Download P.P Chapter 6.3

Binomial & Geometric
Random Variables
Section 6.3
Reference Text:
The Practice of Statistics, Fourth Edition.
Starnes, Yates, Moore
Objectives
1. Binomial Random Variables and Binomial
Distribution
1.
2.
3.
4.
5.
Requirements to be Binomial- B.I.N.S
Binomial Coefficient: by formula and TI-83
Binomial Probability: by formula and TI-83
Mean and Standard Deviation
“10% condition” sampling w/o replacement
2. Geometric Random Variables and Geometric
Distribution
1. Requirements to be Geometric- B.I.T.S
2. Geometric Probability: by formula and TI-83
3. Mean (expected value)
Consider this:
• Toss a coin 5 times, count the number of heads.
• Spin a roulette wheel 8 times. Record how many
times a ball lands in a red slot.
• Take a random sample of 100 babies born in U.S
hospitals today. Count the number of females.
In each case, we’re preforming repeated trials of the same
chance process. The number of trials is fixed in advance.
(number of trials n) In addition, the outcome of one trial has no
effect on the outcome of any other trial, that is, the trials are
independent. We’re interested in the number of times a specific
event occurs, (we’ll call it a “success”). Our chances of getting a
“success” are the same on each trial. (probability p is the same)
Definition:
B
• Binary? The possible outcomes of each trial can be classified as
“success” or “failure.”
I
• Independent? Trials must be independent; that is, knowing the result
of one trial must not have any effect on the result of any other trial.
N
• Number? The number of trials n of the chance process must be fixed
in advance.
S
• Success? On each trial, the probability p of success must be the
same.
Binomial and Geometric Random Variables
A binomial setting arises when we perform several independent trials of the
same chance process and record the number of times that a particular
outcome occurs. The four conditions for a binomial setting are
Consider the following 3 scenarios
In each case, determine whether the given random variable has a
binomial distribution.
Justify your answer. B.I.N.S
1. Genetics say that children receive genes from each of their parents
independently. Each child of a particular pair of parents have a probability
0.25 of having type O blood. Suppose these parents have 5 children. Let
𝑋 = 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐ℎ𝑖𝑙𝑑𝑟𝑒𝑛 𝑤𝑖𝑡ℎ 𝑡𝑦𝑝𝑒 𝑂.
Binary? “success” = has type O blood, “failure” = does not have type O
B
I
N
S
Independent? – children inherit genes are independent of each parent.
Number? There are 𝑛 = 5 number of trials
Success? The probability of a success is 𝑝 = 0.25
This is a binomial setting, 𝑛 = 5 𝑎𝑛𝑑 𝑝 = 0.25
Consider the following 3 scenarios
2. Shuffle a deck of cards. Turn over the first 10 cards, one at a time. Let Y=
the number of Aces you observe.
B
Binary?
“success” = is an ace, “failure” = is not an ace
Independent?
If the first card that gets turned over is an Ace then there is less of a
probability to get another ace, not independent
I
N
S
This is not a binomial setting since they are not independent.
Consider the following 3 scenarios
3. Shuffle a deck of cards. Turn over the top card. Put the card back in the
deck, and shuffle again. Repeat this process until you get an ace. Let W= the
number of cards required.
B Binary?
“success” = is an ace, “failure” = is not an ace
I Independent?
Since the card is being replaced, the probability is not affected.
N
Number?
There has not been a fix number of trials in advance. You could get an ace
the first time, or many times later.
S
This is not a binomial setting since there was not a fixed
number of trials.
Check Your Understanding
• Determine whether the given random variables has a
binomial distribution. Justify your answer.
1. Shuffle a deck of cards. Turn over the top card. Put the card
back in the deck, and shuffle again. Repeat this process 10
times. Let X = the number of aces you observe.
2. Roll a fair die 100 times. Sometime during the 100 rolls, one
corner of the die chips off. Let W = number of 5’s you roll.
The count X of successes in a binomial setting is a binomial random
variable. The probability distribution of X is a binomial distribution with
parameters n and p, where n is the number of trials of the chance process
and p is the probability of a success on any one trial. The possible values of
X are the whole numbers from 0 to n.
Note: When checking the Binomial condition, be sure to check the
BINS and make sure you’re being asked to count the number of
successes in a certain number of trials!
Binomial and Geometric Random Variables
Definition:
Binomial Coefficient
• Binomial Probabilities
Each child of a particular pair of parents has probability 0.25 of having
type O blood. Genetics says that children receive genes from each of
their parents independently. If these parents have 5 children, the count X
of children with type O blood is a binomial random variable with n = 5
trials and probability p = 0.25 of a success on each trial. In this setting, a
child with type O blood is a “success” (S) and a child with another blood
type is a “failure” (F).
What’s P(X = 2)?
