Transcript APSTAT SECTION IV PROBABILITY

APSTAT SECTION IV
PROBABILITY
CHAPTER 14
From Justin To Kelly
(or from randomness to probability)
Randomenss and Probability Basics
 What is Random?
Individual outcomes are unpredictable in the short
run
In the long run, however, outcomes are regular
AND predictable (LOLN - Law of Large Numbers)
 Lets do a simulation
Flip 10, 100, 1000, 10000 coins
Find % of heads to nearest whole %
sum(randint(0,1,10))
Random Simulation
Short Run
10 Flips
100 Flips
Long Run
999Flips
What is Probability
 Over a HUGE number of trials (probability is LongTerm), the proportion of times an outcome would
occur.
 Typically expressed by P= and a range from 0 to 1
0 being never ever happens
1 being always happens
 We can only ESTIMATE real-world probabilities
 Can be expressed as a %, but not as cool.
Models of Probability
Two Main Thangs
LIST all possible outcomes
ASSIGN a probability to each outcome
ie. Year in school probability @ WPS
FROSH
SOPH
JUNIOR
SENIOR
60/230
60/230
70/230
40/230
.261
.304
.164
P= .261
Should add up to 1.0, but may be a bit off
due to ROUNDING ERROR
Vocab Time
Sample Space – S – Set of all possible
outcomes
Event – Any outcome or set of outcomes
ie. Freshman
ie. Juniors AND seniors
CRAPS! – Roll Them Bones!
 Disclaimer – Gambling can be dangerous and addictive, plus over the LONG RUN,
the casino always wins. So don’t gamble, buy a casino!
 Sample space when 2 die are rolled:
1
2
3
4
5
6
1
1,1
1,2
1,3
1,4
1,5
1,6
2
2,1
2,2
2,3
2,4
2,5
2,6
3
3,1
3,2
3,3
3,4
3,5
3,6
4
4,1
4,2
4,3
4,4
4,5
4,6
5
5,1
5,2
5,3
5,4
5,5
5,6
6
6,1
6,2
6,3
6,4
6,5
6,6
36 potential outcomes
CRAPS! – Roll Them Bones!
Event: Rolling a 7 when pips are added
“ProbSpeak”: P(Roll a 7)
1
2
3
4
5
6
1
1,1
1,2
1,3
1,4
1,5
1,6
2
2,1
2,2
2,3
2,4
2,5
2,6
3
3,1
3,2
3,3
3,4
3,5
3,6
4
4,1
4,2
4,3
4,4
4,5
4,6
5
5,1
5,2
5,3
5,4
5,5
5,6
6
6,1
6,2
6,3
6,4
6,5
6,6
P(Roll 7) = 6/36 = .167
CRAPS! – Roll Them Bones!
Event: Rolling a 8 when pips are added
“ProbSpeak”: P(8)
1
2
3
4
5
6
1
1,1
1,2
1,3
1,4
1,5
1,6
2
2,1
2,2
2,3
2,4
2,5
2,6
3
3,1
3,2
3,3
3,4
3,5
3,6
4
4,1
4,2
4,3
4,4
4,5
4,6
5
5,1
5,2
5,3
5,4
5,5
5,6
6
6,1
6,2
6,3
6,4
6,5
6,6
P(8) = 5/36 = .139
CRAPS! – Roll Them Bones!
Event: Rolling a Hard 8 (two 4’s)
“ProbSpeak”: P(Hard 8)
1
2
3
4
5
6
1
1,1
1,2
1,3
1,4
1,5
1,6
2
2,1
2,2
2,3
2,4
2,5
2,6
3
3,1
3,2
3,3
3,4
3,5
3,6
4
4,1
4,2
4,3
4,4
4,5
4,6
5
5,1
5,2
5,3
5,4
5,5
5,6
6
6,1
6,2
6,3
6,4
6,5
6,6
P(8) = 1/36 = .028
More Sample Space
Same problem can have different “look” at
sample space:
If in craps, if all we care about are “pips”
S = (2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)
P(2) = 1/36 = .028
P(3) = 2/36 = .056
Key Points Independence
One outcome does not affect another outcome
ie. If I roll a 6 with one die, it won’t affect my chances
of rolling a 6 with the second die
WITH Replacement
Example: pick a card from a deck, put it back
and pick another…P(2 Aces)
WITHOUT Replacement
Example: pick two cards from a deck without
replacing the first card…P(2 Aces)
Rules o’ Probability
 Probability of an event is between 0 and 1
PROBSPEAK:
 All possible probabilities add up to 1
PROBSPEAK:
P( S )  1
 Probability of an event NOT happening is 1 minus
the probability of the event happening. The
probability of an event NOT happening is called
the COMPLIMENT of an event and is written Ac
PROBSPEAK:
P( Ac )  1  P( A)
Rules o’ Probability Continued
 If two events are disjoint (mutually exclusive), they
have no outcomes in common. For example, in
craps, rolling a 5 AND a 7 is disjoint, one roll can’t
produce both outcomes.
S
Therefore (for disjoint events):
P( AorB)  P( A)  P( B)
AND (for disjoint events)……..
P( AandB)  0
A
B
Rules o’ Probability Continued Again

