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