Probability Topics: Terminology

Module by: Susan Dean, Barbara Illowsky, Ph.D.

Summary: This module defines several key terms related to Probability.

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Probability measures the uncertainty that is associated with the outcomes of a particular experiment or activity. An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, then the experiment is said to be a chance experiment. Flipping one fair coin is an example of an experiment.

The result of an experiment is called an outcome. A sample space is a set of all possible outcomes. Three ways to represent a sample space are to list the possible outcomes, to create a tree diagram, or to create a Venn diagram. The uppercase letter SS is used to denote the sample space. For example, if you flip one fair coin, S = {H, T}S = {H, T} where HH = heads and TT = tails are the outcomes.

An event is any combination of outcomes. Upper case letters like AA and BB represent events. For example, if the experiment is to flip one fair coin, event AA might be getting at most one head. The probability of an event AA is written P(A)P(A).

The probability of any outcome is the long-term relative frequency of that outcome. For example, if you flip one fair coin from 20 to 2,000 times, the relative frequency of heads approaches 0.5 (the probability of heads). Probabilities are between 0 and 1, inclusive (includes 0 and 1 and all numbers between these values). P(A) = 0P(A) = 0 means the event AA can never happen. P(A) = 1P(A) = 1 means the event AA always happens.

To calculate the probability of an event AA, count the outcomes for event A and divide by the total outcomes in the sample space. For example, if you toss a fair dime and a fair nickel, the sample space is {HH, TH, HT, TT}{HH, TH, HT, TT} where TT = tails and HH = heads. The sample space has four outcomes. If AA denotes the probability of getting one head, then there are two outcomes {HT, TH}{HT, TH}in the event. Thus, P(A) = 2 4 P(A) = 2 4 .

Equally likely means that each outcome of an experiment occurs with equal probability. For example, if you toss a fair, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face.

An outcome is in the event A OR BA OR B if the outcome is in AA or is in BB or is in both AA and BB. For example, let A = {1, 2, 3, 4, 5}A = {1, 2, 3, 4, 5} and B = {4, 5, 6, 7, 8}B = {4, 5, 6, 7, 8}. A OR B = {1, 2, 3, 4, 5, 6, 7, 8}A OR B = {1, 2, 3, 4, 5, 6, 7, 8}. Notice that 4 and 5 are NOT listed twice.

An outcome is in the event A AND BA AND B if the outcome is in both AA and BB at the same time. For example, let AA and BB be {1, 2, 3, 4, 5}{1, 2, 3, 4, 5} and {4, 5, 6, 7, 8}{4, 5, 6, 7, 8}, respectively. Then A AND B = {4, 5}A AND B={4, 5}.

The complement of event AA is denoted A'A' (read "A prime"). A'A' consists of all outcomes that are NOT in AA. Notice that P(A) + P(A') = 1P(A) + P(A') = 1. For example, let S = {1, 2, 3, 4, 5, 6}S = {1, 2, 3, 4, 5, 6} and let A = {1, 2, 3, 4}A = {1, 2, 3, 4}. Then, A' = {5, 6}. P(A) = 4 6 , P(A') = 2 6 , and P(A) + P(A') = 4 6 + 2 6 = 1A' = {5, 6}. P(A) = 4 6 , P(A') = 2 6 , and P(A) + P(A') = 4 6 + 2 6 = 1

The conditional probability of AA given BB is written P(A|B)P(A|B). P(A|B)P(A|B) is the probability that event AA will occur given that the event BB has already occurred. A conditional reduces the sample space. We calculate the probability of AA from the reduced sample space BB. The formula to calculate P(A|B)P(A|B) is

P(A|B)=P(A|B)= P(A AND B) P(B) P(A AND B) P(B)

where P(B)P(B) is greater than 0.

For example, suppose we toss one fair, six-sided die. The sample space S = {1, 2, 3, 4, 5, 6}S = {1, 2, 3, 4, 5, 6}. Let AA = face is 2 or 3 and BB = face is even (2, 4, 6). To calculate P(A|B)P(A|B), we count the number of outcomes 2 or 3 in the sample space B = {2, 4, 6}B = {2, 4, 6}. Then we divide that by the number of outcomes in BB (and not SS).

We get the same result by using the formula. Remember that SS has 6 outcomes.

P(A|B)=P(A|B)= P(A and B) P(B) = (the number of outcomes that are 2 or 3 and even in S) / 6 (the number of outcomes that are even in S) / 6 = 1/6 3/6 = 1 3 P(A and B) P(B) = (the number of outcomes that are 2 or 3 and even in S) / 6 (the number of outcomes that are even in S) / 6 = 1/6 3/6 = 1 3

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Last edited by Susan Dean on Feb 19, 2009 10:02 am US/Central.