If two events are NOT disjoint (not
mutually exclusive) but ARE
independent . For example, roll 2
dice
Event A: Die 1 Shows a 6 P(A)=1/6
Event B: Die 2 Shows a 6 P(B)=1/6
P(A and B)= P(A)*P(B)
= 1/6 * 1/6 = 1/36 = .028ish
S
A
B
Disjoint Events Are NOT Independent
Hurting your brain?
Just think…If I roll two die and add up the
pips, what are the chances that I get a 5
and a 7.
That’s why (in disjoint
S
events)
Roll 5
P(A and B)=0
Roll 7
CHAPTER 15
Probability Goes Crazy with the
Cheez Whiz
Tree Diagrams
OUTCOMES
 ie. Flip 3 Coins,
Count # of
Heads
H 3H, 0T
H
T 2H, 1T
H
H 2H, 1T
T
T 1H, 2T
H 2H, 1T
H
T 1H, 2T
T
H 1H, 2T
T
FLIP 1
T 0H, 3T
FLIP 2
FLIP 3
3 Flip outcomes
3 Heads
1/8
P= .125
2 Heads
3/8
.375
1 Head
3/8
.375
0 Heads
1/8
.125
More (and evilererer) Probability Rules
 Addition Rule – 3 or more disjoint events
S
A
C
B
P( AorBorC...)  P( A)  P( B)  P(C )...
Addition Rule – Non-Disjoint Events
S
A
B
 Find P(A or B)
 If I do P(A)+P(B) the
area A&B gets
counted twice
 To make it work, I do
P(A) + P(B) and then
subtract one P(A&B)
P( AorB)  P( A)  P(B)  P( AandB)
Example – P(Male Or Senior Citizen)
 P(Male)=0.5
 P(65+)=0.2
 Assume
independence for
Maleness & Oldness
S
A
B
P( Maleor65)  P( Male)  P(65)  P( Maleand65)
=
.5
+
.2
=
.7
-
.1
=
.6
-
(.5)(.2)
Joint Events
 Not always independent - Can’t assume
 Example – Survey of music tastes at WPS
Probability of student liking hip-hop (A)
 P(A)=0.5
Probability of student liking rap (B)
 P(B)=0.4
 Think! Isn’t there a decent chance that people
who like hip hop may be more likely to like rap
as well….
Proportion of students who like BOTH rap and hip hop
P(A and B)=0.3
Joint Events - Continued
Probability of student liking
hip-hop (A)
 P(A)=0.5
Probability of student liking
rap (B)
 P(B)=0.4
Proportion of students who
like BOTH rap and hip hop
P(A and B)=0.3
S
Hip
Hop
0.2
0.1
0.3
RAP
0.4
Joint Events – Same thing – Using a table
 P(A)=0.5
 Probability of
student liking
rap (B)
Rap
Hip-Hop
 Probability of
student liking
hip-hop (A)
Like
Dislike
Total
Like
0.3
0.2
0.5
Dislike
0.1
0.4
0.5
Total
0.4
0.6
1.0
 P(B)=0.4
 Proportion of
students who
like BOTH rap
and hip hop
 P(A and
B)=0.3
S
Hip
Hop
0.2
0.1
0.3
RAP
0.4
Conditional Probability
Main Idea:
Probability can change if we know some other
event has occurred
World Poker Championships
Flushes are good – All same suit
You get 2 cards that are secret, then 5 cards
are dealt for the community
You make the best 5-card hand you can
World Poker Championships
My Hand
♠ ♠
Community Cards
♠ ♠ ♦ ♥ ?
 Wow, I’m close to a flush! What is the probability that the
last card (the river) is a ♠?
 Overall, the chance of a ♠ is 13/52 or .25, but we already
know what 6 cards are and that 4 of them are ♠s…
 Find Probability(♠ given that 4 of 6 visible cards are ♠s)
 ProbSpeak: P(Spade 4 of 6 visible spades)
 Think: 13Spades  4 spades 9
52Cards  6Cards

