PROBABILITY MODELS

CHAPTER 17

Ted S

hi, K

evin

Yen

Bette

rs, 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?

• 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…

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 / pσ = √ ( q / p2 )

p = probability of successq = 1 – p = probability of failureX = no. of trials until first success

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 trialsp = probability of successq = 1 – p = probability of failurex = no. of successes in n trials

BINOMIAL MODEL CONT.

Formulas (cont.):μ = ( n )( p )σ = √ ( n )( q )( p )

n = number of trialsp = probability of successq = 1 – p = probability of failurex = no. of successes in n trials

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 ConditionA Binomoial model can be considered Normal if we expect

np ≥ 10 and nq ≥ 10

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

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

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: RightIndependent? No, but less than 10% of population.

QUESTION 17: STILL MORE LEFTIES

A) Find the mean and standard deviation of the number of right-handers in the group.

P (right-handed) = 0.87

μ = ( n )( p ) σ = √ ( n )( q )( p ) = (12)(0.87) = √ (12)(0.13)(0.87) = 10.44 people = 1.164 people

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 )(0.8712)(0.130)

= 0.812

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

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+(121)(0.13)11(0.87)1...+(12

2)

(0.13)10(0.87)2

= 0.475

P (no more than10 righties) = 0.475

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

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 + (128)(0.13)4 (0.87)8 +... (12

12)

(0.13)0 (0.87)12

= 0.998

P (majority righties) = 0.998

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.

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)

= (66)(0.70)6 (0.30)0

= 0.118

P (all six serves in) = 0.118

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

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)

= (64)(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

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 + (61)(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