Section 6.3 Binomial and Geometric Random Variables

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Section 6.3Binomial and Geometric Random
Variables

• Learning Objectives

• After this section, you should be able to
• DETERMINE whether the conditions for a binomial
  setting are met
• COMPUTE and INTERPRET probabilities involving
  binomial random variables
• CALCULATE the mean and standard deviation of a
  binomial random variable and INTERPRET these
  values in context
• CALCULATE probabilities involving geometric
  random variables

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• Binomial Settings
• When the same chance process is repeated several
  times, we are often interested in whether a
  particular outcome does or doesnt happen on each
  repetition. In some cases, the number of repeated
  trials is fixed in advance and we are interested
  in the number of times a particular event (called
  a success) occurs. If the trials in these
  cases are independent and each success has an
  equal chance of occurring, we have a binomial
  setting.

Definition 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
Binary? The possible outcomes of each trial can
be classified as success or failure.
B
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.
I
Number? The number of trials n of the chance
process must be fixed in advance.
N
Success? On each trial, the probability p of
success must be the same.
S
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• Binomial Random Variable
• Consider tossing a coin n times. Each toss gives
  either heads or tails. Knowing the outcome of one
  toss does not change the probability of an
  outcome on any other toss. If we define heads as
  a success, then p is the probability of a head
  and is 0.5 on any toss.
• The number of heads in n tosses is a binomial
  random variable X. The probability distribution
  of X is called a binomial distribution.

Definition 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 youre being
asked to count the number of successes in a
certain number of trials!
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• Binomial Probabilities
• In a binomial setting, we can define a random
  variable (say, X) as the number of successes in n
  independent trials. We are interested in finding
  the probability distribution of X.

• Binomial and Geometric Random Variables

Example
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). Whats
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
SFSFF
SFFSF
SFFFS
FSSFF
SSFFF
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
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• Binomial Coefficient
• Note, in the previous example, any one
  arrangement of 2 Ss and 3 Fs had the same
  probability. This is true because no matter what
  arrangement, wed multiply together 0.25 twice
  and 0.75 three times.
• We can generalize this for any setting in which
  we are interested in k successes in n trials.
  That is,

• Binomial and Geometric Random Variables

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• Binomial Probability
• 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.

• Binomial and Geometric Random Variables

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,
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• Example Inheriting Blood Type
• Each child of a particular pair of parents has
  probability 0.25 of having blood type O. Suppose
  the parents have 5 children
• (a) Find the probability that exactly 3 of the
  children have type O blood.

Let X the number of children with type O blood.
We know X has a binomial distribution with n 5
and p 0.25.
(b) Should the parents be surprised if more than
3 of their children have type O blood?
To answer this, we need to find P(X gt 3).
Since there is only a 1.5 chance that more than
3 children out of 5 would have Type O blood, the
parents should be surprised!
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• Mean and Standard Deviation of a Binomial
  Distribution
• We describe the probability distribution of a
  binomial random variable just like any other
  distribution by looking at the shape, center,
  and spread. Consider the probability distribution
  of X number of children with type O blood in a
  family with 5 children.

• Binomial and Geometric Random Variables

xi 0 1 2 3 4 5
pi 0.2373 0.3955 0.2637 0.0879 0.0147 0.00098
Shape The probability distribution of X is
skewed to the right. It is more likely to have 0,
1, or 2 children with type O blood than a larger
value.
Center The median number of children with type O
blood is 1. Based on our formula for the mean
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• Mean and Standard Deviation of a Binomial
  Distribution
• Notice, the mean µX 1.25 can be found another
  way. Since each child has a 0.25 chance of
  inheriting type O blood, wed expect one-fourth
  of the 5 children to have this blood type. That
  is, µX 5(0.25) 1.25. This method can be used
  to find the mean of any binomial random variable
  with parameters n and p.

