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Geometric Distributions

Geometric Distributions. Created by: K Wannan Edited by: N. Gysbers and K Stewart. Dice Activity. Everyone stand. Roll one dice “Success” is rolling a 3. If you get a 3, you sit down. The person standing the longest is the winner. . What is the probability of getting a 3?

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Geometric Distributions

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  1. Geometric Distributions Created by: K Wannan Edited by: N. Gysbers and K Stewart

  2. Dice Activity Everyone stand. Roll one dice “Success” is rolling a 3. If you get a 3, you sit down. The person standing the longest is the winner.  What is the probability of getting a 3? Does the probability change from one roll to the next? Repeat the experiment until there is a winner (last man standing). What was the greatest number of failures before getting a success (rolling a 3)?

  3. Similar to a Binomial Distribution: Success or Failure Probabilities do not change (independent events) Different than a Binomial Distribution: Observing the number of failures until a success. Random Variable, X, is the number of failed trials before a successful event occurs A Geometric Distribution is

  4. Probability for Geometric Distribution Where: p is the probability of a success in each trial q is the probability of a failure in each trial x is the number of failures The expectation converges to a simple formula

  5. Expectation for Geometric Distribution As with any distribution, the expected value is a weighted average. Since the number of failures before the first success is limitless, we cannot calculate

  6. Expectation for Geometric Distribution Fortunately, with some advanced algebraic manipulations, the Expected Value formula for a geometric distribution simplifies to:

  7. Ex: Jamaal has a success rate of 68% for scoring on free throws in basketball. What is the expected number of times he will shoot before he misses the basket on a free throw? The random variable is the number of trials before he misses a free throw A success is Jamaal missing a shot on a free throw. P(score) is 68%, so q = 0.68 P(miss) is 100% - 68%, sop = 1-0.68 = 0.32

  8. Ex: Suppose that an intersection you pass on your way to school has a traffic light that is green for 40 s and then amber or red for a total of 60s What is the probability that the light will be green, when you reach the intersection in the morning, at least once in a school week? What is the expected number of mornings before the light is green when you reach the intersection?

  9. a) What is the probability that the light will be green when you reach the intersection at least one morning in a school week? p= light is green = 40/100 = 0.4 q= light not green = 60/100 = 0.6 There are 5 school days so we want the probability that you will wait 0 mornings, 1 morning, 2 mornings, 3 mornings or 4 mornings before the first time it is green.

  10. Geometric Probability Distribution

  11. To find the probability that you will get a green light at least once in a week, you have to add the probabilities of the outcomes from morning1 to morning 5.

  12. b) What is the expected number of mornings before the light is green when you reach the intersection? The expected waiting time before catching a green light is 1.5 mornings.

  13. Homework! Pg 394 #1,2,3,7,9,10 All Done!! (Kind of...)

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