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 Long Run

10 Flips 100 Flips 999Flips

What is Probability

Over a HUGE number of trials (probability is Long-Term), 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

P= .261 .261 .304 .164

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 1PROBSPEAK:

All possible probabilities add up to 1PROBSPEAK:

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:

( ) 1P S =

( ) 1 ( )cP A 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.

Therefore (for disjoint events):

AND (for disjoint events)……..

( ) ( ) ( )P AorB P A P B= +

( ) 0P AandB =

S

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

BA&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

events)

P(A and B)=0

S

Roll 5

Roll 7

CHAPTER 15

Probability Goes Crazy with the Cheez Whiz

Tree Diagrams ie. Flip 3 Coins,

Count # of Heads

H

T

H

T

H

T

H

T

H

T

H

T

H

TFLIP 1

FLIP 3FLIP 2

OUTCOMES

3H, 0T

2H, 1T

2H, 1T

1H, 2T

2H, 1T

1H, 2T

1H, 2T

0H, 3T

3 Flip outcomes

3 Heads 2 Heads 1 Head 0 Heads

1/8 3/8 3/8 1/8

P= .125 .375 .375 .125

More (and evilererer) Probability Rules

Addition Rule – 3 or more disjoint events

S

A

B

C

( ...) ( ) ( ) ( )...P AorBorC P A P B P C= + +

Addition Rule – Non-Disjoint Events

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)

S

A

B

A&B

Example – P(Male Or Senior Citizen)

P(Male)=0.5 P(65+)=0.2 Assume

independence for Maleness & Oldness

P( Maleor65+) =P(Male) + P(65+)−P(Maleand65+)

S

A

B

A&B

= .5 + .2 - (.5)(.2)

= .7 - .1

= .6

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

HipHop

RAP

BOTH

0.30.2 0.1

0.4

Joint Events – Same thing – Using a table

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

Like Dislike Total

Like 0.3 0.5

Dislike

Total 0.4 1.0

Rap

Hip

-Hop

0.1 0.5

0.6

0.4

0.2

SHipHop

RAP

BOTH

0.30.2 0.1

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:

?

13 4 90.1956

52 6 46

Spades spades

Cards Cards

−= =

−

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

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

Percent of Customers Purchasing

Of those who purchase, Percentage who

purchase extended warranty

Brand 1 70 20

Brand 2 30 40

Using Decision Trees - Continued

P(B1)=.7 P(B2)=.3 P(E І B1)=.2 P(E І B2)=.4

B1

B2

E

Not E

E

Not E

.7

.2

.3 .4

.8

.6

(.7)(.2)=.14

(.7)(.8)=.56

(.3)(.4)=.12

(.3)(.6)=.18

Using Decision Trees – Continued 2

NOW ANSWER QUESTIONS!!!!

B1

B2

E

Not E

E

Not E

.7

.2

.3 .4

.8

.6

(.7)(.2)=.14

(.7)(.8)=.56

(.3)(.4)=.12

(.3)(.6)=.18

What proportion of DVD purchasers also purchased the warranty?

P(B1 and E) + P(B2 and E)

P(E)=.14+.12=.26

Using Decision Trees – Bayes’s Rule

What is probability of B1 given E

B1

B2

E

Not E

E

Not E

.7

.2

.3 .4

.8

.6

(.7)(.2)=.14

(.7)(.8)=.56

(.3)(.4)=.12

(.3)(.6)=.18

What proportion of DVD purchasers also purchased the warranty?

P(B1 І E) = .14/.26

P(B1 and E) = .14P(E)=.14+.12=.26

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) 2 3 4 5 6 7 8 9 10 11 12

Probability .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(X9) = 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

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

Series1

Really bad Histogram

“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:

0.0 1.0

Uniform Distribution

Density Curve

Area underneath is 1.0

Find Probabilities (Just area of rectangle):

P(X>0.8) =

0.0 1.00.8

0.2

Find Probabilities (Just area of rectangle):

P(.2<X<0.8) =

0.0 1.00.8

0.6

0.2

Check this out!

What is the probability P(X=0.8)?

0.0 1.00.8

0! ZERO! ZIP! NADA!

Why? Area of a straight line is Zero, Yeah?

SO……

P(X>0.8) is the same as P(X0.8) With Continuous Variables, It does not matter

which one you use… Cool, Huh???

0.0 1.00.8

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 (population mean)

x = 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…

2 2i(x )X i Xpσ = −∑

Variance of X

Sum

Outcome Probability

Mean of outcomes

Outcome Value

Means and Variances of Random Variables

2 2i(x )X i Xpσ = −∑

Outcome 1 2 3 4 5

Probabiity .13 .23 .25 .19 .20

2 2 2 2 2 2i i i i i(x ) (x ) (x ) (x ) (x )X i X i X i X i X i Xp p p p pσ = − + − + − + − + −

x = 3.1

2 2 2 2 2 2.13(1 3.1) .23(2 3.1) .25(3 3.1) .19(4 3.1) .20(5 3.1)Xσ = − + − + − + − + −

2Xσ =

Standard deviation would be the square root of this. 1.31ish

1.7903

Law of Large Numbers

How can we find the actual 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 .

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:

Here is what we do with those Canadian heights (1in2.54cm)

a bX Xa b + = +

8 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:

DO IT!

X Y X Y + = +

.5 1.3 1.8 /solo assisted 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

DO IT!

