Probability Exercises

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Probability Exercises
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Experiment

• Something capable of replication under stable
  conditions.
• Example Tossing a coin

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Sample Space

• The set of all possible outcomes of an
  experiment.
• A sample space can be finite or infinite,
  discrete or continuous.

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A set is discrete ifyou can put you finger on
one element after another not miss any in
between.

• Thats not possible if a set is continuous.

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Example discrete, finite sample space

• Experiment Tossing a coin once.
• Sample space H, T.
• This sample space is discrete you can put your
  finger on one element after the other not miss
  any.
• This sample space is also finite. There are just
  two elements thats a finite number.

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Example discrete, infinite sample space

• Experiment Eating potato chips
• Sample space 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
  11, ...
• This sample space is discrete you can put your
  finger on one element after another not miss
  any in between.
• This sample space is infinite, however, since
  there are an infinite number of possibilities.

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Example continuous sample space.

• Experiment Burning a light bulb until it burns
  out. Suppose there is a theoretical maximum
  number of hours that a bulb can burn that is
  10,000 hours.
• Sample space The set of all real numbers
  between 0 10,000.
• Between any two numbers you can pick in the
  sample space, there is another number.
• For example, the bulb could burn for 99.777 hours
  or 99.778. But it could also burn for 99.7775,
  which is in between. You cannot put your finger
  on one element after another not miss any in
  between.
• This sample space is continuous.
• It is also infinite, since there are an infinite
  number of possibilities.
• All continuous sample spaces are infinite.

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Eventa subset of the outcomes of an experiment

• Example
• Experiment tossing a coin twice
• Sample space (all possible outcomes)
  HH, HT, TH, TT
• An event could be that you got at least one head
  on the two tosses.
• So the event would be HH, HT, TH

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Subjective versus objective probability
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Subjective vs. Objective Probability

• Subjective Probability is probability in lay
  terms. Something is probable if it is likely.
• Example I will probably get an A in this
  course.
• Objective Probability is what well use in this
  course.
• Objective Probability is the relative frequency
  with which something occurs over the long run.
• What does that mean?

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Objective Probability Developing the Idea

• Suppose we flip a coin get tails. Then the
  relative frequency of heads is 0/1 0.
• Suppose we flip it again get tails again.Our
  relative frequency of heads is 0/2 0.
• We flip it 8 more times get a total of 6 tails
  4 heads.The relative freq of heads is 4/10
  0.4
• We flip it 100 times get 48 heads.The relative
  freq of heads is 48/100 0.48
• We flip it 1000 times get 503 heads.The
  relative freq of heads is 503/1000 0.503
• If the coin is fair, we could flip it an
  infinite number of times, what would the relative
  frequency of heads be?
• 0.5 or 1/2
• Thats the relative freq over the long run or
  probability of heads.

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Two Basic Properties of Probability

• 1. 0 lt Pr(E) lt 1 for every subset of the sample
  space S
• 2. Pr(S) 1

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Counting Rules

• Well look at three counting rules.
• 1. Basic multiplication rule
• 2. Permutations
• 3. Combinations

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Multiplication Rule Example
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Multiplication RuleExample

• Suppose we toss a coin 3 times examine the
  outcomes.
• (One possible outcome would be HTH.)
• How many outcomes are possible?

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We have 2 possibilities for the 1st toss, H T.

• H T

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We can pair each of these with 2 possibilities.

• H T
• H T H
  T

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That gives 4 possibilities on the 2 tosses HH,

• H T
• H T H
  T

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HT,

• H T
• H T H
  T

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TH

• H T
• H T H
  T

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And TT

• H T
• H T H
  T

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If we toss the coin a 3rd time, we can pair each
of 4 possibilities with a H or T.

• H
  T
• H T H
  T
• H T H T H T H
  T

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So for 3 tosses, we have 8 possibilities HHH,

• H
  T
• H T H
  T
• H T H T H T H
  T

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HHT,

• H
  T
• H T H
  T
• H T H T H T H
  T

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HTH,

• H
  T
• H T H
  T
• H T H T H T H
  T

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HTT,

• H
  T
• H T H
  T
• H T H T H T H
  T

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THH,

• H
  T
• H T H
  T
• H T H T H T H
  T

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THT,

• H
  T
• H T H
  T
• H T H T H T H
  T

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TTH,

• H
  T
• H T H
  T
• H T H T H T H
  T

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and TTT.