46
 0.1956
General Rule for Any Two Events
P (A and B) = P(A)P(BІA)
Example:
Probability of getting 2 aces in two successive
draws (no replacement)
P(Ace on 1st and Ace on 2nd)
=P(Ace on 1st)P(Ace on 2nd І Ace on 1st)
=4/52*3/51=0.0045
 Notice if replacement (independence), the formula still
works since P(Ace on 2nd І Ace on 1st)= P(Ace on 1st)
 Therefore 4/52*4/52=0.0059
Definition for Conditional Probability
 P (A and B) = P(A)P(BІA)
 Take this old Formula and solve for P(BІA)
P(A and B)
P(BA) =
P(A)
Using Decision Trees
 The following information gives information on DVD players sold by a
certain electronics store





Percent of Customers
Purchasing
Of those who purchase,
Percentage who
purchase extended
warranty
Brand 1
70
20
Brand 2
30
40
Let B1 = Event that Brand 1 is purchased
Let B2 = Event that Brand 2 is purchased
Let E = Event that Warranty is purchased
Therefore: P(B1)=.7 P(B2)=.3
AND!!!!!! P(E І B1)=.2 P(E І B2)=.4
Using Decision Trees - Continued
 P(B1)=.7
.7
B1
B2
.3
P(B2)=.3
P(E І B1)=.2
.2
E
.8
Not E
.4
P(E І B2)=.4
(.7)(.2)=.14
(.7)(.8)=.56
(.3)(.4)=.12
E
.6
Not E
(.3)(.6)=.18
Using Decision Trees – Continued 2
 NOW ANSWER QUESTIONS!!!!
.7
B1
B2
.3
.2
E
.8
Not E
.4
(.7)(.2)=.14
P(B1 and E) + P(B2 and E)
(.7)(.8)=.56
(.3)(.4)=.12
E
.6
Not E
What proportion of DVD
purchasers also purchased
the warranty?
(.3)(.6)=.18
P(E)=.14+.12=.26
Using Decision Trees – Bayes’s Rule
 What is probability of B1 given E
.7
B1
.2
E
.8
Not E
(.7)(.2)=.14
(.7)(.8)=.56
What proportion of DVD
purchasers also purchased
the warranty?
P(B1 and E) = .14
P(E)=.14+.12=.26
P(B1 І E) = .14/.26
B2
.3
.4
(.3)(.4)=.12
E
.6
Not E
(.3)(.6)=.18
P(B1 І E) = 0.539
CHAPTER 16
Random Variables
Discrete Random Variables
 Discrete????
Just means that there are a reasonable (countable)
number of options.
 What do we do with them? List outcomes and
then probabilities
Answer questions
Easy as pie…
Example – Rolling 2 dice
Outcomes (X)
Probability
2
3
4
5
6
7
8
9
10
11
12
.027
.055
.083
.121
.139
.167
.139
.121
.083
.055
.027
•Find P(X>9) = P(10)+P(11)+P(12)=.083+.055+.027 = .165
•Find P(X9) = P(9)+P(10)+P(11)+P(12)
=.121+.083+.055+.027 = .286
•Find P(5<X<8) = P(6)+P(7)=.139+.167 = .316
0.18
0.16
0.14
Really bad Histogram
0.12
0.1
Series1
0.08
0.06
0.04
0.02
0
“Thanks Microsoft!”
Continuous Random Variables
 Continuous?
Not countable
Example, think of all possible decimals between
0 and 1…Boy that’s a lot!
 If we threw down a histogram of a gazillion
random numbers between 0 and 1, we’d get
this:
Density
Uniform
Curve
Area
underneath
is 1.0
Distribution
0.0
1.0
Find Probabilities (Just area of rectangle):
0.8
0.0
P(X>0.8) =
0.2
1.0
Find Probabilities (Just area of rectangle):
0.0
0.2
P(.2<X<0.8) =
0.6
0.8
1.0
Check this out!
0.0
0.8
1.0
What is the probability P(X=0.8)?
0! ZERO! ZIP! NADA!
Why? Area of a straight line
is Zero, Yeah?
SO……
0.0
0.8
1.0
 P(X>0.8) is the same as P(X0.8)
