Anoutcomeis the result of a single trial of a probability experiment.Sample space is the set of all possible outcomes of a probability experiment.There are finite as well as infinite sample spaces. While Finite sample spaces have a countable number of possible outcomes, infinite sample spaces have uncountable number of possible outcomes.Finite sample spaces of simple experiments can be displayed by listing all possible outcomes.

Examples:ExperimentSample Space

Tossing a coin{Head, Tail}

Throwing a six faced die{1,2,3,4,5,6}

Answering a true or false question(True, False}

Choosing an odd digit.{0.2.4.6.8}

Infinite sample spaces are written using set builder notation.

Example:Suppose the height requirement for a recruitment is between 62 inches and 80 inches, then the sample space can be written asS = { x | 62 x 80}Back to TopTree diagrams are used to determine the sample space when the probability experiment consists of two or more activities or occurrences.Definition:A tree diagram is a method used to find all possible events of a probability experiment, where the outcomes are connected with their starting points by arrows.Examples:A High School Literary committee of 5 members consists of 3 Seniors and 2 Juniors. Find the number of ways a two member team for a Quiz contest consisting of one Junior and one Senior can be selected.

The above tree diagram depicts the selection process. One Senior can be selected from three. And for each of this possible selection a Junior can be added in two ways. Thus there are 6 outcomes in the sample space, which are listed as ordered pairs in the diagram.

The following tree diagram shows the method to find the genders of children in a family.

Each child in order can be a Boy or Girl. These possibilities are combined in order and the sample space consists of 8 possible outcomes which are listed in the diagram.Back to TopTheoretical or classical probability uses sample space to evaluate the probability of events numerically. Theoretical probability assumes that all outcomes in the sample space are equally likely to occur.Thus the probability of the occurrence of an event E is given by the formula,P(E) =n(E)n(S)Where n(E) is the number of outcomes favorable to E and n(S) is the number of outcomes in the sample space S.

Probability of the sample space is theprobability of the occurrence of any one outcome of the sample space. Extending the above formula the probability of sample space is given by

P(S) =n(S)n(S)= 1Solved ExamplesQuestion1:Find the probability of turning an even number up when a six sided fair die is thrown.Solution:Let the E be the event of turning up an even number. Then the outcomes in the sample space and E are as follows:S = { 1, 2, 3, 4, 5, 6 ) 6 outcomes in sample spaceE = { 2, 4, 6 } 3 outcomes in the event E.

P(E) =n(E)n(S)=36=12.

The number of outcomes in the sample space is often determined using combinatorics formulas like permutation and combination, without actually listing the outcomes.Question2:A card is drawn at random from a pack of 52 cards. Find the number of elements in the sample space and also determine the probability of drawing a red card.Solution:As we need to pick one card from 52, this is a case of combination.Hence the number of elements in the sample space n(S) = 52C1= 52.Let E be the event of picking a red card. As this red card has to come from 26 cards (13 Hearts + 13 Diamonds)Number of elements in the event E n(E) = 26C1= 26.

Probability of picking a red card P(E) =n(E)n(S)=2652=12.Back to Top1. Sometimes the sample space can be written in tabular form, like the outcomes when two dice are rolled. Here each outcome is an ordered pair as follows: Die IDie II

123456

1(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)

2(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)

3(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)

4(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)

5(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)

6(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)

It can be seen there are 36 elements in the sample space.

2. A security system uses a five digit code using the numbers 0 to 9. If the digits can be repeated, write a rule to define the sample space and find number of elements in the sample space. The sample space consists of five digit numbers as follows 00000, 01234, .................. 10756, 11111..................... The sample space can be defined as follows: S = { xxxxx | x {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} As repetition of digits is allowed, each digit can be chosen in 10 ways Hence the number of elements in sample space = 10 x 10 x 10 x 10 x 10 = 100,000.

Back to TopWhat is Permutation?The different arrangements which can be made out of a given number of things by taking some or all at a time are called permutations.

Example:1. What are thetwo digit numbers that can be formed using the digits 2,5,7.Solution:Here out of three digits, we formed two digit numbers, (i.e) we form numbers by taking two digits at a time.The numbers that can be formed are, 25, 27, 52, 57, 72, 75.2. What are the different numbers that can be formed using all the digits, 9, 8 and 5 .Solution: Here we form numbers by taking all at a time.(i. e) we form three digit numbers by taking all the digits at a time.The numbers that can be formed are 589, 598, 859, 895, 958 and 985

Hence we observe that in permutations there will be selection and then arrangements.

What is Combination?Each of the different groups or selections which can be formed by taking some or all of a number of objects irrespective of their arrangements, is called a combination.

Example:1. John as three pens each of blue, red and green. In how many ways two pens can be selected from the three pens.Solution:John as three colored pens each of blue, red and green. The different ways of selecting two pens are,blue and red ; red and green (or) blue and green. Therefore there arethree waysof selecting two pens from three pens.2. A bag has an yellow marble, black marble and a blue marble. In how many ways three marbles can be selected?Solution:Since the bag contains three marbles each of yellow, black and blue, the selection of thee marbles contain all the three colors yellow, black and blue. Therefore, the selection can be doneonly one way.

