Download Discrete Probability Distributions handout

10/13/2014
Midterm Contents
Section 4.1
Tues, Oct. 21st
Introduction
1.1
Data & Measurements
1.2 - 2.2
Descriptive Statistics
2.3 - 2.5
Probability I
3.1 - 3.2
Probability II
3.3 - 3.4
Discrete
Probability
Distributions
A hypochondriac at heart, he thought
(Though symptom free) he had a dire disease,
And after fruitless weeks of worry, sought
Some test to take to set his mind at ease.
He forthwith found one that would do the trick,
And accurate (at oh point nine) to tell
Those having the disease that they were sick,
And just the same, the well that they were well.
One crucial point he failed to note was this:
That of a hundred like him, only one
Had the disease, and this slip made him miss
The implication when the test was done
Probability Distributions 4.1 – 4.3
- approximately 1 hour long
- format short answer, similar to homework
- bring calculator, 1 sheet allowed
Larson/Farber Ch. 4
Why Richard Cory Offed Himself
Or One Reason to Take a Course in Probability
by Jasper D. Memory (1935 - )
And positive! therefore, consumed with dread,
And now convinced his blackest fears were right
(By faulty logic fatally misled),
He shattered silence that calm summer night.
Larson/Farber Ch. 4
Random Variables
Types of Random Variables
A random variable, x, is the numerical outcome of a
probability experiment. Value occurs by chance.
A random variable is discrete if the number of possible
outcomes is finite or countable. Discrete random variables
are determined by a count.
x = number of people in a car
x = gallons of gas bought in a week
x = time it takes to drive from home to school
A random variable is continuous if it can take on any
value within an interval. The possible outcomes cannot be
listed. Continuous random variables are determined by a
measure.
x = number of trips to school you make per week
Larson/Farber Ch. 4
Larson/Farber Ch. 4
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Types of Random Variables
Identify each random variable as discrete or continuous.
x = number of people in a car
Discrete Probability Distributions
A discrete probability distribution lists each possible
value of the random variable, together with its probability.
A survey asks how many
vehicles in each family.
Discrete – count the number of people in a car
x = gallons of gas bought in a week
number of
vehicles
Continuous – measure the gallons of gas. You cannot list
the possible values.
x = time it takes to drive from home to school
Continuous – measure the amount of time.
x
0
1
2
3
P(x)
0.004
0.435
0.355
0.206
Properties of a probability distribution
• Each probability must be between 0 and 1, inclusive.
x = number of trips to school you make per week
Discrete – count the number of trips you make.
• The sum of all probabilities is 1. May have rounding.
Larson/Farber Ch. 4
Larson/Farber Ch. 4
Probability Histogram
Mean, Variance, and Standard Deviation
0.435
.40
The mean of a discrete probability distribution is:
0.355
P(x)
.30
0.206
.20
The variance of a discrete probability distribution is:
.10
0.004
0
00
11
22
33
x
Number of Vehicles
• The height of each bar corresponds to the probability of x.
• When the width of the bar is 1, the area of each bar
corresponds to the probability the value of x will occur.
Larson/Farber Ch. 4
The standard deviation of a discrete probability
distribution is:
Since P(x) = f/n, formulas are equivalent to grouped data formulas in Chptr. 2
Larson/Farber Ch. 4
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Mean (Expected Value)
Calculate the mean
Calculate the Variance and Standard
Deviation
The mean is 1.763 vehicles.
Multiply each value by its probability. Add the
products
x
0
1
2
3
P(x)
0.004
0.435
0.355
0.206
x
0
1
2
3
xP(x)
0
0.435
0.71
0.618
1.763
P(x)
0.004
0.435
0.355
0.206
x- μ
-1.763
-0.763
0.237
1.237
(x -μ )
3.108
0.582
0.056
1.530
variance
=
The expected value (the mean) is 1.763 vehicles.