• H
  T
• H T H
  T
• H T H T H T H
  T

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Again, we had 8 possibilities for 3 tosses.

• 2 x 2 x 2 23 8

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In general,

• If we have an experiment with k parts (such as 3
  tosses)
• and each part has n possible outcomes (such as
  heads tails),
• then the total number of possible outcomes for
  the experiment is
• nk
• This is the simplest multiplication rule.

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As a variation, suppose that we have an
experiment with 2 parts, the 1st part has m
possibilities, the 2nd part has n
possibilities.

• How many possibilities are there for the
  experiment?

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In particular, we might have a coin a die.

• So one possible outcome could be H6 (a head on
  the coin a 1 on the die).
• How many possible outcomes does the experiment
  have?

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We have 2 x 6 12 possibilities.

• Return to our more general question about the
  2-part experiment with m n possibilities for
  each part.
• We now see that the total number of outcomes for
  the experiment is
• mn

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If we had a 3-part experiment, the 1st part has
m possibilities, the 2nd part has n
possibilities, the 3rd part has p
possibilities, how many possible outcomes would
the experiment have?

• m n p

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Permutations Counting Rules
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Permutations

• Suppose we have a horse race with 8 horses
  A,B,C,D,E,F,G, and H.
• We would like to know how many possible
  arrangements we can have for the 1st, 2nd, 3rd
  place horses.
• One possibility would be G D F.(F D G would be a
  different possibility because order matters here.)

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• We have 8 possible horses we can pick for 1st
  place.
• Once we have the 1st place horse, we have 7
  possibilities for the 2nd place horse.
• Then we have 6 possibilities for the 3rd place
  horse.
• So we have 8 7 6 336 possibilities.
• This is the number of permutations of 8 objects
  (horses in this case), taken 3 at a time.

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How many possible arrangements of all the horses
are there?In other words, how many permutations
are there of 8 objects taken 8 at a time?

• 8 7 6 5 4 3 2 1 40,320
• This is 8 factorial or 8!

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Wed like to develop a general formula for the
number of permutations of n objects taken k at a
time.

• Lets work with our horse example of 8 horses
  taken 3 at a time.
• The number of permutations was 8 7 6.

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We can multiply our answer (8 7 6) by 1
still have the same answer.
Multiplying we have

• Note that this is

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So the number of permutations of n objects taken
k at a time is
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Combinations Counting Rules
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Combinations

• Suppose we want to know how many different poker
  hands there are.
• In other words, how many ways can you deal 52
  objects taken 5 at a time.
• Keep in mind that if we have 5 cards dealt to us,
  it doesnt matter what order we get them. Its
  the same hand.
• So while order matters in permutations, order
  does not matter in combinations.

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Lets start by asking a different question.

• What is the number of permutations of 52 cards
  taken 5 at a time?
• nPk 52P5
  

52 51 50 49 48
311,875,200 When we count the
number of permutations, we are counting each
reordering separately .
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However, each reordering should not be counted
separately for combinations.

• We need to figure out how many times we counted
  each group of 5 cards.
• For example, we counted the cards ABCDE
  separately as ABCDE, BEACD, CEBAD, DEBAC, etc.
• How many ways can we rearrange 5 cards?
• We could rearrange 8 horses 8! ways.
• So we can rearrange 5 cards 5! 120 ways.

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That means that we counted each group of cards
120 times.

• So the number of really different poker hands is
  the number of permutations of 52 cards taken 5 at
  a time divided by the 120 times that we counted
  each hand.

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So the number of combinations of 52 cards taken
5 at a time is
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The number of combinations of n objects taken k
at a time is
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The number of combinations of n objects taken k
at a time nCk is also written as

• n
• k
• and is read as n choose k.
• Its the number of ways you can start with n
  objects and choose k of them without regard to
  order.

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Complements, Unions, Intersections

• Suppose A B are events.

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The complement of A is everything in the sample
space S that is NOT in A.

• If the rectangular box is S, and the white circle
  is A, then everything in the box thats outside
  the circle is Ac , which is the complement of A.

S
A
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Theorem

• Pr (Ac) 1 - Pr (A)
• Example
• If A is the event that a randomly selected
  student is male, and the probability of A is 0.6,
  what is Ac and what is its probability?
• Ac is the event that a randomly selected student
  is female, and its probability is 0.4.