Hence we observe that in combinations there will be only selection (and no arrangement).Back to TopFactorial:Factorial notation is used to express the product of first n natural numbers.(i. e)n! = 1.2.3.4 . . . . . . . . . . nExample : 5 ! = 1.2.3.4.5 = 12010 ! = 1.2.3.4.5.6.7.8.9.10 = 3628800

Permutations Formula:Formula 1:The permutations of n objects by taking r at a time is,P (n, r)= nPr = n (n-1) (n-2) . . . . . . . . (n - r + 1)

Formula 2:The above permutation can be expressed using factorial notation as follows.P (n, r)= nPr =n!(nr)!

Example:In how many ways three digits numbers can be formed using the digits, 3, 4, 5, 7 and 9.Solution:We have 5 digits, 3, 4, ,5, 7 and 9.The number of three digit numbers that can be formed is permutation of five things taken three at a time.(i. e),5P3= 5 ( 5 - 1) (5 - 2) [ by Formula 1, nPr = n (n-1) (n-2) . . . . . . (n - r + 1) ]= 5 . 4 . 3=60

Factorial Method:Using the factorial formula, we have nPr =n!(nr)!

P (5, 3) =5P3=5!(53)!

=5.4.3.2.12.1

= 5.4.3=60

There will be 60 three digit numbers that can be formed.

Combinations Formula:Formula 1:Combination of n objects by taking r at a time is nCr = C (n, r) =nPrr!Formula 2:Combination of n objects by taking r at a time is nCr = C (n, r) =n!(nr)!r!

Example:In how many ways 3 balls can be selected from a box containing 10 balls.Solution:Since the above situation is only selected we have combination.Therefore, C (n, r) = C(6, 3)=nPrr!

=6P33!

=6.5.43.2.1

=20(or) C (6, 3) =n!(nr)!r!

=6!3!3!

=1.2.3.4.5.61.2.3.1.2.3

= 20Back to TopSolved ExamplesQuestion1:How many four letter words, with or without meaning can be formed out of the letters of the word, "MATHEMAGIC", if repetition of letters is not allowed.Solution:The word "MATHEMAGIC" contain the letters, M, A, T, H, E, G, I, C which are 8 letters.Here we need to select four letters and arrange them.Hence we have selection and arrangement which is permutation.Therefore the required number of words = Permutations of 8 letters by taking four at a time= P (8, 4)= 8 (8 - 1) (8 - 2) (8 - 3)= 8 . 7. 6. 5= 1680Question2:In how many ways can 10 books be arranged on a shelf so that a particular pair of books shall bea. always together.b. never together.Solution:a.Since a particular pair of books is always together.If we keep these together as one pair, then we have to arrange 9 books on the shelf.This can be done in P (9, 9) ways =9 ! waysSince the pair of books together can be arranged in2 ways,the total number of ways of arranged the ten books so that a particular pair of books is always together =2 x 9 !

b.The ten books can be arranged in P (10, 10) ways = 10 ! waysFrom (a), the ten books can be arranged in 9! . 2 ways by arranging particular pair of books together.Therefore, the number of ways of arranging the 10 books so that a particular pair is never together is,=10 ! - 2 x 9 != 10 x 9 ! - 2 x 9!= (10 - 2) x 9 != 8 x 9 !the number of ways of arranging the 10 books so that a particular pair is never together is = 8 x 9 !Question3:How many diagonals are there in an octagon?Solution:A polygon of 8 sides has 8 vertices.By joining any two of these vertices, we obtain either a side or a diagonal of the polygon.Here we have only a selection (and no arrangement) , hence we have combination.Number of all straight lines obtained by joining 2 vertices at a time= C (8 , 2)=8!6!.2!

=8.7.6.5.4.3.2.11.2.3.4.5.6.1.2

= 28Since the number of sides = 8,Number of diagonals of the octagon = 28 - 8= 20Back to TopPractice ProblemsQuestion1:In how many ways 7 boys and 5 girls be arranged for a group photograph, if the girls are to sit on chairs in a row and the boys are to stand in a row behind them?Question2:Find the number of ways in which the letters of the word, "LAPTOP", can be arranged such that the vowels occupy only even position.Question3:How many three digit numbers can be formed with the digits, 3, 4, 5, 6, 7, 8, when the digits may be repeated any number of times in any arrangement.Question4:Out of 6 men and 4 women a committee of 54 is to be formed containing at least one woman. In how many ways the committee can be formed.Question5:A code word is to consists of two distinct alphabets followed by two distinct numbers between 1 and 5. How may such code words are there?Sample Space Tree diagram

Sample Space Formula

Sample Space Examples

Permutations and Combinations Formula

Permutations and Combinations Examples

Permutation vs Combination

Permutations and Combinations Practice