P(SSFFF) = (0.25)(0.25)(0.75)(0.75)(0.75) = (0.25)2(0.75)3 = 0.02637
However, there are a number of different arrangements in which 2 out of
the 5 children have type O blood:
SSFFF
SFSFF
SFFSF
SFFFS
FSSFF
FSFSF
FSFFS
FFSSF
FFSFS
FFFSS
Verify that in each arrangement, P(X = 2) = (0.25)2(0.75)3 = 0.02637
Therefore, P(X = 2) = 10(0.25)2(0.75)3 = 0.2637
Binomial Coefficient: With TI-83
Binomial Probability
• If X has the binomial distribution with n trials
and probability p of success on each trial, the
possible values of X are 0,1,2…n. If k is any
one of these values,
• With our formula in hand, we can now
calculate any binomial probability!
• Lets take a look at those 5 kids and blood
type O!
•
Binomial Probability
Binomial Probability
the possible values of X are 0, 1, 2, …, n. If k is any one of these values,
æ nö k
P(X = k) = ç ÷ p (1- p) n-k
è kø
Number of
arrangements
of k successes
Probability of k
successes
Probability of
n-k failures
Binomial and Geometric Random Variables
The binomial coefficient counts the number of different
ways in which k successes can be arranged among
n trials. The binomial probability P(X = k) is this
count multiplied by the probability of any one
specific arrangement of the k successes.
Each child of a particular pair of parents has probability 0.25 of having type
O blood. Genetics says that children receive genes from each of their
parents independently. If these parents have 5 children, the count X of
children with type O blood is a binomial random variable with n = 5 trials and
probability p = 0.25 of a success on each trial. In this setting, a child with type
O blood is a “success” (S) and a child with another blood type is a “failure”
(F).
a.What’s P(X = 2)? The chance parents have 2 type O Blood kids
b.What’s P(X < 2)? The chance parents have at most 2 type O Blood kids
c.What’s P(X> 2)? The chance Parents have more than 2 type O Blood kids
æ nö
n!
ç ÷=
è k ø k!(n - k)!
æ nö k
P(X = k) = ç ÷ p (1- p) n-k
è kø
Binomial Probability On The Calculator
There are two handy commands on the TI-83/84 for finding binomial probabilities:
binompdf(n,p,k) computes P(X = k)
binomcdf(n,p,k) computes P(X ≤ k)
For the parents having n = 5 children, each with probability p = 0.25 of type
O blood:
a.What’s P(X = 2)? The chance parents have 2 type O Blood kids
b.What’s P(X < 2)? The chance parents have at most 2 type O Blood kids
c.What’s P(X>2)? The chance Parents have more than 2 type O Blood kids
d.Should a parent be surprised if they have more than three children with
type O Blood?
Binomial Probability with TI-83
For example we want to find probability of exactly 3
children with blood type O.
P(X = 3), n = 5, p =0.25, k=3
TI-83:
• 2nd > VARS >binompdf(> fill in the
information> binompdf(5,0.25,3)
TI-89:
In the CATALOG under Flash Apps
What if I wanted to find out P(X>3)???
What if I wanted to find out
P(X>3)???
Check Your Understanding
• To introduce her class to binomial distributions, Mrs. Desai
gives a 10-item, multiple choice quiz. The catch is, students
must simply guess an answer (A through E) for each question.
Mrs. Desai uses her computer’s random number generator to
produce the answer key, so that each possible answer as an
equal chance to be chosen. Patti is one of the students in this
class. Let 𝑋 = 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑃𝑎𝑡𝑡𝑖’𝑠 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 𝑔𝑢𝑒𝑠𝑠𝑒𝑠
1) Show that X is a binomial random variable.
2) Find 𝑃(𝑋 = 3), explain what this result means.
3) To get a passing score on the quiz, a student must guess
correctly at least 6 times. Would you be surprised if Patti
earned a passing score? Compute an appropriate probability
to support your answer.
Check Your Understanding
1) Show that X is a binomial random variable.
1) Binary?
Independent?
Number?
Success?
2) Find P(X=3), explain what this result means.
1) Binompdf(10, .2, 3) = .2013. there is a 20.13% chance that
Patti will answer exactly 3 questions correctly.
3) To get a passing score on the quiz, a student must
guess correctly at least 6 times. Would you be
surprised if Patti earned a passing score? Compute
an appropriate probability to support your answer.
1) 1 – binomcdf(10, .2, 5) = .0064, since there is only a .64%
chance that a student will pass, we would be quite
surprised if Patti was able to pass.
Mean and Standard Deviation of
Binomial Dist.
• Mean and Standard Deviation of a Binomial
Distribution
If a count X has the binomial distribution with number of trials n and
probability of success p, the mean and standard deviation of X are
m X = np
s X = np(1- p)
Note: These formulas work ONLY for binomial distributions.
