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CHAPTER 17. Ted Shi, Kevin Yen. Betters, 1st. WHY DO WE USE PROBABILITY?. Instead of simulations, we can use actual probabilities to determine the chance of an event occurring.. WHAT IS A . BERNOULLI TRIAL. ID: 330003 Download Presentation

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## PROBABILITY MODELS

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Slide1

PROBABILITY MODELS

CHAPTER 17

Ted Shi, Kevin Yen

Betters, 1st

Slide2

WHY DO WE USE PROBABILITY?

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

Slide3

WHAT 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…

Slide4

WHY/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

Slide5

WHY/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

Slide6

BINOMIAL 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

Slide7

APPLYING 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

Slide8

CONTINUOUS 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

Slide9

QUESTION 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

Slide10

QUESTION 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.

Slide11

QUESTION 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

Slide12

QUESTION 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

Slide13

QUESTION 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

Slide14

QUESTION 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

Slide15

QUESTION 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

Slide16

QUESTION 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.

Slide17

QUESTION 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

Slide18

QUESTION 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

Slide19

QUESTION 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

Slide20

QUESTION 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

Slide21

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