2PLINKOOur goal is to determine the probability that a ball will land in slot “D”. A win occurs when the ball falls to the Right 3 times and to the Left 3 times in any order.
3Let the random variable X = number of times the ball falls to the right. Our goal is to find P (X=3)
4RandInt (1,2,6) where 1 = Left and 2 = Right Use the random number generator on your calculator to generate six numbers representing the positions A, B, C, E, F, and G._ A__ ___B___ ___C___ ___D__ ___E___ ___F___ ___G___RandInt (1,2,6) where 1 = Left and 2 = RightLands in D __________________________Not in D _____________________________
5Lands in D _____________________________ Not in D ______________________________CLASS TOTALS in D _________Not in D _________
6LLLLLL RLLLLL LRLLLL RRLLLL LLRLLL RLRLLL LRRLLL RRRLLL Determine the total number of possible outcomes for Plinko…list them systematicallyHINT: How many positions to left and right?Use powers of 2 to find to combinations ofR’s and L’s I’ll get you started…see the boardLLLLLLRLLLLLLRLLLLRRLLLLLLRLLLRLRLLLLRRLLLRRRLLL
7Complete the probability distribution for the random variable X = # times the ball falls to the right (i.e.) 0 R’s, 1 R, 2 R’s, # R’s etc.X123456P(X)
8The features of this experiment are as follows: There are two outcomes (L, R) or heads/tails, evens/odds (success/failure)6 digits are drawn for a single trialThe flips or draws are independent(one outcome has no influence on thenext)The probability of success (falling to right) is the same each trial
9A situation in which these four conditions are satisfied is called a binomial setting To reiterate these characteristics…
10The Binomial Settingeach observation falls into one of two categories (success and failure)2. there is a fixed number n of observations3. The n observations are all independent.4. the probability of success, p, is the same for each observation
11The distribution of the count X of successes in the binomial setting is the binomial distribution with parameters n and p. The parameters n is the number of observations, and p is the probability of a success on any one observation. The possible values of X are the whole numbers from 0 to n. B ( n, p)
12Continuing with PLINKO Suppose 5 PLINKO balls are dropped down the board in succession. Find the probability that all of them will land in slot “D”.Find the probability that exactly 2 of them land in slot “D”.
13PLINKO P(lands in slot D) = .3125 So all 5 in slot D would be P(all in slot D) = (.3125)5=P(2 in slot D) = 2 in slot D & 3 not in D= (.3125)2( )3 = .0317
14PLINKOIf a ball landing in slot “A” or “G” pays $50, a ball landing in “B” or “F” pays $25, a ball in “C” and “E” pay $10, and a ball landing in “D” pays $5, find the expected winnings(mean)when 5 balls are dropped.What is the standard deviation of the total amount won?
15PLINKOPayout$50$25$10$5X123456P(X).0156.0938.2344.3125Expected value = 50(.0156)+25(.0938)+10(.2344)+5(.3125)+10(.2344)+25(.0938)+50(.0156)= $12.50
17Example 1Suppose Dolores is a 65% free throw shooter. If we assume that the repeated shots are independent “what is the probability that Dolores makes exactly 7 of her next 10 free throws?” If X is the binomial random variable that gives us the count of successes for the experiment, we say X has B(10,.65)The question is to find P(X=7)use n C r under MATH PRB where r = x10 C 7 (.65)7(.35)3 = .252
18(b) What is Dolores’ probability that she makes no more than 5 free throws? This time the question is asking what isP (x ≤ 5)?P (x ≤ 5) = P(x=0) + P(x=1) + P(x=2) + P(x=3) + P(x=4) + P(x =5)= 10 C 0 (.65)0(.35) C 1 (.65)1(.35) C 2 (.65)2(.35) C 3 (.65)3(.35) C 4 (.65)4(.35) C 5 (.65)5(.35)5= .249 about 25% chance
19What is the probability that Dolores will make at least 6 free throws? Do we have to redo all of the calculations or is there another way? If she makes at least 6 how many will she miss? Hmmm…????? How is this related to the previous problem?P(x ≥ 6) is the same as P( 1- P(x ≤ 5)) === .751
20Mean and Standard Deviation Clearly we can calculate the mean and standard deviation of a binomial random variable using the methods from Chapter 7 but there is another way…μx = np
21Normal Approximation for Binomial Distribution When np and n(1-p) are SUFFICIENTLY LARGE i.e. both are ≥ 10, the binomial random variable X has an approximately normal distribution.The mean μ = np and
22Community College Problem Nationally, 15% of community college students live more than 6 miles from campus. Data from a simple random sample of 400 students at one community college is analyzed.What are the mean and standard deviation for the number of students in the sample trial?X has B(400,.15)μ = np = 400(.15) = 60
24Community College Problem (b) Use a normal approximation to calculate probability that at least 65 of the students in the sample live more than 6 miles from campus. Because 400(.15) = 60 ≥10 and 400(.85) = 340 ≥ 10, we can use the normal approximation to the binomial with N(60,7.14)
25Community College Problem Find P (x ≥ 65)P (z ≥ (65-60)/7.14 = .70From Table A the P (z < .70) = soP (z ≥ .70 ) = ( )= .242
26Credit Card ExampleSuppose 60% of adults have credit card debt. If we survey 2500 adults, what is the probability more than 1520 would have credit card debt?X = # adults who have credit card debt out of 2500X is B(2500,.60)We want to find P(X > 1520)
27Credit Card Example Is np ≥ 10 ? 2500(.60) = 1500 ≥ 10 Yes, it approximates a normal distribution.μ = np = 1500
28Credit Card ExampleWe want to find P (x >1520)P (z >
39Chapter 8 Section 8.2 The Geometric Distribution
40Hawaiian Villager Problem On the island of Oahu in the village of Nankuli, 80% of the residents are of Hawaiian ancestry. If you visit Nanakuli, what is the probability the first village you meet is Hawaiian?X = # villagers you must meetP(X = 1)P(X = n) = (1 – p)n-1p -- probability it is (p) * probability it is not (1-p)P (X = 1) = (1-.8)1-1(.8) = .8
41RULE FOR CALCULATING GEOMETRIC PROBABILITIES If X has a geometric distribution with probability p of success and (1-p) of failure on each observation, the possible values of X are 1,2,3,…If n is any one of these, the probability that the first success occurs on the nth trial is
42Hawaiian Villager Problem What is the P( you don’t meet a Hawaiian until the 2nd villager?)P (X =2) (1-.8)2-1(.8)= .16Let’s extend this concept for third, fourth, fifth villagers…
44Hawaiian Villager Problem When this data is graphed what do you notice?
