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Chapter 8
The Geometric Distributions
PLINKO
Our 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.
• Let the random variable X = number of times the ball falls to the right.
• Our goal is to find P (X=3)
• 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 = Right
Lands in D __________________________
Not in D _____________________________
Lands in D _____________________________
Not in D ______________________________
CLASS TOTALS in D _________
Not in D _________
Determine the total number of possible outcomes for Plinko…list them systematically
HINT: How many positions to left and right?
Use powers of 2 to find to combinations of
R’s and L’s I’ll get you started…see the board
LLLLLL
RLLLLL
LRLLLL
RRLLLL
LLRLLL
RLRLLL
LRRLLL
RRRLLL
• Complete 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.
X 0 1 2 3 4 5 6
P(X)
The 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 trial
• The flips or draws are independent
(one outcome has no influence on the
next)
• The probability of success (falling to right) is the same each trial
A situation in which these four conditions are satisfied is called a binomial setting
To reiterate these characteristics…
The Binomial Setting
1. each observation falls into one of two categories (success and failure)
2. there is a fixed number n of observations
3. The n observations are all independent.
4. the probability of success, p, is the same for each observation
The 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)
Continuing 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”.
PLINKO
• P(lands in slot D) = .3125
• So all 5 in slot D would be
P(all in slot D) = (.3125)5= .00298
• P(2 in slot D) = 2 in slot D & 3 not in D
= (.3125)2(1-.3125)3 = .0317
PLINKO
• If 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?
PLINKO
Payout
$50 $25 $10 $5 $10 $25 $50
X 0 1 2 3 4 5 6P(X) .0156 .0938 .2344 .3125 .2344 .0938 .0156
Expected value = 50(.0156)+25(.0938)+10(.2344)+5(.3125)+10(.2344)+25(.0938)+50(.0156) = $12.50
PLINKO
Payout
$50 $25 $10 $5 $10 $25 $50
X 0 1 2 3 4 5 6P(X) .0156 .0938 .2344 .3125 .2344 .0938 .0156
VAR (X) = (50-12.50)2(.0156) + (25-12.5)2(.0938) + (10-12.50)2(.2344) + (5- 12.50)2(.3125) + (10-12.50)2(.2344) + (25-12.5)2 (.0938) + (50-12.50)2(.0156)VAR (X) = 93.7499STD DEV(X) = $9.68
Example 1
Suppose 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 = x
10 C 7 (.65)7(.35)3 = .252
(b) What is Dolores’ probability that she makes no more than 5 free throws?
This time the question is asking what is
P (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)10 + 10 C 1 (.65)1(.35)9 + 10 C 2 (.65)2(.35)8 + 10 C 3 (.65)3(.35)7 + 10 C 4 (.65)4(.35)6 + 10 C 5 (.65)5(.35)5
= .249 about 25% chance
What 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)) = = 1 - .249
= .751
Mean 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 =x np p( )1
Normal 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 = np p( )1
Community 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.
(a) 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
= 400 15 85 714(. )(. ) .
Community College Problem
= np p( )1
Community 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)
Community College Problem
Find P (x ≥ 65)
P (z ≥ (65-60)/7.14 = .70
From Table A the P (z < .70) = .7580 so
P (z ≥ .70 ) = (1-.7580)= .242
Credit Card Example
Suppose 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 2500
X is B(2500,.60) We want to find P(X > 1520)
Credit Card Example
Is np ≥ 10 ? 2500(.60) = 1500 ≥ 10
Is n(1-p) ≥ 10? 2500(.40) = 1000 ≥ 10
Yes, it approximates a normal distribution.
μ = np = 1500
= np p( )1
= 1500 1 6
24 49
( . )
.
Credit Card Example
We want to find P (x >1520)
P (z > 1520 1500
24 498167
..
Do problems 8.2, 8.8 and 8.16
Technology ToolboxExploring binomial
distributions
Chapter 8 Section 8.2 The Geometric Distribution
Hawaiian 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 meet P(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
RULE 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
P n p pnX = b g ( )1 1
Hawaiian Villager Problem
What is the P( you don’t meet a Hawaiian until the 2nd villager?)
P (X =2) (1-.8)2-1(.8)= .16
Let’s extend this concept for third, fourth, fifth villagers…
Hawaiian Villager Problem
P (X =1) (1-.8)1-1(.8)= .8
P (X =2) (1-.8)2-1(.8)= .16
P (X =3) (1-.8)3-1(.8)= .032
P (X =4) (1-.8)4-1(.8)= .0064
P (X =5) (1-.8)5-1(.8)= .00128
Hawaiian Villager Problem
When this data is graphed what do you notice?
00.10.20.30.40.50.60.70.8
Villager
Hawaiian Villager
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.
Characteristics of a Geometric Distribution
Hawaiian 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
Hawaiian Villager Problem
Find the average number of villagers it will take to meet a native Hawaiian.
x =1
p
x =1
8125
..
Hawaiian Villager Problem
How much variability is there in the number of villagers required to meet a Hawaiian?
x = 1-.8
.82
.
.. .
20
643125 5590
x = 1- p
p2
The Geometric Setting
• 1. each observation falls into one of two categories, success or failure
• 2. the probability of a success is p
• 3. the observations are all independent
• 4. the variable of interest is the number of trials required to obtain the first success
HANDY 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 =
4
52
1
p
P(X > n) = (1- p)n x 2
=- p
p
1
How 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 independent• The distinguishing characteristic is: A binomial probability model is appropriate
for a random variable that counts the # of successes in a fixed number of trials.
How 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.)
Which 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?
Which are these?(2) Sophie is a dog who loves to play catch.
Unfortunately, she isn’t very good, and the P(catches a ball) = 0.1. Define X=# tosses required until Sophie to catches the ball.
Is this binomial or geometric?
Which 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?
Which 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?
Let’s Explore the Sophie problem and the cereal problem in more
depth.
The Sophie problem(2) Sophie is a dog who loves to play catch.
Unfortunately, she isn’t very good, and the P(catches a ball) = 0.1. 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)n
P(X ≥ 3) = (1- .1)3 = .729
Sophie(2c) calculate and interpret the mean and
standard deviation of X
μx =
x 2
=- p
p
1x
2=
-.9
.9
1351.
1 1
110
p =x
.
Cereal Problem• Suppose 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)
Cereal 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)n P (X ≤ 2) = (1-.5)2 = .25
Cereal Problem• (b) What is the probability you will
have to buy exactly 4 boxes? ( X = 4) P(X = n) = (1- p)n-1p
P(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)n
P(X ≥ 4) = (1- .5)4 = .0625
#37 in the textbookWhich are binomial or
geometric?
# 37 in the book.(a) yes, geometric X = success (tail)
failure (head)
a trial is one flip of the coin
P(tail) = .5
(b) Not independent
(c) X = success of getting a Jack
a trial is drawing a card with replacement
P(J) = 4
52
Dolores 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)= .028• Using the TI 83/84 geometpdf (p,n)
geometpdf (.65,4) = .028
Technology ToolboxExploring geometric
distributions
