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Geometric Distributions for AP Statistics

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By — McGraw-Hill Professional
Updated on Feb 4, 2011

Practice problems for these concepts can be found at:

A binomial setting is an experiment in which the following conditions are present:

  • The experiment consists of a fixed number, n, of identical trials.
  • There are only two possible outcomes: success (S) or failure (F).
  • The probability of success, p, is the same for each trial.
  • The trials are independent (that is, knowledge of the outcomes of earlier trials does not affect the probability of success of the next trial).
  • Our interest is in a binomial random variable X, which is the count of successes in n trials. The probability distribution of X is the binomial distribution.

There are times we are interested not in the count of successes out of n fixed trials, but in the probability that the first success occurs on a given trial, or in the average number of trials until the first success. A geometric setting is defined as follows.

  • There are only two possible outcomes: success (S) or failure (F).
  • The probability of success, p, is the same for each trial.
  • The trials are independent (that is, knowledge of the outcomes of earlier trials does not affect the probability of success of the next trial).
  • Our interest is in a geometric random variable X, which is the number of trials necessary to obtain the first success.

Note that if X is a binomial, then X can take on the values 0, 1, 2,…, n. If X is geometric, then it takes on the values 1, 2, 3,…. There can be zero successes in a binomial, but the earliest a first success can come in a geometric setting is on the first trial.

If X is geometric, the probability that the first success occurs on the nth trial is given by P(X = n) = p(1 - p)n-1. The value of P(X = n) in a geometric setting can be found on the TI-83/84 calculator, in the DISTR menu, as geometpdf(p, n) (note that the order of p and n are, for reasons known only to the good folks at TI, reversed from the binomial). Given the relative simplicity of the formula for P(X = n) for a geometric setting, it's probably just as easy to calculate the expression directly. There is also a geometcdf function that behaves analogously to the binomcdf function, but is not much needed in this course.

example: Remember Dolores, the basketball player whose free-throw shooting percentage was 0.65? What is the probability that the first free throw she manages to hit is on her fourth attempt?

solution: P(X = 4) = (0.65) (1–0.65)4–1 = (0.65) (0.35)3 = 0.028. This can be done on the TI-83/84 as follows: geometpdf(p,n) = geometpdf(0.65,4) = 0.028.

example: In a standard deck of 52 cards, there are 12 face cards. So the probability of drawing a face card from a full deck is 12/52 = 0.231.

  1. If you draw cards with replacement (that is, you replace the card in the deck before drawing the next card), what is the probability that the first face card you draw is the 10th card?
  2. If you draw cards without replacement, what is the probability that the first face card you draw is the 10th card?

solution:

  1. P(X = 10) = (0.231) (1 – 0.231)9 = 0.022. On the TI-83/84: geometpdf(0.231,10)= 0.0217).
  2. If you don't replace the card each time, the probability of drawing a face card on each trial is different because the proportion of face cards in the deck changes each time a card is removed. Hence, this is not a geometric setting and cannot be answered the by techniques of this section. It can be answered, but not easily by the techniques of the previous chapter.

Rather than the probability that the first success occurs on a specified trial, we may be interested in the average wait until the first success. The average wait until the first success of a geometric random variable is 1/p. (This can be derived by summing (1)·P(X = 1) + (2)·P(X = 2) + (3)·P(X = 3) +... = 1p + 2p(1 – p) + 3p(1 – p)2 +…, which can be done using algebraic techniques for summing an infinite series with a common ratio less than 1.)

example: On average, how many free throws will Dolores have to take before she makes one (remember, p = 0.65)?

solution: 1/0.65 = 1.54.

Since, in a geometric distribution, P(X = n) = p(1–p)n–1 the probabilities become less likely as n increases since we are multiplying by 1 – p, a number less than one. The geometric distribution has a step-ladder approach that looks like this:

Practice problems for these concepts can be found at:

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