Larson/Farber Ch. 4
P(x)(xP(x)
- )
0.012
0.253
0.020
0.315
0.601
The standard deviation is 0.775 vehicles.
Larson/Farber Ch. 4
Binomial Experiments
Section 4.2
Binomial Distributions
Characteristics
• Fixed number of trials (n).
• n trials are independent and identically repeated.
• Each trial has 2 outcomes, S = Success or F = Failure.
• Probability of success on a single trial is p. P(S) = p
• Probability of success is the same in each trial
• Probability of failure is q. P(F) =q where p + q = 1
• Challenge: find the probability of x successes out of n trials.
Where x = 0 or 1 or 2 … n.
x is a count of the number of successes in n trials
Examples: coin tosses, basketball free throws
Larson/Farber Ch. 4
Larson/Farber Ch. 4
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Exercise Example: Guess the Answers
1. What is the 11th digit after the decimal point for the irrational number e?
(a) 2
(b) 7
(c) 4
(d) 5
2. What was the Dow Jones Average on February 27, 1993?
(a) 3265
(b) 3174
(c) 3285
(d) 3327
3. How many Sri Lankan students studied at U.S. universities from 1990-91?
(a) 2320
(b) 2350
(c) 2360
(d) 2240
4. How many kidney transplants were performed in 1991?
(a) 2946
(b) 8972
(c) 9943
(d) 7341
Results
The correct answers to the quiz are:
1. d
2. a
3. b
4. c
5. b
Count the number of correct answers. Let the
number of correct answers = x.
Why is this a binomial experiment? success or failure
What are the values of n, p and q? n=5, p=.25, q=.75
5. How many words are in the American Heritage Dictionary?
(a) 60,000
(b) 80,000
(c) 75,000
(d) 83,000
Larson/Farber Ch. 4
What are the possible values for x? X{0,1,2,3,4,5}
Larson/Farber Ch. 4
Binomial Experiments
Example: A multiple choice test has 8 questions each of which
has 3 choices, one of which is correct. You want to know the
probability that you guess exactly 5 questions correctly.
Find n, p, q, and x.
n=8
p = 1/3
q = 2/3
p = 0.80
q = 0.20
Find the probability of getting exactly 3 questions correct
out of 5 on a quiz (with 4 choices for each question).
1. Write the first 3 correct and the last 2 wrong as SSSFF
P(SSSFF) = (.25)(.25)(.25)(.75)(.75) = (.25)3(.75)2 = 0.0087954
x=5
Example: A doctor tells you that 80% of the time a certain type of
surgery is successful. If this surgery is performed 7 times, find the
probability exactly 6 surgeries will be successful. Find n, p, q, and x.
n=7
Binomial Probabilities
x=6
2. Since order does not matter, you could get any combination
of three correct out of five questions.
SSSFF
FFSSS
SSFSF
FSFSS
SSFFS
FSSFS
SFFSS
SFSSF
SFSFS How to calculate
FFSSF the number?
Each of these 10 ways has a probability of 0.00879.
What is the probability of 3 out of 5?
P(x = 3) = 10(0.25)3(0.75)2 = 10(0.00879) = 0.0879
Larson/Farber Ch. 4
Larson/Farber Ch. 4
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Binomial Probabilities
Combination of n values, choosing x
In a binomial experiment, the probability of exactly x successes
in n trials is:
There are
ways.
Calculate the probability of getting 0, 1, 2, 3, 4, or all 5 correct
Each of these 10 ways has a probability of 0.00879.
P(x = 3) = 10(0.25)3(0.75)2= 10(0.00879)= 0.0879
P(3) = 0.088
Larson/Farber Ch. 4
x
0
1
2
3
4
5
Binomial Histogram
Probability (x)
.30
.396
.264
.237
Probabilities
P(x)
0.237
0.396
0.264
0.088
0.015
0.001
1. What is the probability of answering
either 2 or 4 questions correctly?