45Characteristics of a Geometric Distribution Graphs of Geometric Distributions have a ‘step ladder” appearance since you are multiplying the height of each bar by a number less than 1. Each bar will be shorter than the previous bar. The histogram is ALWAYS right skewed.
46Hawaiian Villager Problem Find the probability it will take more than 4 villagers to meet a native Hawaiian.P(x > 4) = (1-p)n = (1-.8)4 = (.2) 4 = .0016
47Hawaiian Villager Problem Find the average number of villagers it will take to meet a native Hawaiian.
48Hawaiian Villager Problem How much variability is there in the number of villagers required to meet a Hawaiian?
49The Geometric Setting 2. the probability of a success is p 1. each observation falls into one of two categories, success or failure2. the probability of a success is p3. the observations are all independent4. the variable of interest is the number of trials required to obtain the first success
50HANDY DANDY FORMULAS If X is a Geometric random variable with P(success) = p these formulas apply:P(X=n) = (1- p)n-1(p) μx =P(X > n) = (1- p)n
51How Can You Tell Geometric from Binomial? Both of these models must meet the 3 conditions often called the Bernoulli trials.(1) there are two possible outcomes(2) the probability of a success is constant(3) the trials are independentThe distinguishing characteristic is:A binomial probability model is appropriate for a random variable that counts the # of successes in a fixed number of trials.
52How Can You Tell Geometric from Binomial? While a geometric probability model is appropriate for a random variable that counts the # of trials until the first success. (there could be an unlimited number of trials.)
53Which are these?(1) The Los Angeles Times reported that 80% of airline passengers prefer to sleep on long flights rather than watch movies, read, etc. Consider randomly selecting 25 passengers from a particular long flight. Defind a random variable X , calculate P( X=12).Is this binomial or geometric?
54Which are these?(2) Sophie is a dog who loves to play catch. Unfortunately, she isn’t very good, and the P(catches a ball) = Define X=# tosses required until Sophie to catches the ball.Is this binomial or geometric?
55Which are these?(3) You are to take a multiple choice exam of 100 questions with five possible responses (A,B,C,D,E). Suppose you have not studied and decide to guess randomly on each question. Let X = # correct responses.Is this binomial or geometric?
56Which are these?(4) Suppose 5% of cereal boxes contain a prize. You are determined to buy cereal boxes until you win a prize.Is this binomial or geometric?
57Let’s Explore the Sophie problem and the cereal problem in more depth.
58The Sophie problem(2) Sophie is a dog who loves to play catch. Unfortunately, she isn’t very good, and the P(catches a ball) = Define X=# tosses required until Sophie to catches the ball.(a) calculate and interpret P(X=2)P (X = n) = (1-p)n-1(p)P (X =2) (1-.1)2-1(.1)= .09(b) calculate and interpret P(X ≥ 3)P(X > n) = (1- p)nP(X ≥ 3) = (1- .1)3 = .729
59Sophie(2c) calculate and interpret the mean and standard deviation of Xμx =
60Cereal ProblemSuppose 5% of cereal boxes contain a prize. You are determined to buy cereal boxes until you win a prize.(a) What is the probability you will have to buy at most 2 boxes? (X ≤ 2)(b) What is the probability you will have to buy exactly 4 boxes? ( X = 4)(c) What is the probability you will have to buy more than 4 boxes? (X ≥ 4)
61Cereal Problem(a) What is the probability you will have to buy at most 2 boxes?X = # boxes you will buy until you win a prize.Find P (X ≤ 2) is the same as P (1- complement) P( X > 2)P(X > n) = (1- p)nP (X ≤ 2) = (1-.5)2 = .25
62Cereal Problem(b) What is the probability you will have to buy exactly 4 boxes? ( X = 4)P(X = n) = (1- p)n-1pP(X = 4) = (1- .5)4-1(.5) = .0625(c) What is the probability you will have to buy more than 4 boxes? (X ≥ 4)P(X > n) = (1- p)nP(X ≥ 4) = (1- .5)4 = .0625
63#37 in the textbook Which are binomial or geometric?
64# 37 in the book. yes, geometric X = success (tail) failure (head) a trial is one flip of the coinP(tail) = .5(b) Not independent(c) X = success of getting a Jacka trial is drawing a card with replacementP(J) =
65Dolores the Basketball Player Remember Dolores the basketball player whose free throw shooting percentage was .65? What is the probability that the first free throw she hits is on her 4th attempt?P(X = 4) (1-.65)4-1(.65)= (.35)3 (.65)= .028Using the TI 83/84 geometpdf (p,n)geometpdf (.65,4) = .028