# PROBABILITY MODELS

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PROBABILITY MODELS

CHAPTER 17

Ted Shi, Kevin Yen

Betters, 1st

Slide2WHY DO WE USE PROBABILITY?

Instead of simulations, we can use actual probabilities to determine the chance of an event occurring.

Slide3WHAT IS A BERNOULLI TRIAL?

Two possible outcomes on each trial:

Success

Failure

Probability,

p

, is the same on each trail

Each trial is independent of one another

Or, if not independent, then sample must be < 10% of pop.

Not usually that interesting…

Slide4WHY/HOW DO WE USE A GEOMETRIC PROBABILITY MODEL?

Tell us how long, and with what probability, we can achieve success, denoted by “Geom(p)”Follows rules of Bernoulli’s trials.Formulas: P ( X = x ) = ( q x – 1 )( p ) μ = 1 / p σ = √ ( q / p2 )p = probability of success q = 1 – p = probability of failureX = no. of trials until first success

Slide5WHY/HOW DO WE USE A BINOMIAL PROBABILITY / MODEL?

Interested in number of successes (usually > 1) in a specific number of trials

Not as easy as it seeeeems.

Formulas:

P ( X = x ) = ( n C x ) ( p

x

)( q

n - x

)

n = number of trials

p = probability of success

q = 1 – p = probability of failure

x = no. of successes in n trials

Slide6BINOMIAL MODEL CONT.

Formulas (cont.): μ = ( n )( p ) σ = √ ( n )( q )( p )n = number of trialsp = probability of success q = 1 – p = probability of failurex = no. of successes in n trials

Slide7APPLYING THE NORMAL MODEL

Normal models extend indefinitely – need to consider 3 standard deviations

Useful when dealing with

large number of trials

involving binomial models

Works well if we expect at least 10 success/failures

Success/Failure Condition

A Binomoial model can be considered Normal if we expect

np ≥ 10

and

nq ≥ 10

Slide8CONTINUOUS RANDOM VARIABLES

The Binomial Model is discrete (success = 1, 2, etc.)

The Normal Model is continuous (any random variable at any value :o )

Continuity Correction

“For continuous random variables we can no longer list all the possible outcomes and their probabilities”

e.g. donating ≥ 1850 pints of blood vs donating exactly 1850 pints of blood

Slide9QUESTION 17: STILL MORE LEFTIES

Original premise:

Question 13.

Assume that 13% of people are left-handed. If we select 5 people at random, find the probability of each outcome described below

Slide10QUESTION 17: STILL MORE LEFTIES

Our Situation:

Question 17.

Suppose we choose 12 people instead of the 5 chosen in Exercise 13.

Conditions:

Is this a Bernoulli trial? Yes.

Success: LEFT, Failure: Right

Independent? No, but less than 10% of population.

Slide11QUESTION 17: STILL MORE LEFTIES

A) Find the mean and standard deviation of the number of right-handers in the group.P (right-handed) = 0.87μ = ( n )( p ) σ = √ ( n )( q )( p ) = (12)(0.87) = √ (12)(0.13)(0.87) = 10.44 people = 1.164 people

Slide12QUESTION 17: STILL MORE LEFTIES

B)What’s the probability that they’re not all right- handed?

P (all not right-handed) = 1 – P(all right-handed)

= 1 – P (X = x) = 1 – P( X = 12 right handed ppl)

= 1 – (

12

C

12

)(0.87

12

)(0.13

0

)

= 0.812

p (right handed) = 0.87, p (left handed) = 0.13

Slide13QUESTION 17: STILL MORE LEFTIES

C) What’s the probability that there are no more than 10 righties?

P (no more than10 righties)

= P (X ≤ 10) = P (X =0) + P (X = 1) + P (X =2) ...+P (X = 10)

= (

12

0

)(0.13)

12

(0.87)

0

+(

12

1

)(0.13)

11

(0.87)

1

...+(

12

2

)(0.13)

10

(0.87)

2

= 0.475

P (no more than10 righties) = 0.475

Slide14QUESTION 17: STILL MORE LEFTIES

D) What’s the probability that there are exactly 6 of each?

P (exactly 6 of each) = P(Y =6)

= (

12

6

) (0.13)

6

(0.87)

6

= 0.00193

P (exactly 6 of each) = 0.00193

Slide15QUESTION 17: STILL MORE LEFTIES

E) What’s the probability that the majority is right- handed?

P (majority righties)

= P(Y ≥ 7)

= P (X ≤ 10) = P (X = 7) + P (X = 8) + P (X = 9) ...+P (X = 12)

= (

12

7

)(0.13)

5

(0.87)

7

+ (

12

8

)(0.13)

4

(0.87)

8

+... (

12

12

)(0.13)

0

(0.87)

12

= 0.998

P (majority righties) = 0.998

Slide16QUESTION 19: TENNIS, ANYONE?

Question 19.

A certain tennis player makes a successful first serve 70% of the time. Assume that each serve is independent of the others.

Slide17QUESTION 19: TENNIS, ANYONE?

A) If she serves 6 times, what’s the probability she gets

all 6 serves in?

P (all six serves in) = P (X = 6)

= (

6

6

)(0.70)

6

(0.30)

0

= 0.118

P (all six serves in) = 0.118

Slide18QUESTION 19: TENNIS, ANYONE?

B) If she serves 6 times, what’s the probability she gets

exactly 4 serves in?

P (exactly four serves in) = P (X = 4)

= (

6

4

)(0.70)

4

(0.30)

2

= 0.324

P (exactly four serves in) = 0.324

Slide19QUESTION 19: TENNIS, ANYONE?

C)If she serves 6 times, what’s the probability she gets

at least 4 serves in?

P (at least four serves in) = P (X = 4) + P (X = 5) = P (X = 6)

= (

6

4

)(0.70)

4

(0.30)

2

+ (

6

5

)(0.70)

5

(0.30)

1

+ (

6

6

)(0.70)

6

(0.30)

0

= 0.744

P (at least four serves in) = 0.744

Slide20QUESTION 19: TENNIS, ANYONE?

D) If she serves 6 times, what’s the probability she gets

no more than 4 serves in?

P (no more than four serves in)

= P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

= (

6

0

)(0.70)

0

(0.30)

6

+ (

6

1

)(0.70)

1

(0.30)

5

+ (

6

2

)(0.70)

2

(0.30)

4

+ (

6

3

)(0.70)

3

(0.30)

3

+ (

6

4

)(0.70)

4

(0.30)

2

= 0.580

P (no more than four serves in) = 0.580

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## PROBABILITY MODELS

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