before death that is several hundred million years long. Yet I do not conclude that I probably am not alive. The actual result of each trial will be either occurrence or nonoccurrence of a specific outcome, but our interest is in proportions for a large number of trials.
The most important theorem of probability is this: when dealing with several independent events, the probability of all of them happening is the product of the individual probabilities. For example, the probability of flipping a coin twice and getting heads both times is 1/2•1/2=1/4; the chance of flipping a coin and a die and getting a head plus a two is 1/2•1/6=1/12. If one has already flipped a coin twice and gotten two heads, the probability of getting heads on a third trial and thus making the winning streak three heads in a row is 1/2, not 1/2•1/2•1/2=1/8. The third trial is independent of previous results.
Though simple, this theorem of multiplicative probabilities is easy to misuse. For example, if the probability of getting a speeding ticket while driving to and from work is 0.05 (i.e., 5%) per round trip, what is the probability of getting a speeding ticket sometime during an entire week of commuting? The answer is not .05•.05•.05•.05•.05=.0000003; that is the probability of getting a speeding ticket on every one of the five days. If the question is expressed as “what is the probability of getting at least one speeding ticket”, then the answer is 1-0.955=0.226, or 1 minus the probability of getting no speeding tickets at all.
Often the events are not completely independent; the odds of one trial are affected by previous trials. For example, the chance of surviving one trial of Russian roulette with a 6-shot revolver is 5/6; the chance of surviving two straight trials (with no randomizing spin of the cylinders between trials) is 5/6•4/5=2/3.
Independence is the key to assuring that an undesirable outcome is avoided, whether in a scientific research project or in everyday life. The chance of two independent rare events occurring simultaneously is exceedingly low. For example, before a train can crash into a station, the engineer must fail to stop the train (e.g., fall asleep) and the automatic block system must fail. If the chance of the first occurring is 0.01 and the chance of the second occurring is 0.02, then the chance of a train crash is the chance of both occurring together, or 0.01•0.02=0.0002. The same strategy has been used in nuclear reactors; as I type this I can look out my window and see a nuclear reactor across the Hudson River. For a serious nuclear accident to occur, three ‘independent’ systems must fail simultaneously: primary and secondary cooling systems plus the containment vessel. However, the resulting optimistic statements about reactor safety can be short-circuited by a single circumstance that prevents the independence of fail-safes (e.g., operator panic misjudgments or an earthquake).
Entire statistics books are written on probability, permitting calculation of probabilities for a wide suite of experimental conditions. Here our scope is much more humble: to consider a single ‘probability distribution function’ known as the normal distribution, and to determine how to assess the probabilities associated with a single variable or a relationship between two variables.
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Sampling Distribution for One Variable
Begin with a variable which we will call X, for which we have 100 measurements. This dataset was drawn from a table of random normal numbers, but in the next section we will consider actual datasets of familiar data types. Usually we have minimal interest in the individual values of our 100 (or however many) measurements of variable X; these measurement values are simply a means to an
