.

Therefore probability of doublet = 6/36 = 1/6.

(iii) Two favourable outcomes {(5, 6), (6, 5)}

Therefore probability = 2/36 = 1/18

(iv) A sum of 7 can occur in the following cases {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} which are 6 in number. Therefore probability = 6/36 = 1/6


Example -: Seven accidents occur in a week. What is the probability that they happen on the same day.

Sol:                  Total no. of cases=Total no. of ways in which 7 accidents can happen in a week (or be distributed)= 77

Favourable no. of cases out of these = number of those in which all 7 happen on one day (any day of the week) = 7

\ Required. probability = 7/77

Example -:       From a set of 17 cards 1, 2, 3, ……16, 17, one is drawn at random. Find the probability that number on the drawn card would be divisible by 3 or 7.

Sol:                   Numbers which are divisible by three are 3, 6, 9, 12, 15.  Similarly numbers which are divisible by 7 are 7, 14.

=>  Probability of the number written on the card to be divisible by 3 or 7 is

= 7/17

Click Here for Next Topic »

CBSE Class 12 Maths Probability All Topic Notes CBSE Class 12 Maths All Chapters Notes

MIT Pune University Admission Apply Now!!

Leave a Reply