• Binomial and Geometric Random Variables

Note These formulas work ONLY for binomial
distributions. They cant be used for other
distributions!
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• Example Bottled Water versus Tap Water
• Mr. Bullards 21 AP Statistics students did the
  Activity on page 340. If we assume the students
  in his class cannot tell tap water from bottled
  water, then each has a 1/3 chance of correctly
  identifying the different type of water by
  guessing. Let X the number of students who
  correctly identify the cup containing the
  different type of water.

Find the mean and standard deviation of X.
Since X is a binomial random variable with
parameters n 21 and p 1/3, we can use the
formulas for the mean and standard deviation of a
binomial random variable.
If the activity were repeated many times with
groups of 21 students who were just guessing, the
number of correct identifications would differ
from 7 by an average of 2.16.
Wed expect about one-third of his 21 students,
about 7, to guess correctly.
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• Binomial Distributions in Statistical Sampling
• The binomial distributions are important in
  statistics when we want to make inferences about
  the proportion p of successes in a population.
• Suppose 10 of CDs have defective copy-protection
  schemes that can harm computers. A music
  distributor inspects an SRS of 10 CDs from a
  shipment of 10,000. Let X number of defective
  CDs. What is P(X 0)? Note, this is not quite a
  binomial setting. Why?

• Binomial and Geometric Random Variables

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• Normal Approximation for Binomial Distributions
• As n gets larger, something interesting happens
  to the shape of a binomial distribution. The
  figures below show histograms of binomial
  distributions for different values of n and p.
  What do you notice as n gets larger?

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• Example Attitudes Toward Shopping
• Sample surveys show that fewer people enjoy
  shopping than in the past. A survey asked a
  nationwide random sample of 2500 adults if they
  agreed or disagreed that I like buying new
  clothes, but shopping is often frustrating and
  time-consuming. Suppose that exactly 60 of all
  adult US residents would say Agree if asked the
  same question. Let X the number in the sample
  who agree. Estimate the probability that 1520 or
  more of the sample agree.

1) Verify that X is approximately a binomial
random variable.
B Success agree, Failure dont agree I
Because the population of U.S. adults is greater
than 25,000, it is reasonable to assume the
sampling without replacement condition is met. N
n 2500 trials of the chance process S The
probability of selecting an adult who agrees is p
0.60
2) Check the conditions for using a Normal
approximation.
Since np 2500(0.60) 1500 and n(1 p)
2500(0.40) 1000 are both at least 10, we may
use the Normal approximation.
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The number X of switches that fail inspection has
approximately the binomial distribution with n
10 and p 0.1.

• What is the probability that no more than one
  switch fails?
• P(X 1) P(X 0) P(X 1)
• (0.1)0(0.9)10 (0.1)1(0.9)9
• 0.7361

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The mailing list of an agency that markets
scuba-diving trips to the Florida Keys contains
70 males and 30 females. The agency calls 30
people chosen at random.

• ?What is the probability that 20 of the 30 are
  men?
• ?What is the probability that the first woman is
  reached on the fourth call? (That is, the first
  4 calls give MMMF.)

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• Let X the number of men called. X is B(30,
  0.7)
• What is the probability that 20 of the 30 are
  men?
• P(X 20) (0.7)20(0.3)10
• What is the probability that the first woman is
  reached on the fourth call? (That is, the first
  4 calls give MMMF.)
• P(MMMF) (0.7)3(0.3)1 0.1029

0.1416
a)
b)
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Binomial probabilities on the calculator

• P(X k) binompdf (n, p, k)
• pdf ? probability distribution function ?
• Assigns a probability to each value of a discrete
  random variable, X.
• P(X lt k) binomcdf (n, p, k)
• cdf ? cumulative distribution function ?
• for R.V. X, the cdf calculates the sum of the
  probabilities for 0, 1, 2 up to k.

