#### Transcript Chapter 8: The Binomial Distribution and The Geometric Distribution

Chapter 8: The Binomial Distribution and The Geometric Distribution “In God we trust. All others must bring data.” Robert Hayden, Plymouth State College 8.1 The Binomial Distributions (pp. 414 – 432) The Binomial distribution is frequently useful in situations where there are two outcomes of interest, such as SUCCESS or FAILURE. It is often used to model real-life situations, and it finds its way into many extremely useful and important statistical applications and computations. The Binomial Setting 1. 2. 3. 4. Each observation is in one of two categories: success or failure There is a fixed number, N, of observations. Observations are independent. Knowing the result of one observation tells you nothing about the other observations. The probability of success is the same for each observation. If a count, X, has a binomial distribution with N number of observations and p of success, then: M ean X X N p standard deviation X X N p 1 p T he probability that one w ill get exactly k successes is C k p 1 p k N N k Example: Die is rolled 60 times. X = number of times a “3” is rolled: X X 1 60 10 6 1 5 6 0 2 .8 9 6 6 T h e p ro b ab ility th at ex actly ten "3 's" w ill b e ro lled : 60 C10 1 6 10 5 6 50 .1 3 7 0 1 3 .7 % 1 O n th e T I8 3 + , b in o m p d f 1 0 0, ,1 0 .1 3 7 0 1 3 .7 % 6 Another example: small sample size from large population. Use of binomial distribution is appropriate. Assumption: 30% of a population is Hispanic Random sample of size 4 is chosen from the population If X is the number of Hispanics in the sample then: X 4 0 .3 1 .2 X 4 .3 .7 . 9165 P r X 0 4 C 0 . 3 . 7 0 . 2401 0 4 P r X 1 4 C 1 . 3 . 7 0 . 4116 1 3 Pr X 2 Pr X 3 Pr X 4 T he probability tha a sam ple w ould contain tw o or few er H ispanics is P r X 0 P r X 1 P r X 2 _________ Using the TI83+: The probability that the sample contains exactly 2 Hispanics is Binompdf(4, .3, 2) = .2646 The probability that the sample contains 2 or fewer Hispanics is Binomcdf(4, .3, 2) = .9163 It is important to understand when one has a binomial setting and when one does not. Consider a shuffled deck of 52 playing cards: Example #1: random card is selected; suit is noted (is it a heart or not); then card is replaced Cards shuffled with random card selected; suit is noted (is it a heart or not); then card is replaced Entire process is repeated 8 more times for a total of 10 random selections X = total number of hearts obtained in 10 trials Binomial setting? Why? X = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} N =10 p = 0.25 AND each observation is INDEPENDENT of the other It is important to understand when one has a binomial setting and when one does not. Consider a shuffled deck of 52 playing cards: Example #2: random card is selected; suit is noted (is it a heart or not); then card is NOT replaced Cards shuffled with random card selected; suit is noted (is it a heart or not); then card is NOT replaced Entire process is repeated 8 more times for a total of 10 random selections X = total number of hearts obtained in 10 trials Binomial setting? Why NOT? 8.2 The Geometric Distributions pp. 434 – 444 The AP Syllabus states that you only need to know how to obtain geometric probabilities through simulation. The geometric setting is somewhat similar to that of the binomial. The basic difference is that the geometric setting DOES NOT HAVE A FIXED NUMBER OF OBSERVATIONS. The Geometric Setting 1. 2. 3. 4. Each observation is in one of two categories: success or failure The probability of success is the same for each observation. Observations are independent. Knowing the result of one observation tells you nothing about the other observations. The variable of interest is the number of trials required to obtain the first success. Example: How many times would you expect to have to roll a single die to get a “6” ? Simulate 10 trials using TI83+: randint(1, 6, 10) Trial 1: Trial 2: Trial 3: Trial 4: Trial 5: Trial 6: Trial 7: Trial 8: Trial 9: Trial 10: The mean of rolls for the 10 trials is _________ If p is the probability of success, and q =1 – p is the probability of failure, then: p = probability of success on first trial qp = probability of success on second trial q^2(p)=probability of success on the third trial, etc. If X is a variable representing the number of trials until the first success, the expected value of X is X 1 p 2 qp 3 q p 4 q p .... 2 3 p 1 2 q 3 q 4 q ... 2 3 Observe that: 1 2 q 3 q 4 q ... 2 3 1 q q q ... 1 q q q ... 2 3 2 3 Also, for 0<q<1, the sum of the infinite series 1 q q 2 q ... 3 1 1 q X p 1 q q q ... 1 q q q ... 2 3 2 1 1 p p 1 p 2 2 p p 1 q 1 q 1 q 3 The probability of rolling a “6” is 1/6. The expected number of rolls before the first success is 1/(1/6) = 6. California Lottery: You choose 6 numbers for {1, 2, 3, …,49, 50, 51} The state randomly selects 6 numbers You win $1 million if your 6 match the 6 selected by the state Your probability of matching all six is 1 51 C6 1 18009460 So you would expect your first success after playing 18,009,460 times or after 346,336 weeks (6660 years) if you play once a week.