 With Continuous Variables, It does not matter
which one you use…
 Cool, Huh???
What’s the sassiest density curve?
NORMAL DISTRIBUTION. YEAH!!!
Male Height N(68,2)
Let X=Ht in Inches
Find P(X>71)
Normalcdf(71,100000000,68,2)
P(X>71)=_____
Means and Variances of Random
Variables
 Example: 2005 AP Stat Scores
Outcome
1
2
3
4
5
Probabiity
.13
.23
.25
.19
.20
Remember x-bar is a sample mean, but we are
talking about the entire population of AP Stat
test takers, so we must use m (population
mean).
mx = 1(.13)+2(.23)+3(.25)+4(.19)+5(.20) = 3.1
Variances of Random Variables
 Recall: Variance is (Standard Deviation)2
 Here is the formula, It looks icky, but it’s pretty
easy to use…
   pi (xi  mX )
2
X
Variance
of X
Sum
2
Outcome
Value
Outcome
Probability
Mean of
outcomes
Means and Variances of Random
Variables
Outcome
1
2
3
4
5
Probabiity
.13
.23
.25
.19
.20
mx = 3.1
 X2   pi (xi  mX )2
 X2  pi (xi  mX )2  pi (xi  mX )2  pi (xi  mX )2  pi (xi  mX )2  pi (xi  mX )2
 X2  .13(1  3.1)2  .23(2  3.1)2  .25(3  3.1)2  .19(4  3.1)2  .20(5  3.1)2
 X2 
1.7903
Standard deviation
would be the square
root of this. 1.31ish
Law of Large Numbers
 How can we find the actual m of men’s heights?
 Not really realistic to measure every man in the
world
 Use x-bar as a reasonable estimate
 Gets more reasonable as the sample size
increases – That’s the LAW OF LARGE
NUMBERS.
 The larger the sample size, the more likely x-bar
will approach the m.
Rules for Means
If I taught in Canada, they would not dig
the average height of males in inches,
they like centimeters. Plus, all men there
measure their heights while wearing 8cm
high pumps. Very stylish!
How does that change the mean???
Rules for Means
 Here is the rule:
mabX  a  bm X
 Here is what we do with
those Canadian heights
(1in2.54cm)
m82.54(68)  8  2.54(68)  180.72cm
Rules for Means 2
 In volleyball there are two main blocking statistics, solo
blocks (by self) and assisted blocks (with a buddy). If
Chrissa Trudelle averaged .5 solo blocks and 1.3
assisted blocks per match, how many total blocks did she
average?
 Rule:
m X Y  m X  mY
 DO IT!
msoloassisted  .5  1.3  1.8 blocks / match
Rules For Variances
OK, the last rule was ridiculously easy, but
this next stuff is a bit rough.
Think about the men’s height and the
changes in Canada with the centimeters
and 8cm pumps.
How would these change the variance?
Rules For Variances
If we add the same value (8cm) to every
height, how does variance (and standard
deviation) change?
Right, variance does not change if I add
the same value to each height…
Rules For Variances
If we multiply each observation by the
same amount what will that do?
Right, multiplying the variance by a factor
will change the variance. Greater if >1 or
if <-1. Less if between 1 and -1
Rules For Variances – Linear Transform
 If given N(68,2) for average male height, and
we transform it again with 8+2.5X, what
happens to the variance?
 Rule
2
2 2
a bX
X

