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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
P(X)
0
1
2
3
4
5
6
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
Pay
out
X
P(X)
$50 $25
0
1
$10
2
$5
3
$10
$25
$50
4
5
6
.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
Pay
out
X
P(X)
$50 $25
0
1
$10
2
$5
3
$10
$25
$50
4
5
6
.0156 .0938 .2344 .3125 .2344 .0938 .0156
VAR (X) = (50-12.50)2(.0156) + (25-12.5)2(.0938) + (1012.50)2(.2344) + (5- 12.50)2(.3125) + (10-12.50)2(.2344) + (25-12.5)2
(.0938) + (50-12.50)2(.0156)
VAR (X) = 93.7499
STD 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
7(.35)3 = .252
C
(.65)
10
7
(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(1 p)
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(1 p)
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
Community College Problem
 = np(1 p)
 = 400(.15)(.85)  714
.
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(1 p)
 = 1500(1.6)
 24.49
Credit Card Example
We want to find P (x >1520)
1520  1500
P (z >
.8167
24.49
Do problems 8.2, 8.8 and 8.16
Technology Toolbox
Exploring 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
b g
n 1
P X = n  (1  p) p
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?
Hawaiian Villager
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Villager
Characteristics 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.
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.
1
x =
p
1
x =  125
.
.8
Hawaiian Villager Problem
How much variability is there in the number
of villagers required to meet a Hawaiian?
 =
1- p
p2
 =
1-.8
.20

 .3125 .5590
2
.8
.64
x
x
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:
4
52
P(X=n) = (1-
p)n-1(p)
P(X > n) = (1- p)n
1
μx =
p
1- p
x =
2
p
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
1
1
μx =  x =  10
p
1- p
x =
2
p
.1
1-.9
x =
.351
2
.9
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 (1complement) 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 textbook
Which 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 Toolbox
Exploring geometric
distributions