They can’t be used for other distributions!
Binomial and Geometric Random Variables
Mean and Standard Deviation of a Binomial Random Variable
For the parents having
children, each with
probability
of type O blood:
Find the mean and standard deviation of X.
Since X is a binomial random variable with
parameters 𝑛 = 4 and 𝑝 = .25, we can use the formulas
for the mean and standard deviation of a binomial
random variable.
m X = np
s X = np(1- p)
We’d expect about one fourth of
there 4 children, about 1 to have
type O blood.
If parents who had 4 children were
repeatedly observed, the number of
children with type O blood would
differ from 1 by an average of .866
“10% condition” sampling w/o replacement
Geometric Random Variables
• In a binomial setting, the number of trials n
is fixed in advance, and the binomial
random variable X counts the number of
successes. The possible values of X are
0,1,2,….,n. In other situations, the goal is
to repeat a chance process until success
occurs.
Consider this:
• Roll a pair of dice until you get doubles.
• In basketball, attempt a three-point shot until you
make one.
• Keep placing $1 bet on the number 15 in roulette
until you win.
These are all examples of a geometric setting. Although the
number of trials isn’t fixed in advance, the trials are independent
and the probability of success remains constant.
Geometric Random Variables and
Geometric Distribution
• The number of trials Y that it takes to get a
success in geometric setting is a
geometric random variable. The probability
distribution of Y is a geometric distribution
with parameters p, the probability of a
success on any trial. The possible values
of Y are 1,2,3…
• Geometric Settings
Definition:
B
• Binary? The possible outcomes of each trial can be classified as
“success” or “failure.”
I
• Independent? Trials must be independent; that is, knowing the result
of one trial must not have any effect on the result of any other trial.
T
• Trials? The goal is to count the number of trials until the first success
occurs.
S
• Success? On each trial, the probability p of success must be the
same.
Binomial and Geometric Random Variables
A geometric setting arises when we perform independent trials of the same
chance process and record the number of trials until a particular outcome
occurs. The four conditions for a geometric setting are
Requirements to be Geometric- B.I.T.S
• A geometric setting arises when we preform independent
trials of the same chance process and record the number
of trials until a particular outcome occurs.
The four conditions for a geometric setting are:
• Binary? The possible outcomes of each trial can be classified as
“success” or “failure”
• Independent? Trials must be independent; that is, knowing the
result of one trial must not have any effect on the result of any other
trial
• Trials? The goal is the count the number of trials until the first
success occurs.
• Success? On each trial, the probability p of success must be the
same.
The Birth Day Game
• Your teacher is planning to give you 10 problems for homework. As
an alternative, you can agree to play the Birth Day Game. Here’s
how it works. A student will be selected at random from your class
and asked to guess the day of the week (for instance, Thursday) on
which one of your teacher's friends was born. If the student guesses
correctly, then the class will have only one homework problem. If the
student guesses the wrong day of the week, your teacher will once
again select a student from the class at random. The chosen student
will try to guess the day of the week on which a different one of your
teacher’s friends was born. If this student gets it right, the class will
have two homework problems. The game continues until a student
correctly guesses the day on which one of your teacher's many
friends was born. Your teacher will assign a number of homework
problems that is equal to the total number of guesses made by
members of your class. Are you ready to play the Birth Day Game?
• Geometric Settings
Birthday Game Geometric?
I
• Independent? student guess has no influence on another guess, since
it’s a different person’s birthday.
T
• Trials? We are counting the number of trials up and including the first
correct guess.
S
• Success? The probability of a success 𝑖𝑠 𝑝 = 7
1
Binomial and Geometric Random Variables
B
• Binary? “success” = correct guess, “failure” = incorrect guess
Geometric Probability
The Birth Day Game
Geometric Probability with TI-83
Geometric Probability with TI-83
For example we want to find probability that the class
gets exactly 10 homework problems
P(Y = 10), p =1/7, k=10
TI-83:
• 2nd > VARS >geometpdf(> fill in the
information> geometpdf(1/7, 10)
TI-89:
In the CATALOG under Flash Apps
What if I wanted to find out P(Y<10)???
Mean of Geometric Dist.
Objectives
1. Binomial Random Variables and Binomial
Distribution
1.
2.
3.
4.
5.
Requirements to be Binomial- B.I.N.S
Binomial Coefficient: by formula and TI-83
Binomial Probability: by formula and TI-83
Mean and Standard Deviation
“10% condition” sampling w/o replacement
2. Geometric Random Variables and Geometric
Distribution
1. Requirements to be Geometric- B.I.T.S
2. Geometric Probability: by formula and TI-83
3. Mean (expected value)
Homework
6.3 homework worksheet
Finish Chapter 6 Reading Guide