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Binomial and Geometric Distributions: Special Discrete Distributions

Learn about the properties and applications of binomial and geometric distributions in probability. Understand how to calculate probabilities and expected values for binomial and geometric random variables.

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Binomial and Geometric Distributions: Special Discrete Distributions

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  1. Chapter 7Lesson 7.5 Random Variables and Probability Distributions 7.5: Binomial and Geometric Distributions

  2. Special Distributions Two Discrete Distributions: Binomial and Geometric One Continuous Distribution: Normal Distributions

  3. Properties of a Binomial Experiment • There are a fixed number of trials • Each trial results in one of two mutually exclusive outcomes. (success/failure) • Outcomes of different trials are independent • The probability that a trial results in success is constant. The binomial random variable x is defined as x = the number of successes when a binomial experiment is performed We use n to denote the fixed number of trials.

  4. Are these binomial distributions? • Toss a coin 10 times and count the number of heads Yes • Deal 10 cards from a shuffled deck and count the number of red cards No, probability does not remain constant • The number of tickets sold to children under 12 at a movie theater in a one hour period No, no fixed number

  5. At a particular hospital, let’s record the gender of the next 5 newborns at this hospital. Is this a binomial experiment? Yes, if the births were not multiple births (twins, etc). Define the random variable of interest. x = the number of females born out of the next 5 births What are the possible values of x? x 0 1 2 3 4 5 What is the probability of “success”?

  6. Newborns Continued . . . What is the probability that exactly 2 girls will be born out of the next 5 births? What is the probability that less than 2 girls will be born out of the next 5 births?

  7. Since this is a discrete distribution, we could use: Newborns Continued . . . Let’s construct the discrete probability distribution table for this binomial random variable: What is the mean number of girls born in the next five births? Notice that this is the same as multiplying n × p mx = 0(.03125) + 1(.15625) + 2(.3125) + 3(.3125) + 4(.15625) + 5(.03125) =2.5

  8. Formulas for mean and standard deviation of a binomial distribution

  9. Newborns Continued . . . How many girls would you expect in the next five births at a particular hospital? What is the standard deviation of the number of girls born in the next five births?

  10. The Binomial Model (cont.) Binomial probability model for Bernoulli trials: n= number of trials p = probability of success q = 1 – p = probability of failure X = # of successes in n trials P(X = x) = nCxpxqn-x

  11. Homework • Pg.344: #7.43 (π=.9 means p=.9) 44, 46, 51, 52

  12. Newborns Revisited . . . Suppose we were not interested in the number of females born out of the next five births, but which birth would result in the first female being born? How is this question different from a binomial distribution?

  13. Properties of Geometric Distributions: • There are two possible outcomes: a success or failure • Each trial is independent of the others • The probability of success is constant for all trials. A geometricrandom variable x is defined as x = the number of trials UNTIL the FIRST success is observed ( including the success). x 1 2 3 4 So what are the possible values of x To infinity How far will this go? . . .

  14. Probability Formula for the Geometric Distribution Let p = constant probability that any trial results in a success Where x = 1, 2, 3, …

  15. Suppose that 40% of students who drive to campus at your school or university carry jumper cables. Your car has a dead battery and you don’t have jumper cables, so you decide to stop students as they are headed to the parking lot and ask them whether they have a pair of jumper cables. Let x = the number of students stopped before finding one with a pair of jumper cables Is this a geometric distribution? Yes

  16. Jumper Cables Continued . . . Let x = the number of students stopped before finding one with a pair of jumper cables p = .4 What is the probability that third student stopped will be the first student to have jumper cables? What is the probability that at most three student are stopped before finding one with jumper cables? P(x = 3) = (.6)2(.4) = .144 P(x< 3) = P(1) + P(2) + P(3) = (.6)0(.4) + (.6)1(.4)+ (.6)2(.4) = .784

  17. The Geometric Model (cont.) Geometric probability model for Bernoulli trials: p = probability of success q = 1 – p = probability of failure X = # of trials until the first success occurs P(X = x) = qx-1p

  18. Homework • Pg.344: #7.56-58

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