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• binompdf( n,p,k)
• Allows to find the probability of an individual
  outcome.
• - specify n and p ( calculator 2nd
  Dist)
• Ex Suppose a player's picture is in 20 of the
  cereal boxes. You want to know the probability of
  finding that player is exactly twice among 5
  boxes. Then,
• n 5, p 0.2, k 2
• binompdf( 5, 0.2, 2) 0.2048
• ( means 20 chance)

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• binomcdf ( n, p, k)
• Allows to find the total probability of getting k
  or fewer successes among the n trials
• Ex Suppose a player's picture is in 20 of the
  cereal boxes. We have 10 boxes of cereal and now
  we wonder about the probability of finding ---
• up to 4 pictures of the Player.
• So, thats the probability of 0, 1, 2, 3 or 4
  successes. ( at most 4 pictures)
• binomcdf( 10, 0.2, 4) 0.9672065025

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• What is the probability of getting at least 4
  pictures of the player in 10 boxes?
• at least 4 means not 3 or fewer. That is
  the complement of 0,1,2 or 3 successes
• 1-binomcdf( 10, 0.2, 3) 0.1208738816

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Using the calculator for calculating binomial
probabilities

• An engineer chooses an SRS of 10 switches from a
  shipment of 10,000 switches. Suppose that
  (unknown to the engineer) 10 of the switches in
  the shipment are bad. What is the probability
  that no more than 1 of the 10 switches in the
  sample fail inspection?

The calculator can calculate the probabilities
of each value X using the command binompdf(n,
p, X)
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Using the calculator

• Whats the probability that no more than 1 fails
  ( or at most 1 fails) ?
• P(X 0) P(X 1)
• binompdf(10, 0.1, 0) binompdf(10, 0.1, 1)
• 0.3487 0.3874
• 0.7361
• The suffix pdf stands for probability
  distribution function.

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• Corrine makes 75 of her free throws over the
  course of a season. In a key game, she shoots 12
  free throws and only makes 7 of them. The fans
  think she failed because she is nervous. Is it
  unusual for Corinne to perform this poorly?
  (Assume that each shot is independent of one
  another)
• What is the probability of making a basket on at
  most 7 free throws?
• On calculator binomcdf(n, p, x)
• B(12, 0.75)
• P(X 7) binomcdf(12, 0.75, 7) 0.1576436

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Cumulative Binomial Probability

• In applications we frequently want to find the
  probability that a random variable takes a range
  of values. The cumulative binomial probability
  is useful in these cases.
• On calculator binomcdf(n, p, x)
• B(12, 0.75)
• P(X 7) binomcdf(12, 0.75, 7)

0.1576
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Cumulative Binomial Probability
It is very important that you understand the
difference between binomcdf and binompdf
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Among employed women, 25 have never been
married. Select 10 employed women at random.

1. What is the probability that exactly 2 of the 10
   women in your sample have never been married?
2. What is the probability that 2 or fewer have
   never been married?

binompdf(10, 0.25, 2) 0.2816
binomcdf(10, 0.25, 2) 0.5256
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Gretchen is a 60 free throw shooter. In a
season, she shoots an average of 75 free throws.
What is the probability that Gretchen

• makes exactly 50 out of 75 free throws?
• makes more than 50 free throws?
• makes no more than 40 free throws?

P(X 50) binompdf(75, 0.6, 50) 0.0479
P(X gt 50) 1 - binomcdf(75, 0.6, 50) 0.0963
P(X 40) binomcdf(75, 0.6, 40) 0.1446
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Caution..

• On the AP Exam you are not allowed to just write
  calculator talk.
• You must write out the formula for each problem,
  but feel free to use your calculator for the
  calculations
• For cumulative, use to show the first two
  and last part of the formula.

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• Geometric Distribution

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Getting Started

• Suppose Sam is bashful and has trouble getting
  girls to say Yes when he asks them for a date.
  Sadly, in fact, only 10 of the girls he asks
  actually agree to go out with him. Suppose that
  p 0.10 is the probability that any randomly
  selected girl will agree to go out with Sam when
  asked.
• What is the probability that the first four girls
  he asks say No and the fifth says Yes?
• How many girls can he expect to ask before the
  first one says Yes?

