An experiment, in probability, is any process that generates a set of outcomes. One simple example of an experiment is tossing a coin. This is considered an experiment because it generates two possible outcomes, head and tail. Another example of an experiment is rolling a die. This is an experiment because it generates six possible outcomes, 1 through 6. A sample space is defined as the set of all possible outcomes. We write the sample space for tossing a coin as S = {Head, Tail} and the sample space for rolling a die as S = {1,2,3,4,5,6}. A sample point is an element of a sample space. Since the sample space is made up of outcomes, sample point is another word for ourcome.
A multiple-step experiment is several experiments rolled up into a single experiment. We call each of the individual experiment a step in the multiple-step experiment. For example, tossing 2 coins is a multiple-step experiment because it consists of two individual experiments, or steps. The first step is tossing the first coin and the second step is tossing the second coin. The counting rule for multiple-step experiments can be used to calculate the number of outcomes in a multiple-step experiment. It says that in a multiple-step experiment with k steps and n1 possible outcomes in the first step, n2 possible outcomes in the second step, and so on, the total number of possible outcomes is (n1)(n2)...(nk). A multiple-step experiment can be visualized with a graph known as a tree diagram.
The counting rule for combinations can be used to calculate the number of outcomes when an experiment involves selecting n objects from a larger set of N objects. For example, suppose you are selecting 2 objects from a set of 5 objects. If the objects are labeled A, B, C, D and E, there are 10 possible ways of doing this: AB, AC, AD, AE, BC, BD, BE, CD, CE and DE. The counting rule says that the number of different ways of selecting n objects from a set of N objects is equal to N!/n!(N-n)!. Plugging in 2 for n and 5 for N, we get N!/n!(N-n)! = 5!/2!(5-2)! = 10. We can see that using the counting rule gives the same number of outcomes as you get when you write out all the possible outcomes.
The counting rule for permutations is similar to the one for combinations. It can be used to calculate the number of outcomes when an experiment involves selecting n objects from a larger set of N objects, when the order of selection matters. For example, suppose you are selecting 2 objects from a set of 5 objects. If the objects are labeled A, B, C, D and E, there are 20 possible ways of doing this: AB, AC, AD, AE, BA, BC, BD, BE, CA, CB, CD, CE, DA, DB, DC, DE, EA, EB, EC AND ED. The counting rule says that the number of different ways of selecting n objects from a set of N objects, when order matters, is equal to N!/(N-n)!. Plugging in 2 for n and 5 for N, we get N!/(N-n)! = 5!/(5-2)! = 20. We can see that using the counting rule gives the same number of outcomes as you get when you write out all the possible outcomes.
When assigning probabilities, there are two requirements and three methods. The first requirement is that the probability of each outcome is between zero and one. The second requirement is that the sum of the probabilities of the outcomes is equal to one. The classical method is appropriate when all outcomes are equally likely. The relative frequency method is appropriate when the experiment has been repeated many times and there is data on the outcomes of the repetitions. The subjective method is appropriate when the outcomes are not equally likely and there is little to no data available.
An event is simply a collection of sample points (outcomes). An even can be defined any way that one likes. For example, consider the experiment of selecting a card from a deck of 52 playing cards. The event of getting an Ace is all the sample points corresponding to getting an ace. Since there are four suits in a deck of cards, there will be four sample points. Those sample points are ace of clubs, ace of hearts, ace of spades and ace of diamonds. To compute the probability of an event, simply sum the probabilities of the sample points in the event. So if you want to compute the probability of selecting an ace, sum up the probability of the sample points. Also, since each card has an equally likely change of being selected, then the probability of each card is 1/52. So P(ace) = P(Ac) + P(Ah) + P(As) + P(Ad) = 1/52 + 1/52 + 1/52 + 1/52 = 4/52 = 1/13.
Consider an event A. The complement of A, written Ac, is the even containing all sample points that are not in A. For example, consider the experiment of rolling a die. Recall that the sample points (outcomes) in this experiment are 1,2,3,4,5 and 6. Let's define event A as getting an odd number (1, 3, 5). Then the complement of A will be the even numbers (2, 4, 6). Since an event and its complement make up the entire sample space, the sum of their probabilities is equal to one. This relationship is useful because given the probabilitiy of the complement of an event, we can calculate the probability of the event itself. The formula for this is as follows P(A) = 1 - P(Ac).
Consider two events A and B. The union of A and B, written AUB, is the event containing all sample points in A or B or both. For example, consider an experiment with five possible outcomes: E1, E2, E3, E4, E5. Suppose A = {E1, E3} and B = {E2, E3, E5}. Then the union of A and B is AUB = {E1, E2, E3, E5}. The intersection of A and B is the event containing all sample points in both A and B. Consider the same experiment as before with events A and B defined in the same way. Then the intersection of A and B is {E3}. Union and intersection can be visualized clearly in a graph known as a venn diagram.
The addition law can be used to calculate the probability of the union of two events. It says that the probability of the union of two events is the sum of their probabilities minus the probability of their intersection. The reason that the intersection is subtracted in the addition law is that when summing the probabilities, the intersection is counted twice. Events are said to be mutually exclusive if they have no sample points in common. If they have no sample points in common, then there intersection is empty and thus the probability of their intersection is zero. So the addition law for mutually exclusive events will simple be the sum of the probabilities of the events.