Sample space
Set of all possible outcomes of an experiment or trial is called sample space and is denoted by S
Consider the following examples
 Tossing of a coin
The outcome will be either heads or tails
so S = {H , T}
n(S) = 2
 Tossing of 2 coins
If 2 coins are tossed each can give two results
so S = { HT , HH , TT, TH}
n(S) = 4
 Tossing of a dice
S = { 1,2,3,4,5,6 }
n(S) = 6
Events
Consider example 2 stated above
Now we may be interested in some happening. Consider the outcome that the pair has one heads and one tails
so E = { HT , TH ) so n(E) = {2}
Thus the chance that an event happens is called Probability
Formula P(E) = n(E) / n(S)
Solved Examples
 7 books on physics and 4 on mathematics are to be arranged at random on a shelf. Find the probability that the books on maths do not stand next to each other
Solution
11 books can be selected in ^{11}P_{11} ways
n(S) = 11!
Let B be the event that no 2 maths book stand next to each other
Look at the following arrangement
M P M P M P M P M P M P M P M
7 books on physics can be arranged in ^{7}P_{7}= 7!ways
The 4 maths books can take any of the 8 positions shown above.
n(B) = ^{7}P_{7}x^{8}P_{4}
P(B) = n(B)/n(S) = 7!x8x7x6x5/11!
P(B) = 7/33
 The letters of the word 'E Q U A T I O N' are arranged at random . Find the probability that the arrangement
a)Starts with a vowel and ends with a consonant.
b)Starts with a vowel and ends with a vowel 5/14.
Solution
n(s) = ^{8}P_{8}= 8!
Consider the first case(a)
Let A be the event that the word starts with a vowel and ends with a consonant.
As there are 5 vowels in the word the first position can be filled in ^{5}P_{1}=5 ways.
As there are 3 consonants in the word the last position can be filled in ^{3}P_{1}=3 ways.
So far we have taken care of 2 positions.
*       *
Now the 6 spaces inbetween can be filled in ^{6}P_{6}=6! ways.
So n(A) = 5x6!x3
So P(A) = 5x6!x3/8! = 15/56.
Consider the second case (b)
Let B be the event that the word starts with a vowel and ends with a vowel.
As there are 5 vowels in the word the first position can be filled in ^{5}P_{1}=5 ways.
With 4 vowels remaining the last the last position can be filled in sup>4P_{1}=4 ways.
*       *
Now the 6 spaces inbetween can be filled in ^{6}P_{6}=6! ways.
So n(B) = 5x6!x4/8!
So P(B) = 5x6!x4/8!
 Tom and Dick play with 2 dice. If Tom throws 9 what is Dick's chance of throwing a higher number?
Solution
Total number of outcome of throwing 2 dice = 6x6 = 36.
Dick must throw either 10,11,12. This can be done in 5 ways.
(4+6,5+6,6+4,6+5,6+6)
Dicks chance of throwing a higher number is 5/36.
 A 3 digit number is to be formed by using the digits 1,2,3,4,5,6,7,8,9.(Repetions not allowed). What is the probability that the digit formed is greater than 600.
Solution
The possibility of forming 3 digit numbers is ^{9}P_{3} = 504.
n(S)= 504.
Let A be the event that the number is greater 600.
n(A)= 4x^{8}P_{2} = 224.
P(A)= 224/504.
 A bag contains 3 white , 4 black and 2 red balls. 2 balls are chosen at random. What is the probability that 1 white and 1 red ball is chosen.
Solution
2 balls can be drawn from 9 balls in ^{9}C_{2} ways = 36 ways
n(S)= 36
n(A)= ^{3}C_{1}.^{2}C_{1} = 6
P(A) = 6/36 = 1/6
