 random experiment statisticians use the word experiment to describe any process that generates a set of data a simple example of a statistical experiment is tossing of a coin in this experiment there are only two outcomes heads and tails another experiment might be the launching of a missile and observing its velocity at specified times the opinions of voters concerning a new sales tax can also be considered as observations of an experiment another very common experiment is throwing a die we are particularly interested in the observations obtained by repeating the experiment several times in most cases the outcomes will depend on chance and therefore cannot be predicted with certainty when a coin is tossed repeatedly we cannot predict whether a given toss will be head or tail but we know the entire set of possibilities for each toss next I would like to talk about sample space the set of all possible outcomes of a statistical experiment is called the sample space and is represented by the symbol capital S if the sample space has finite number of elements we may list the members separated by commas and enclosed by branches thus if we choose the experiment of tossing a coin the sample space will be written as capital S1 here it is like this capital S1 equal to the set containing H, T where H and T correspond to heads and tails respectively consider the experiment of throwing a die the sample will be capital S2 which is set of 6 numbers 1 2 3 4 5 6 we are interested in the number that shows on the top face if we are interested only in whether the number is even or odd the sample space is simply if the set containing even and odd this example shows that for particular experiment more than one sample space is possible so the sample space for a particular experiment depends on the situation or some particular problem now let us take one example here an experiment of flipping a coin and then flipping it second time if a head occurs if a tail occurs on the first flip then a die is tossed once if we see the figure 1 below that is this one we see that the top branch moving along the first path gives the sample point H H indicating the possibility of that head occurs on two successive flips of the coin so let us see the figure here here the first outcome is head so if the coin is tossed we have two outcomes head and tail so this if head occurs then the coin is tossed second time so if the first one is head then from this node we have two branches head and tail so finally we will have two sample points H H and HT now according to the problem if tail occurs for the first time then a die is thrown so from this tail node we will have six possibilities 1 2 3 4 5 and 6 so finally we will have six sample points T 1 T 2 T 3 T 4 T 5 and T 6 so for this particular experiment we will have the sample space S which is the set containing the sample points H H HT T 1 T 2 T 3 T 4 T 5 and T 6 another example suppose three items are selected at random from a manufacturing process the items are identified as defective which is denoted by capital T or non defective which is denoted by capital N here also the possible outcomes can be represented by a tree diagram which is given here so once the first item is chosen we will have two outcomes D and N now if it is D then the second item is chosen and it will have two branches D and N again so we will have DD or DN now suppose the second item is also defective so we will follow this path so third item is chosen again we will have two outcomes D and N so if we have the first item defective second item also defective then finally the sample sample point will be DDD DDN now if we go along this path that means the first item is defective second item is non defective so the third item is chosen so it will have two outcomes D and N so ultimately we will have two sample points D and D DNN in this way if the first item is non defective then the second item is chosen and we will have two outcomes D and N first and first one is non defective and second one is defective so we are going along this path we will have again two outcomes D and N so finally we will have two sample points N DD and N DNN now the last case that is the first item is non defective second item is also non defective so we are going along this path third item is selected it will have two outcomes D and N so finally we will have two outcomes again N N D N N N so from this experiment we have the sample space is which is the set of the sample points DDD DDN DND DNN NDD NDN NND NNN with large or infinite number of sample points are best described by a statement or a rule for example if the possible outcomes of an experiment are the set of pupil in India with an age more than 65 years our sample space is written as capital S which is the set of X such that X is a person with an age more than 65 years similarly if S is the set of all points XY inside the circle with radius 2 and center at the origin we write is that is the sample space is the set of the points XY such that X square plus Y square less than 4 next is events what is an event an event is a subset of sample space it is denoted by capital A capital B etc so we can define random experiment sample space events in this way also an experiment is a process that yields outcome an event is an outcome or combination of outcomes from an experiment of course and what is the sample space the sample space is an event consisting of all possible outcomes so sample space is the event consisting of all possible outcomes now in the previous example 2 if we consider the event A that the number of defectives is greater than 1 that means the event A can be written as D the set containing DDD DNT NDD DDN so this is an event because this is a subset of the sample space