2 2 2a bX Xbσ σ+ =

2 2 28 2.5 (2.5) 2 (6.25)(4) 25Xσ + = = =

Standard deviation would be the

square root of this. 5

Rules For Variances – Add/Subtract

Here are the formulas:

2 2 2

2 2 2

2

2

X Y X Y X Y

X Y X Y X Y

p

p

σ σ σ σ σ

σ σ σ σ σ

+

−

= + +

= + −

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

Don’t these look familiar???

Check this out

Rearrange the formulas a bit…..

2 2 2

2 2 2

2

2

X Y X X Y Y

X Y X X Y Y

p

p

σ σ σ σ σ

σ σ σ σ σ

+

−

= + +

= − +

Perfect square trinomials???

2 2 2

2 2 2

( ) 2

( ) 2

X Y X XY Y

X Y X XY Y

+ = + +

− = − +

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 2 2

2 2 2

X Y X Y

X Y X Y

σ σ σ

σ σ σ+

−

= +

= +

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 minutesNow let’s find the standard deviation…

Let’s use it now!

2 2 2

2 2 2

2

2

2 3

4 9

13

13 3.606

X Y X Y

X Y

X Y

X Y

X Y

σ σ σ

σ

σ

σ

σ

+

+

+

+

+

= +

= +

= +

=

= ≈

REMEMBER! You can not add standard deviations, you must square them to get variances, add the variances and then square root the sum!

But if X and Y had p = .5

2 2 2

2 2 2

2

2

2

2 2(.5)(2)(3) 3

4 6 9

19

19 4.359

X Y X X Y Y

X Y

X Y

X Y

X Y

pσ σ σ σ σ

σ

σ

σ

σ

+

+

+

+

+

= + +

= + +

= + +

=

= ≈

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) firstTalk about any assumptions you are

makingDon’t forget that standard deviation is the

square root of varianceHave 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 observationsProbability 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

THEKEY

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(,σ) 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?

NOW – Binomial PDFCumulative Distribution Function

Probability Distribution Function

Binomial PDF – 10 FT @ 70%

B(10,0.7)X = 0 to 10

0 1 2 3 4 5 6 7 8 9 10

Binompdf(10,0.7,0) Binompdf(10,0.7,1)

.3

.2

.1

Cumulative Distribution Function

Cumulative It accumulates, adds up… EXAMPLE – 70% FT Shooter, 10 FTs

0 1 2 3 4 5 6 7 8 9 10X =

.000 .000 .001 .009 .037 .103 .200 .267 .233 .121 .029PDF

.000 .000 .001 .010 .047 .150 .350 .617 .850 .971 1.00CDF

Graph the CDF

0 1 2 3 4 5 6 7 8 9 10

.25

1.0

.75

.50

Formulae for Binomial Distribution

Mean For Binomial Distribution = np Makes sense yeah? Example, I flip a coin 16 times, how many

heads? = 16(.5) = 8

Formulae for Binomial Distribution

Standard Deviation For Binomial Distributionσ= Why? Sausage. Just deal and know where

it is on the Formulae Sheet. Ex. Find SD of 10 FT Problemσ=σ= 1.449

(1 )np p−

10(.7)(1 .7)−

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 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

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

5 Choose 0

5 Choose 2

5 Choose 1

5 Choose 3

Formula Way

Formula:

Use it! 5 Choose 3

!

!( )!

n n

k k n k

⎛ ⎞=⎜ ⎟ −⎝ ⎠

5 5! 5 4 3 2 1 12010

3 3!(5 3)! 3 2 1 2 1 12

⎛ ⎞= = = =⎜ ⎟ −⎝ ⎠

g gg gg gg g

“n choose k”

Binomial Probability

Recall 3 coins flipped, X = # of Heads

H

T

H

T

H

T

H

T

H

TH

TH

TFLIP 1

FLIP 3

FLIP 2

3H, 0T

2H, 1T

2H, 1T

1H, 2T

2H, 1T

1H, 2T1H, 2T

0H, 3T

X 0 1 2 3

1 3 3 1

P(X=0) = 1*P(HC)3 = .125

P(X=1) = 3*P(HC)2 P(H) = .375

P(X=2) = 3*P(H)2 P(HC) = .375

P(X=3) = 1*P(H)3 = .125

Imagine doing P(5 heads in 9 flips)

What we need is a formula…

( ) (1 )k n knP X k p p

k−⎛ ⎞

= = −⎜ ⎟⎝ ⎠

Insert binomial coefficient here…

Let’s do 3 flips P(2 heads)

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

( ) (1 )k n knP X k p 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 and σ of the number of 7’s that would be rolled in 10 attempts

=

σ =

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 observationsProbability 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?

THEKEY

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

Not 7

P=1/6

P=5/6

P(X=1) = 1/6 = .167

7

Not 7

P=1/6

P=5/6

P(X=2) = (5/6)(1/6) = .139

7

Not 7

P=1/6

P=5/6

P(X=3) = (5/6)2(1/6) = .116

7

Not 7

P=1/6

P=5/6

P(X=4)= (5/6)3(1/6)= .097FORMULA FOR

GEOMETRIC PROBABILITIES

1( ) (1 ) ( )nP X n p p−= = −

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”

1 2 3 4 5

1( ) (1 ) ( )nP X n p p−= = −

Let’s try it!

#2 – Make a CDF of the data from #1

1 2 3 4 5

1.0

0.0

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)

In Riebhoff Date Example: geometpdf(0.2,1) = geometpdf(0.2,2) = geometpdf(0.2,3) = geometpdf(0.2,4) =

Number if trials ‘till success

Mean of Geometric Random Variable

Common SenseGuess 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?= 1/p