 DO IT!2
b
 82.5 X  (2.5) 2  (6.25)(4)  25
2
2
Standard deviation would be the
square root of this.
5
Rules For Variances – Add/Subtract
 Here are the formulas:

2
X Y
     2 p X  Y
2
X
2
Y
Don’t these look familiar???

2
X Y
p (rho) is like r, it shows
the correlation between X
and Y and is between -1
and 1. Should be stated
unless X and Y are
independent
     2 p X  Y
2
X
2
Y
Check this out
 Rearrange the formulas a bit…..

   2 p X  Y  
2
2
( X  Y )  X  2 XY  Y
2
X Y
2
X
2
Y
2
Perfect square trinomials???

   2 p X  Y  
2
2
( X  Y )  X  2 XY  Y
2
X Y
2
X
2
Y
2
Speaking of Rho
 That little p only affects things if there is some
correlation between the variables
If the problem lists a rho, you gotta use it
If it doesn’t list a rho, but it should have, do the problem
without it, but talk about how there could be some
correlation which would affect the variance (or standard
deviation)
If no correlation p = 0. Therefore:

2
X Y
2
X
  
2
Y

2
X Y
  
2
Y
2
X
Let’s use it now!
Coach Boff sweeps the gym floor in
N(10,2) minutes and mops the floor in
N(15,3) minutes. Assume that the time
sweeping and mopping are independent.
Find the mean and standard deviation of
the combined time.
Mean is easy. 10+15= 25 minutes
Now let’s find the standard deviation…
Let’s use it now!

2
X Y
  

2
X Y
2 3

2
X Y
 49

2
X Y
 13
2
X
2
2
2
Y
REMEMBER! You can
not add standard
deviations, you must
square them to get
variances, add the
variances and then
square root the sum!
 X Y  13  3.606
But if X and Y had p = .5

