Transcript ST_PP_17_ProbabilityModelsx

Probability Models
Statistics 17
• The basis for the probability models we will examine in this
chapter is the Bernoulli trial.
• We have Bernoulli trials if:
– there are two possible outcomes (success and failure).
– the probability of success, p, is constant.
– the trials are independent.
Bernoulli Trials
Bernoulli Trials
Binomial Variables
Binomial Variables
Binomial Variables
Binomial Variables
Binomial Variables
Binomial Variables
• A Binomial model tells us the probability for a
random variable that counts the number of
successes in a fixed number of Bernoulli
trials.
• Two parameters define the Binomial model:
n, the number of trials; and, p, the probability
of success. We denote this Binom(n, p).
The Binomial Model
• In n trials, there are
n!
n Ck 
k ! n  k !
ways to have k successes.
– Read nCk as “n choose k.”
• Note: n! = n  (n – 1)  …  2  1, and n! is
read as “n factorial.”
The Binomial Model
Binomial probability model for Bernoulli trials: Binom(n,p)
n = number of trials
p = probability of success
q = 1 – p = probability of failure
X = # of successes in n trials
x
n–x
P(X = x) = nCx p q
  np
  npq
The Binomial Model
Example
Binomial Coefficient
Binomial Coefficient
Binomial Coefficient
Binomial Probability
Binomial Distribution
Example
Example
Example
Example
Example
Example
Example
Example
Example
Example
Calculators
• A single Bernoulli trial is usually not all that interesting.
• A Geometric probability model tells us the probability for a
random variable that counts the number of Bernoulli trials
until the first success.
• Geometric models are completely specified by one
parameter, p, the probability of success, and are denoted
Geom(p).
The Geometric Model
Geometric probability model for Bernoulli trials: Geom(p)
p = probability of success
q = 1 – p = probability of failure
X = number of trials until the first success occurs
x-1
P(X = x) = q p
1
E(X )   
p
 
q
2
p
The Geometric Model
• One of the important requirements for Bernoulli trials is that
the trials be independent.
• When we don’t have an infinite population, the trials are not
independent. But, there is a rule that allows us to pretend
we have independent trials:
– The 10% condition: Bernoulli trials must be independent.
If that assumption is violated, it is still okay to proceed as
long as the sample is smaller than 10% of the population.
Independence
• A new sales gimmick has 30% of the M&M’s
covered with speckles. These “groovy”
candies are mixed randomly with the normal
candies as they are put into the bags for
distribution and sale. You buy a bag and
remove candies one at a time looking for the
speckles.
Practice
• A new sales gimmick has 30% of the M&M’s covered with
speckles. These “groovy” candies are mixed randomly with
the normal candies as they are put into the bags for
distribution and sale. You buy a bag and remove candies
one at a time looking for the speckles.
• Does this situation involve Bernoulli trials?
Practice
• Does this situation involve Bernoulli trials?
• There are two outcomes: success = speckles, failure
= ordinary. The probability of success, based on the
information from the candy company, is 30%. Trials
can be assumed independent – there is no reason to
believe that finding a speckled candy reveals
anything about whether the next one out of the bag
will have speckles.
•
This situation involves Bernoulli trials
Practice
• What’s the probability that the first speckled one we
see is the fourth candy we get?
• P(First speckled is fourth candy) = (0.7)3 (0.3) = 0.1029
Practice
• What’s the probability that the first speckled one is
the tenth one?
• P(First speckled is tenth candy) = (0.7)10 (0.3) ≈ 0.0085
• General Formula: P(X = x) = qx−1 p
Practice
• What’s the probability that the first speckled candy is one of
the first three we look at?
•
P(First speckled is among first three) = (0.3) + (0.7)(0.3) + (0.7)2 (0.3) ≈ 0.657
Practice
•
How many do we expect to check, on average to find a speckled one?
Practice
•
What’s the probability that we’ll find two speckled ones in a handful of five
candies?
Students need to show the setup on the AP Exam, even though
they will use technology to get
the answer.
Practice
Mean and Standard Deviation
Example
Example
Example
Example
Example
Example
Geometric Probability Model
Geometric Probability Model
Number of successes in a
fixed number of trials
Number of trials required
for one success
binomPdf( )
binomCdf( )
geometPdf( )
geometCdf( )
Binomial or Geometric
Binomial and Geometric
Example
Example
Example
Example
• When dealing with a large number of trials in
a Binomial situation, making direct
calculations of the probabilities becomes
tedious (or outright impossible).
• Fortunately, the Normal model comes to the
rescue…
The Normal Model to the Rescue!
• As long as the Success/Failure Condition
holds, we can use the Normal model to
approximate Binomial probabilities.
– Success/failure condition: A Binomial model is
approximately Normal if we expect at least 10
successes and 10 failures:
np ≥ 10 and nq ≥ 10
The Normal Model to the Rescue
• When we use the Normal model to
approximate the Binomial model, we are
using a continuous random variable to
approximate a discrete random variable.
• So, when we use the Normal model, we no
longer calculate the probability that the
random variable equals a particular value,
but only that it lies between two values.
Continuous Random Variables
• Be sure you have Bernoulli trials.
– You need two outcomes per trial, a constant
probability of success, and independence.
– Remember that the 10% Condition provides a
reasonable substitute for independence.
• Don’t confuse Geometric and Binomial models.
• Don’t use the Normal approximation with small
n.
– You need at least 10 successes and 10 failures to
use the Normal approximation.
What Can Go Wrong?
• Bernoulli trials show up in lots of places.
• Depending on the random variable of
interest, we might be dealing with a
– Geometric model
– Binomial model
– Normal model
What have we learned?
– Geometric model
• When we’re interested in the number of Bernoulli trials
until the next success.
– Binomial model
• When we’re interested in the number of successes in
a certain number of Bernoulli trials.
– Normal model
• To approximate a Binomial model when we expect at
least 10 successes and 10 failures.
What have we learned?
• Pages 398 - 401;
• 2, 3, 5, 7, 9, 13, 15, 17, 20, 23, 25, 32, 33, 35
Homework