Ch. 14 Notes

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Ch. 14 Notes

• MATH 2400
• Mr. J. Dustin Tench

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Recap Ch. 13

• P(A or B)
• P(A) P(B) P(AnB)
• If A and B are disjoint, P(AnB) 0, so we get
• P(A) P(B).

• P(A and B)
• P(AB) P(B)
• If A and B are independent, P(AB) P(A), so we
  get P(A) P(B)

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When are A and B disjoint?

• When the sample spaces for A and B contain no
  common elements, they are considered disjoint.
• Ex Rolling a 2 or a 4 on a die.
• Ex Drawing a black ace or a red ace out of a
  standard deck of cards.
• Ex Selecting a senior citizen or a teenager at
  random for a prize from a collection of surveys.

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Venn Diagram of Disjoint Events
P(AnB)0
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When are A and B Non-Disjoint?

• When A and B contain common elements in their
  sample space.
• Ex Drawing a red card or a king from a standard
  deck of cards.
• Ex Rolling a 4 or an even number.
• Ex Buying a used car or buying a Toyota.

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Venn Diagram of Disjoint Events
P(AnB) ? 0
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When are A and B Independent?

• Events A and B are independent when the outcome
  of A cannot affect the outcome of B.
• Ex Rolling a 6 and flipping Heads.
• Ex Having twins and the first being a boy and
  the second being a girl.

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Conditional Probability

• Conditional probability exists when we are only
  considering a subset of the original sample
  space. Generally, the word given is used, but
  not always.
• Ex Of the freshmen, whats the probability they
  are a science major?
• Ex Given a student has a job, whats the
  probability they have above a 3.5 GPA?

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Example

• The Clemson University Fact Book for 2007 shows
  that 123 of the universitys 338 assistant
  professors were women, along with 76 of the 263
  associate professors and 73 of the 375 full
  professors.
• What is the probability that a randomly chosen
  Clemson professor is a woman?
• Given a selected individual is a full professor,
  what is the probability that the person is a
  woman?
• Are the rank and gender of Clemson professors
  independent?

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Example

• A recent study has shown that 75 of teenagers
  own cell phones. The study also indicates that
  65 of teenagers own a phone and text on their
  phones.
• What is the probability that a randomly selected
  teen is a texter if we know they own a phone.
• Of the teens that own phones, 15 send more than
  6,000 texts a month. What percent of all teens
  own a phone, are texters, and send more than
  6,000 texts a month?

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Ch. 14 Notes FINALLY!!!

• Binomial Distributions are simply situations in
  which only two different outcomes are possible
  for each trial.
• Ex Shooting free throws in a basketball game.
• Ex Having a baby.
• Ex Computerized telephone survey successfully
  calling a residence.

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Combinatorics
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Permutations Combinations

• A permutation is an arrangement of objects in
  which the order they are arranged can be counted
  differently than a different arrangement of the
  same objects
• ABC is a different arrangement than ACB.
• A combination is a grouping of objects in which
  order does not matter and the group is counted as
  the same group if it contains that same elements
• ABC is the same group as ACB.

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Permutations

• Consider assigning three people 3 different roles
  in an organization.
• A President
• B Vice-President
• C Secretary
• We can arrange these same 3 people 6 different
  ways.
• ABC, ACB, BAC, BCA, CAB, CBA

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Permutations

• Consider choosing the president, we have 3
  choices.
• Then, consider choosing the vice-resident, we
  have 2 choices.
• Finally, consider choosing the secretary, we have
  1 choice. So, the number of arrangements of 3
  people is 321 3! 6.
• Factorial n! n(n-1)(n-2)(n-3)321

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Permutations

• What if we dont use all the objects in a group.
  Consider the case where a President and
  Vice-President are to be chosen from a group of 4
  people. How many different ways can we assign
  these two positions?
• For the president, we have 4 choices. For the
  vice-president, we have 3 choices. So, the
  number of ways we can assign these positions is
  43 12.

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Permutations A Formula
There is a formula, but it is much easier just to
think about the problem and count the number of
ways each part can happen.
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Combinations

• Consider forming a team where each person has the
  same role. If 3 people are to be used from a
  pool of3people..then there is only 1 group
  that can be formed.
• But, what if the pool was larger?
• Consider a terrible situation in which there are
  4 people and a team of 3 is to be formed. How
  many ways can this happen?
• ABC ACD
• ABD BCD There are 4 ways.

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Combinations

• Combinations, unlike Permutations, have an
  accepted notation used.
• For the situation we just looked at, n4 and r3,
  so 4.

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On The Ti-84

• Permutations MATH, PRB tab, 2 nPr
• Combinations MATH, PRB tab, 3 nCr
• Type in what you want to calculate like
• 4 3 and hit ENTER.

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Okay, Back to Binomial Distributions

• We use combinations to count the number of ways
  an event can occur.
• For example, if Jordan makes 70 of his free
  throws, what is the probability he makes 4 of the
  6 free throws he attempts in a game?
• The number of ways he can make 4 of the 6 is
• 15.

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Jordan

• There are 15 different ways Jordan can make 4 of
  his 6 free throws. The probability Jordan hits
  any one of the free throws is .7. The
  probability Jordan misses is .3.
• Lets consider the case where Jordan makes the
  first four free throws, but misses the next two.
• The probability of this happening is
  (.7)(.7)(.7)(.7)(.3)(.3) .021609
• Consider the case where Jordan makes the first
  three, misses two, but makes the last.
• (.7)(.7)(.7)(.3)(.3)(.7) .021609
• They are exactly the same!

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Jordan continued

• So, there are 15 ways of making 4 out of 6 free
  throws, so the probability that Jordan makes
  exactly 4 out of the 6 free throws is.
• 15(.7)(.7)(.7)(.7)(.3)(.3)
• 15 (.7)4(.3)2 .324135

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More Jordan

• 1. Calculate the probability that Jordan makes
  at least 2 of the 6 free throws.
• P(X2)
• 2. Calculate the probability that Jordan makes
  at most 4 of the 6 free throws.
• P(X4)
• 3. Calculate the probability that Jordan makes
  2, 3, or 4 of the 6 free throws.
• P(2X4)

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On the Calculator

• 2nd, DISTR
• 0binompdf(n,p,r), for P(x r)
• Abinomcdf(n,p,r), for P(x lt,,gt, or r)
• So, for Jordan
• P(X4) binompdf(6,0.7,4)
• P(X2) 1 - binomcdf(6,0.7,4)
• P(X4) binomcdf(6,0.7,4)
• P(2X4) binomcdf(6,0.7,4) binomcdf (6,0.7,2)

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µ and s for Binomial Distributions

•  

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An Application

• Delta has been collecting data regarding its
  flights. Data indicates that 82 of the flights
  from Atlanta to Las Vegas depart on time.
• If there are 227 flights from Atlanta to Las
  Vegas this year, what is the expected number
  (mean) of flights that should depart on time?
• What is the standard deviation using this data?

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HW

• 13.1, 13.2, 13.3, 13.4, 13.5, 13.24, 13.26