2
X Y
   2 p X  Y  

2
X Y
 2  2(.5)(2)(3)  3

2
X Y
 469
2
X Y
 19

2
X
2
 X Y  19  4.359
2
Y
2
Why more? If rho is
positive, X more likely
to be higher if Y is
also higher. Variation
moving in the same
way will increase the
variance.
Now you try!
Mr. Riebhoff and Mr. Marsheck are the
nation’s 1030th best partner biathlon team.
Mr. Riebhoff will do the running leg which
is a 10k road race where he has
historically had a time of N(48,5) in
minutes. Marsheck will do the bike ride of
50k where he has historically had a time of
N(106, 10) in minutes. What are the mean
and standard deviation of their combined
finish times?
Remember…
Show formula(ae) first
Talk about any assumptions you are
making
Don’t forget that standard deviation is the
square root of variance
Have fun!
CHAPTER 17
Probability Models
Binomial Distribution
 4 Requirements for a Binomial Distribution
2 outcomes – Success/Failure
I.e. – Heads or Tails, Boy or Girl Baby, Make or
Miss a Shot
Independent observations
Probability does not change when you learn the
result of a previous event
Probability for success (p) is constant for all
observations
FIXED NUMBER OF OBSERVATIONS!!!!!!!!!
5 Free throws, 17 exam questions, 20 Students
Important parts
n = # of Observations
Fixed # for a binomial distribution
p = Probability of success
Defined by you or the question
x = # of successes
Can be from 0 to n
Do you remember…
Normal Distribution:
N(m,) ex. N(68,2)
NOW! Binomial Distribution:
B(n,p)
Example:
A 70% free throw shooter shoots
10 Free throws
B(10,0.7)
Which of these would be Binomial?
 Flip a fair coin and count number of flips
until a head appears.
 350 students at WPS. 10% are 6th
graders. Choose 10 names at random
with no replacement and count # of 6th
graders.
 Shaq is a 52% free throw shooter.
Observe next 10 free throws and count #
of makes.
Binomial PDF
Remember Normal CDF?
Cumulative Distribution Function
NOW – Binomial PDF
Probability Distribution Function
Binomial PDF – 10 FT @ 70%
B(10,0.7)
X = 0 to 10
.3
.2
.1
0
1
Binompdf(10,0.7,0)
2
3
4
5
Binompdf(10,0.7,1)
6
7
8
9
10
Cumulative Distribution Function
 Cumulative
It accumulates, adds up…
EXAMPLE – 70% FT Shooter, 10 FTs
X=
0
1
2
3
4
5
6
7
8
9
10
PDF .000
.000
.001
.009
.037
.103
.200
.267
.233
.121
.029
CDF .000
.000
.001
.010
.047
.150
.350
.617
.850
.971
1.00
Graph the CDF
1.0
.75
.50
.25
0
1
2
3
4
5
6
7
8
9
10
Formulae for Binomial Distribution
Mean For Binomial Distribution
m = np
Makes sense yeah?
Example, I flip a coin 16 times, how many
heads?
m = 16(.5) = 8
Formulae for Binomial Distribution
 Standard Deviation For Binomial Distribution
 = np(1  p)
Why? Sausage. Just deal and know where
it is on the Formulae Sheet.
Ex. Find SD of 10 FT Problem
 = 10(.7)(1  .7)
 = 1.449
LET’S DO IT!
 Find Mean and Standard Deviation on 20 Free
Throws if…
 p=0.7
 p=0.8
 p=0.9
 p=0.99
 What happens to m as p approaches 1.0?
Math Attack
 Remember Factorials? -- n!
 Examples:
5! = 5*4*3*2*1 = 120
3! = 3*2*1 = 6
 Now the crazy stuff:
0! = 1
Kinda Like a0 = 1, yah?
 We’ll need these in a minute, you’ll see why.
Binomial Coefficient
# of ways I can get k successes in n tries.
Example: How many ways can I get three
tails in 5 flips?
Old Skool Way: (easy to mess up)
TTTHH TTHTH TTHHT THTTH
THTHT THHTT HTHTT HTTTH
HTTHT HHTTT
Impress your friends at the next math
party way…
Pascal’s Triangle
5 Choose 0
5 Choose 1
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
5 Choose 2
5 Choose 3
Formula Way
 Formula:
“n choose k”
n
n!
 k   k !(n  k )!
 
