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Lecture Comments (2)

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Post by Ken Mullin on January 24, 2012

RE: the marble example
What might aid a student here is seeing that this type computation yields the same answer as if the calculation was performed by simply doing the multiplication of each event (w/0 replacement)--> (5/10)*(4/9)*(3/8) = .08333..

I wouldn't have thought to use the hypergeometric prob. distrib here.
What would be the advantage in doing do, if any?

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Post by Malinda Holland on November 9, 2010

Is the answer in the Marbles example correct ? I think 0.083 should be 0.0083 (the probability reduces to 1/120 not 10/120)

Geometric and Hypergeometric Probability Distributions

  • The geometric distribution is used to find the probability that the first success occurs on the xth trial.

  • For the geometric distribution, the trials are independent and have two outcomes: “success” or “failure.”

  • The hypergeometric distribution is used when sampling without replacement.

Geometric and Hypergeometric Probability Distributions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Geometric Probability Distribution 0:08
  • Example: Engine Malfunction 3:00
  • Example: Interviews 5:45
  • Hypergeometric Probability Distribution 7:36
  • Example: Engineers 10:16
  • Example: Marbles 12:55
  • Extra Example 1
  • Extra Example 2