P(x = 2 or x = 4) = 0.264 + 0.015 = 0. 279
x
0
1
2
3
4
5
P(x)
0.237
0.396
0.264
0.088
0.015
0.001
2. What is the probability of answering at least 3 questions correctly?
.20
P(x  3) = P( x = 3 or x = 4 or x = 5) = 0.088 + 0.015 + 0.001 = 0.104
.088
.10
0
0
P(5) = 0.001
Larson/Farber Ch. 4
Binomial Distribution
.40
P(4) = 0.015
1
2
3
.015
.001
4
5
3. What is the probability of answering at least one question correctly?
x
P(x  1) = 1 - P(x = 0) = 1 - 0.237 = 0.763
Table 2 in text (pg. A 8) has binomial probability distribution values
Larson/Farber Ch. 4
Larson/Farber Ch. 4
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Binomial Distribution
Parameters for a Binomial Experiment
Mean:
Variance:
Standard deviation:
Find the mean, variance, and standard deviation for the
distribution of correct answers on the quiz.
Find: P(3 heads in 5 coin tosses); P(3 out of 5 correct with 4 choices for each question)
0.312 (p=0.5),
0.088 (p=0.25)
Larson/Farber Ch. 4
Larson/Farber Ch. 4
The Geometric Distribution
Section 4.3
More Discrete
Probability Distributions
A geometric distribution is a discrete probability
distribution of the random variable x that satisfies:
1. A trial is repeated until a success occurs.
2. Repeated trials are independent of each other.
3. Probability of success p is the same for each trial.
The probability that the first success will occur on trial
number x is
P(x) = (q)x – 1p
where q = 1 – p
Larson/Farber Ch. 4
Larson/Farber Ch. 4
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The Geometric Distribution
The probability that a person who enters a store will make a purchase is 0.30
• The probability the first purchase will be made by the first person
who enters the store is 0.30.
P(1) = 0.30.
• The probability the first purchase will be made by the second
person who enters the store is (0.70) ( 0.30).
P(2) = (0.70) ( 0.30) = 0.21.
• The probability the first purchase will be made by the third person
who enters the store is (0.70)(0.70)( 0.30).
P(3) = (0.70) (0.70)(0.30) = 0.147.
The probability the first purchase will be made by
person number x is P(x) = (.70)x - 1(.30)
Larson/Farber Ch. 4
Application
Example: A cereal maker places a game piece in its boxes. The
probability of winning a prize is 1 in 4. Find the probability you
a) Win your first prize on the 4th purchase
P(4) = (.75)3(.25) = 0.1055
b) Win your first prize on your 2nd or 3rd purchase
P(2) = (.75)1(.25) = 0.1875 and
P(3) = (.75)2(.25) = 0.1406
So P(2 or 3 ) = 0.1875 + 0.1406 = 0.3281
c) Do not win your first prize in your first 4 purchases.
1 – (P(1) + P(2) + P(3) + P(4))
1 – ( 0.25 + 0.1875 + 0.1406 + 0.1055)
= 1 – .6836 = 0.3164
Larson/Farber Ch. 4
Application
The Poisson Distribution
The Poisson distribution is a discrete probability distribution of
the random variable x that satisfies:
1. The experiment consists of counting the number of times, x, an event
occurs in an interval of time, area, or space.
Example: Sharks kill approximately 10 people
each year. Find the probability that:
a) Three people are killed by sharks this year
2. The probability that an event will occur is the same for each interval.
3. The number of occurrences in one interval is independent of the
number of occurrences in other intervals.
The probability of exactly x occurrences in an interval is:

where e is the irrational number approximately 2.71828
 is the mean number of occurrences per interval.
Larson/Farber Ch. 4
b) Two or three people are killed by sharks this
year
P(3) = 0.0076
P(2 or 3) = 0.0023 + 0.0076 = 0.0099
Larson/Farber Ch. 4
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