 Use it! 5 Choose 3
 5
5!
5 4 3 2 1 120
 3  3!(5  3)!  3 2 1 2 1  12  10
 
Binomial Probability
Recall 3 coins flipped, X = # of Heads
H 3H, 0T
H
T 2H, 1T
H
H 2H, 1T
T
H
T
FLIP
1
T
FLIP
2
0
1
2
3
1
3
3
1
T 1H, 2T
H 2H, 1T
P(X=0) = 1*P(HC)3 = .125
T 1H, 2T
H 1H, 2T
P(X=2) = 3*P(H)2 P(HC) = .375
T 0H, 3T
FLIP
3
X
P(X=1) = 3*P(HC)2 P(H) = .375
P(X=3) = 1*P(H)3 = .125
Imagine doing P(5 heads in 9 flips)
 What we need is a
formula…
Insert binomial
coefficient here…
n k
n k
P( X  k )    p (1  p)
k 
Let’s do 3 flips P(2 heads)
 3 2
32
P( X  2)    (.5) (1  .5)
 2
3!
2
1
P( X  2) 
(.5) (.5)
2!1!
3
P( X  2)  3(.5)  3(.125)  .375
Now You Try!
In a previous chapter, we found that the
probability of rolling a 7 (craps!) with two
fair die is 0.167. Let X be the number of
7’s rolled in a series of 10 rolls
Now You Try!
 #1 – Find the probability that 3 7’s will be rolled
in the 10 attempts
n k
n k
P( X  k )    p (1  p)
k 
Now You Try!
 #2 – Use your TI-83 and find the distribution of X
 binompdf(trials,p,x)
0
1
2
3
4
5
6
7
8
9
10
Now You Try!
 #3 – Find the m and  of the number of 7’s that
would be rolled in 10 attempts
m=
=
Geometric Distribution
 4 Requirements for a Geometric Distribution
2 outcomes – Success/Failure
I.e. – Heads or Tails, Boy or Girl Baby, Make or
Miss a Shot
Independent observations
Probability does not change when you learn the
result of a previous event
Probability for success (p) is constant for all
observations
Looking for # of trials needed for 1 success!!!!!!!
Flip a coin, how many flips until 1st Head?
Geometric vs. Binomial
Binomial
Shoot 10 FT’s with p(make)=0.7 find p(8
makes)
Geometric
With p(make) = 0.7, Shoot until 1st make, count
the number of attempts
Identify the Geometric Distributions
 A – Flip a coin until you get a head
 B – Record the number of times a player makes
both shots in a one-and-one foul-shooting
situation. (In this situation, you get to attempt a
second shot only if you make the first)
 C – Draw a card from the deck, observe it and
replace it into the deck. Count the number of
times you draw a card in this manner until you
observe a jack.
Identify the Geometric Distributions
 D – Buy a “pick 6” lottery ticket every week until
you win the lottery. Count the # of weeks it
takes for you to win.
 E – There are 10 red marbles and 5 blue
marbles in a jar. You reach in, and without
looking, select a marble. You want to know how
many marbles you need to draw (without
replacement), on average, in order to be sure
that you have 3 red marbles.
CRAPS! Roll till a 7 shows up
7
P(X=1) = 1/6 = .167
P=1/6
7
P=1/6
P(X=2) = (5/6)(1/6) = .139
P=5/6
Not 7
7
P=1/6
P(X=3) = (5/6)2(1/6) = .116
P=5/6
Not 7
7
P=1/6
FORMULA FOR
GEOMETRIC
PROBABILITIES
P=5/6
Not 7
n 1
P( X  n)  (1  p) ( p)
P=5/6
Not 7
P(X=4)=
(5/6)3(1/6)=
.097
Let’s try it!
Mr. Riebhoff is U-G-L-Y (he ain’t got no
alibi…). In college, he had only a 20%
chance of a randomly selected woman (he
used a random # table) agreeing to meet
him for a soda.
Let’s try it!
 #1 – Find a probability distribution from x = 1
to x = 5 that shows x = the # of females he
would ask before getting a “yes”
n 1
P( X  n)  (1  p) ( p)
1
2
3
4
5
Let’s try it!
 #2 – Make a CDF of the data from #1
1.0
0.0
1
2
3
4
5
Let’s try it!
 #3 – What is the probability that after 5 girls
asked, Riebhoff would still be dateless?
Using the TI-83
geometpdf(p,x)
Number if trials ‘till success
In Riebhoff Date Example:
geometpdf(0.2,1) =
geometpdf(0.2,2) =
geometpdf(0.2,3) =
geometpdf(0.2,4) =
Mean of Geometric Random Variable
Common Sense
Guess what the mean number of rolls I
would need to roll a 5 on a fair die?
Guess the mean number of flips I would
need to get a head on a fair coin?
 m= 1/p