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The Geometric Distribution

• There are situations in which the goal is to
  obtain a fixed number of successes.
• If the goal is to obtain one success, a random
  variable X can be defined that counts the number
  of trials needed to obtain that first success.
• Geometric distribution a distribution of a
  random variable that counts the number of trials
  needed to obtain a success.

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• Geometric Settings
• In a binomial setting, the number of trials n is
  fixed and the binomial random variable X counts
  the number of successes. In other situations, the
  goal is to repeat a chance behavior until a
  success occurs. These situations are called
  geometric settings.

Definition 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
Binary? The possible outcomes of each trial can
be classified as success or failure.
B
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.
I
Trials? The goal is to count the number of
trials until the first success occurs.
T
Success? On each trial, the probability p of
success must be the same.
S
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All four of these conditions must be met in order
to use a geometric distribution.

1. Each observation falls into one of two
   categories success or failure
2. The n observations are independent
3. The probability of success, call it p, is the
   same for each observation.
4. The variable of interest is the number of trials
   required to obtain the first success.

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The Geometric Distribution

• Let X number of trials required for the first
  success
• X is a geometric random variable
• How is this distribution different than a
  binomial distribution?

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• Be aware of the key differences between binomial
  and geometric distributions.
• Binomial
• Finds the probability that k success will
  occur for in n number of attempts.
• Geometric  
• Finds the probability that a success will
  occur for the first time on the nth try.
• equation  P(xn) ( 1 - p )n-1 p

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Is this a Geometric Setting?

• A game consists of rolling a single die. The
  event of interest is rolling a 3 this event is
  called a success. The random variable is defined
  as X the number of trials until a 3 occurs.
• ? Yes.
• ?Success 3, Failure any other number.
• ? p 1/6 and is the same for all trials.
• ?The observations are independent.
• ?We roll until a 3 is obtained.

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Geometric Setting?

• Suppose you repeatedly draw cards without
  replacement from a deck of 52 cards until you
  draw an ace. There are two categories of
  interest ace success not ace failure.
• No. It fails the independence requirement. If
  the first card is not an ace then the 2nd card
  will have a greater chance of being an ace.

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Using example
1, lets calculate some probabilitiesA game
consists of rolling a single die. The event of
interest is rolling a 3 this event is called a
success. The random variable is defined as X
the number of trials until a 3 occurs.
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Example 1 continued
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• Example The Birthday Game
• Read the activity on page 398. The random
  variable of interest in this game is Y the
  number of guesses it takes to correctly identify
  the birth day of one of your teachers friends.
  What is the probability the first student guesses
  correctly? The second? Third? What is the
  probability the kth student guesses corrrectly?

Verify that Y is a geometric random variable.
B Success correct guess, Failure incorrect
guess I The result of one students guess has no
effect on the result of any other guess. T Were
counting the number of guesses up to and
including the first correct guess. S On each
trial, the probability of a correct guess is 1/7.
Calculate P(Y 1), P(Y 2), P(Y 3), and P(Y
k)
Notice the pattern?
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Understanding Probability

• In rolling a die, it is possible that you will
  have to roll the die many times before you roll a
  3.
• In fact, it is theoretically possible to roll the
  die forever without rolling a 3 (although the
  probability gets closer and closer to zero the
  longer you roll the die without getting a 3)
• P(X 50) (5/6)49(1/6) 0.0000

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Probability distribution table for the geometric
random variable never ends.

• This should remind you of an infinite geometric
  sequence from Algebra II
• In order for it to be a true pdf, the sum of the
  second row must add up to 1.