is given in example 2 that is this one let us consider another example here given the sample space is that is the set containing X such that X is a person with an age more than 65 years the event A that the person is an woman so the event is subset A that is the set containing X such that X is a woman with an age more than 65 years so this is the event here this is also a subset of the sample space S which is this one here we can easily say that an event may be a subset that includes the entire sample space S or a subset of S called the null set and denoted denoted by the symbol Phi which contains no elements at all so let us consider more examples here describe a sample space that might be appropriate for an experiment in which we roll a pair of dice one red one green so the solution will be like this the sample space that provides the most information consists of the 36 points given by S1 which is the set containing the points XY such that X will take the value from 1 to 6 Y will take the value from 1 to 6 so the sample space is like this the sample space will contain 36 points because X is from 1 to 6 and Y is from 1 to 6 where X represents the number turned up by the red die and Y represents the number turned up by the green die now for this particular experiment let us describe the event B that the total number of points rolled with the pair of dice is 7 so in this case among 36 possibilities only 1 6 2 5 3 4 4 3 5 2 and 6 1 so this 6 cases are favorable so the event B will contain this 6 points this 6 sample points another example is like this if someone takes three shots a target we care whether each shot is a hit a miss describe suitable sample space for this case so let us try the solution in this way if we let 0 and 1 represent miss a hit respectively the 8 possibilities 0 0 0 1 0 0 0 1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1 0 1 0 1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 1, 1, 1 may be displayed as in the following figure, so this axis is corresponding to first short this one is corresponding to second short and this one is corresponding to third short. So these are the sample points, so the sample points here can be displayed by the following figure, so this is the sample space for this particular experiment. Now let us find the event M that the person will miss the target 3 times in a row and the event N that the person will hit the target once and miss it twice, so the event M will be, so it will contain only the sample point 0, 0, 0 because it will miss, so it is corresponding to the event that the person will miss the target 3 times, so that is why it will be 0, 0, 0. Now the next event N will be 1, 0, 0, 0, 1, 0, 0, 0, 1, so the set containing this 3 points because it is the event that the person will hit the target once and miss it twice. Let us consider one more example in this context construct a sample space for length of the useful life of a certain electronic components and indicate the subset that represents the event F that the component fails before the end of the 6th year, so here the sample space we have to find, so let us write this in this way, if is the length of the components useful life in years, the sample space can be written as S that is the set of T such that T is greater than equal to 0 and the subset F will be the set of T such that 0 less than equal to T less than 6, so this is the event that the component fails before the end of the 6th year. So in this way we can find the sample space and the events and some events for a particular experiment, now let us consider some more events, the complement of an event A with respect to S is the subset of all elements of S that are not in A. We denote the complement of A by the symbol A prime it is like this, so in this context let us discuss one example, let R be the event that a red card is selected from an ordinary deck of 52 playing cards and let S be the entire deck. Then R prime is the event that the card selected from the deck is not a red but black card. Because we know that in a deck of cards there are only 2 possibilities either it will be red or black, so that is why the complement of the event R will be a set of cards, so it will be the event that the card selected from the deck is not a red but black card. Let us consider a sample space S that is the set of 6 integers 1, 2, 3, 4, 5, 6 now consider 2 subsets A that is the set of containing 2, 4, 6 and B containing 4, 5, 6 so B is the set 4, 5 containing 4, 5, 6 so here the subset 4, 6 so 4, 6 this is a subset of A this is a subset of B, so this subset 4, 6 that is the set containing 4 and 6 is the intersection of A and B, so it is common to A as well as B, so this is called the intersection of A and B, so if we have a sample space S and we have 2 events A and B then A intersection B is another event which contains all elements that are common to A and B, so in this way combining various events we can form many events. Let us consider another example here let P be the event that a person selected at random while dining at a popular restaurant is a smoker and let Q be the event that the person is a smoker is sorry let Q be the event that the person is over 55, 50 years of age. So P be the event that a person selected at random while dining at a popular restaurant is a smoker and Q be the event that the person is over 50 years of age. Then the event P intersection Q is the set of all smokers of the restaurant who are over 50 years of age, so P intersection Q is it should be in P as well as in Q, so that is why it will have both the cases that is they should be smokers they should be over 50 years of age. Consider the random experiment of throwing a die the event A is of occurring odd numbers on the top face and