Sunday, December 1, 2013

Complements of events

This post explains the concepts of sample space and the complement of an event.

The question in this post involves data from all of Cal Ripken's at bats in the 1996 regular season.   The sample space for the experiment is all possible outcome from an at bat, which differs of course from a plate appearance.

Data for Cal Ripken 1996 Season
Home Runs
At Bats that Did not end in a hit
At Bats
Batting Average

The sample space of an experiment is the list of possible outcomes.  The Cal Ripken batting experiment illustrates three properties of probabilities that must hold in any experiment.  These three properties are:

o The probability of each outcome of the experiment must be at least zero.  
o No event can have a probability greater than one.
o The sum of the probabilities for all events in the sample space must equal one.   

An event has an occurrence probability of zero if it is impossible for it to occur.  By contrast, an event with a probability of one is certain to occur.  The Cal Ripken at-bat experiment satisfies all three of these properties.  There are five possible outcomes each of which has a probability between 0 and 1.0.  The five probabilities sum to one.     

Questions:  Describe the complement of the following three events: Cal Ripken gets a hit, Cal Ripken gets a single, and Cal Ripken gets an extra base hit for a particular at bat.

Answers:  The complement of the event that Cal Ripken gets a hit is simply the event that he does not get a hit which is simply 0.722 (1-0.278).

The complement of Cal Ripken hitting a single is the event that Cal Ripken hits a double, triple, home run or gets out.  This probability can be obtained by adding up the probabilities of these mutually exclusive events or by subtracting the probability of getting a single from 1.0.  Both techniques give us the answer of 0.827.

An extra base hit occurs if Cal Ripken gets a double, triple or home run.  The probability of the complement can be obtained either by adding these three probabilities and subtracting from 1.0 or by adding the probability of a single to the probability of an out.  Both techniques give us the answer of 0.895.         

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