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Verifying it is a valid pdf

• Recall the infinite sum formula
• In this case, a1 p and r 1 p

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Calculating Geometric Probabilities

• P(X k) (1 p)k 1p
• P(X k) ? geometpdf (p, k)
• Probability that the first success occurs on the
  kth trial
• P(X lt k) ? geometcdf (p, k)

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The Expected Value

• The mean or expected value of the geometric
  random variable is the expected number of trials
  needed for the first success.
• µ 1/p

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• The formula for Standard Deviation of a Geometric
  series in not on the AP Statistics course
  syllabus. In case you are wondering, the formulas
  is
• The variance of X is s2 (1 - p)/p2

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The State Fair

• Glenn likes the game at the state fair where you
  toss a coin into a saucer. You win if the coin
  comes to a rest in the saucer without sliding
  off. Glenn has played the game many times and
  has determined that on average he wins 1 out of
  every 12 times he plays. He believes that his
  chances of winning are the same for each toss.
  He has no reason to think that his tosses are not
  independent. Let X be the number of tosses until
  a win. Find his expected value.

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The State Fair

• This is a geometric distribution
• p 1/12
• µ 1/(1/12) 12

This means that he can expect to play 12 games
before he wins
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Sometimes we are interested in the probability
that is takes more than a certain number of
trials to achieve success.

• The probability that it takes more than n trials
  to reach the first success is
• P(X gt n) (1 - p)n

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Roll a die until a 3 occurs.

• What is the probability that it takes more than 6
  rolls to observe a 3?

P(X gt 6) (1 1/6)6 0.3349
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Game of chance

• Three friends each toss a coin. The odd man
  wins that is, if one coin comes up different
  from the other two, that person wins the round.
  If the coins all match, then no one wins and they
  toss again. Were interested in the number of
  times the players will have to toss the coins
  until someone wins.

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Game of chance.

• What is the probability that no one will win on a
  given coin?
• Out of 8 outcomes, HHH and TTTdo not
• produce winners. So,
• Define a success as someone wins a given coin
  toss. What is the probability of success?

P(no winner) 2/8 or 0.25
P(winner) 1 P(no winner) 1 0.25 0.75
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Game of chance.

• Define the random variable of interest
• X number of tosses until someone wins
• Is X binomial or geometric?
• What is the probability that is takes no more
  than 2 tosses for someone to win?

geometric
P(X 2) 1 P(X gt 2) 1 (0.25)2 0.9375
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Game of chance.

• What is the probability that is takes more than 4
  tosses for someone to win?
• What is the expected number of tosses needed for
  someone to win?

P(X gt 4) (0.25)4 0.0039
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Section 6.3Binomial and Geometric Random
Variables

• Summary

• In this section, we learned that
• A binomial setting consists of n independent
  trials of the same chance process, each resulting
  in a success or a failure, with probability of
  success p on each trial. The count X of successes
  is a binomial random variable. Its probability
  distribution is a binomial distribution.
• The binomial coefficient counts the number of
  ways k successes can be arranged among n trials.
• If X has the binomial distribution with
  parameters n and p, the possible values of X are
  the whole numbers 0, 1, 2, . . . , n. The
  binomial probability of observing k successes in
  n trials is

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Section 6.3Binomial and Geometric Random
Variables

• Summary

• In this section, we learned that
• The mean and standard deviation of a binomial
  random variable X are
• The Normal approximation to the binomial
  distribution says that if X is a count having the
  binomial distribution with parameters n and p,
  then when n is large, X is approximately Normally
  distributed. We will use this approximation when
  np 10 and n(1 - p) 10.

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Section 6.3Binomial and Geometric Random
Variables

• Summary

• In this section, we learned that
• A geometric setting consists of repeated trials
  of the same chance process in which each trial
  results in a success or a failure trials are
  independent each trial has the same probability
  p of success and the goal is to count the number
  of trials until the first success occurs. If Y
  the number of trials required to obtain the first
  success, then Y is a geometric random variable.
  Its probability distribution is called a
  geometric distribution.
• If Y has the geometric distribution with
  probability of success p, the possible values of
  Y are the positive integers 1, 2, 3, . . . . The
  geometric probability that Y takes any value is
• The mean (expected value) of a geometric random
  variable Y is 1/p.

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Looking Ahead