event B is of occurring even numbers on the top face. So A that is A is the set event A that will contain 3 points 1, 3 and 5 and B will contain 2, 4 and 6 if we observe we see that there is no common element because we have considered 2 cases one is the first one is of odd numbers second one is of even numbers. So that is why there is 2 common element in A and B and that is why we can say that A intersection B is equal to 5 in this case the events A and B are called as mutually exclusive events. So we can define the mutually exclusive events in this way 2 events A and B are mutually exclusive events if they do not have any common element or another way we can say it in this way that if A intersection B is equal to 5 the union of the 2 events A and B denoted by the symbol A union B is the event containing all the elements that belong to A or B or both. So this is another kind of event another kind of combination so we have 2 events A and B so we are combining these 2 events in this way that the resulting event contains all the elements that belong to A or B or both so this is denoted as A union B so union between this 2 events. So let us consider this example here let A equal to the set ABC and B the set containing ABCDE then A union B will be ABCDE because it will contain all the elements that belong to A or B or both so that is why it will be the set containing ABCDE the relationship between events and the corresponding sample space can be represented by figure 3 and figure 4 let us consider figure 3 first here we have a sample space S and in this sample space we have 3 events A B and C so we have specified the regions different regions this is one region which is denoted by 1 this is another region which is denoted by 2 3 4 5 6 7 like this. So A intersection B if we have to find this particular event so this event corresponds to the regions 1 and 2 because A intersection B is the common part between A and B so that is why if we say this here this is the common portion between A and B so that is why it will correspond to regions 1 and 2 if we have to find B intersection C so we have to find the common portion between B and C and this is the common region here so this will be regions 1 and 3 in this way if we have to obtain A union C so the event A union C will correspond to the regions 1 2 3 4 5 and 7 because A union C so this whole portion we have to consider so it will be 1 2 3 4 5 and 7 now let us find B prime intersection A what is B prime this is complement of the event B so complement of event B will be this portion this whole portion so B prime intersection A so we have to take the intersection of this with A so B prime intersection A so this will correspond to the regions 4 and 7 so this is the common between B prime and A now let us find A intersection B intersection C so that means we have to find the common region between A B and C between A and B we have this common region between B and C we have this common region between A and C we have this common region so if we take this 3 this is the common between A B and C so this region 1 is the region corresponding to A intersection B intersection C this event now let us find another event that is A union B intersection C prime so A union B is this whole region and we have to take the intersection with C prime so C prime is this portion so A union B intersection C prime will correspond to regions 2 6 and 7 in figure 4 we see that events A B and C are all subsets of the sample space S so this is the sample space S and A B C this 3 events are subsets of this sample spaces it is also clear that event B is a subset of the event A so B is the subset so it is inside A it is a subset of A event B intersection C has no elements they are disjoint and hence B and C are mutually exclusive event A intersection C has at least one element and A union B is equal to A because B is a subset of A figure 4 might therefore depict a situation where we select a card at random from an ordinary deck of 52 playing cards and observe whether the following events occur the event A is the event that the card is rate B is the event that the card is the jack queen or king of diamonds C is the event that the card is an S so the event A intersection C consists only of the two red asses so finally I would like to discuss some results which are related to this kind of combinations if we have two events A and Phi then A intersection Phi will be equal to Phi so because this is the common portion between A and Phi and Phi is the null set so that is why the resulting event will be a null set again if we combine A and Phi in this way that is by union that is A union Phi again Phi is the null set so A union Phi will become A only so the resulting event will be A A intersection A prime so A intersection A complement so if we consider the common portion between A and A prime naturally the resulting event will be the null set Phi A union A prime union between A and complement of A so we will have the whole sample space as the resulting event S prime that is the complement of the sample space so it will be the null set now if we consider Phi prime so it will be S now if we consider A complement of the complement of A that is A prime prime so first we are taking the complement of A then again we are taking the complement of that so we will get back A A intersection B prime A intersection B whole it is an event A intersection B and then we take the complement of A intersection B so this will be same as A complement union B complement so complement of A intersection B will be same as the event A complement union B complement A union B complement so complement of the event A union B so this is same as A complement intersection B complement so the resulting event will be