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The union of A B (denoted A U B) is everything
in the sample space that is in either A or B or
both.
S
A
B

• The union of A B is the whole white area.

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The intersection of A B (denoted AnB) is
everything in the sample space that is in both A
B.
S
B
A

• The intersection of A B is the pink
  overlapping area.

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Example

• A family is planning to have 2 children.
• Suppose boys (B) girls (G) are equally likely.
• What is the sample space S?
• S BB, GG, BG, GB

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Example continued

• If E is the event that both children are the same
  sex, what does E look like what is its
  probability?
• E BB, GG
• Since boys girls are equally likely, each of
  the four outcomes in the sample space S BB,
  GG, BG, GB is equally likely has a
  probability of 1/4.
• So Pr(E) 2/4 1/2 0.5

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Example contd Recall that E BB, GG
Pr(E)0.5

• What is the complement of E and what is its
  probability?
• Ec BG, GB
• Pr (Ec) 1- Pr(E) 1 - 0.5 0.5

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Example continued

• If F is the event that at least one of the
  children is a girl, what does F look like what
  is its probability?
• F BG, GB, GG
• Pr(F) 3/4 0.75

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Recall E BB, GG Pr(E)0.5F BG, GB,
GG Pr(F) 0.75

• What is EnF?
• GG
• What is its probability?
• 1/4 0.25

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Recall E BB, GG Pr(E)0.5F BG, GB,
GG Pr(F) 0.75

• What is the EUF?
• BB, GG, BG, GB S
• What is the probability of EUF?
• 1
• If you add the separate probabilities of E F
  together, do you get Pr(EUF)? Lets try it.
• Pr(E) Pr(F) 0.5 0.75 1.25 ? 1 Pr (EUF)
• Why doesnt it work?
• We counted GG (the intersection of E F) twice.

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A formula for Pr(EUF)

• Pr(EUF) Pr(E) Pr(F) - Pr(EnF)
• If E F do not overlap, then the intersection is
  the empty set, the probability of the
  intersection is zero.
• When there is no overlap, Pr(EUF) Pr(E)
  Pr(F) .

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Conditional Probability of A given BPr(AB)

• Pr(AB) Pr (AnB) / Pr(B)

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ExampleSuppose there are 10,000 students at a
university.2,000 are seniors (S). 3,500 are
female (F).800 are seniors female.

• Determine the probability that a randomly
  selected student is (1) a senior, (2) female, (3)
  a senior female.
• 1. Pr(S) 2,000/10,000 0.2
• 2. Pr(F) 3,500/10,000 0.35
• 3. Pr(SnF) 800/10,000 0.08

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Use the definition of conditional probability
Pr(AB) Pr(AnB) / Pr(B) the previously
calculated information Pr(S) 0.2 Pr(F)
0.35 Pr(SnF) 0.08 to answer the
questions below.

• 1. If a randomly selected student is female,
  what is the probability that she is a senior?
• Pr(SF) Pr(SnF) / Pr(F)
• 0.08 / 0.35 0.228
• 2. If a randomly selected student is a senior,
  what is the probability the student is female?
• Pr(FS) Pr(FnS) / Pr(S)
• 0.08 / 0.2 0.4
• Notice that SnF FnS, so the numerators are the
  same, but the denominators are different.

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Joint Probability Distributions Marginal
Distributions
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Example Suppose a firm has 3 departments. Of
the firms employees, 10 are male in dept. 1,
30 are male in dept. 2, 20 are male in
dept. 3, 15 are female in dept. 1, 20 are
female in dept. 2, 5 are female in dept.
3. Then the joint probability distribution of
gender dept. is as in the table below.
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Example contd What is the probability that a
randomly selected employee is male?
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Example contd What is the probability that a
randomly selected employee is male?
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Example contd What is the probability that a
randomly selected employee is female?
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Example contd What is the probability that a
randomly selected employee is female?
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Example contd What is the probability that a
randomly selected employee is in dept. 1?
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Example contd What is the probability that a
randomly selected employee is in dept. 1?
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Example contd What is the probability that a
randomly selected employee is in dept. 2?
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Example contd What is the probability that a
randomly selected employee is in dept. 2?
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Example contd What is the probability that a
randomly selected employee is in dept. 3?
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Example contd What is the probability that a
randomly selected employee is in dept. 3?
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Example contd The marginal distribution of
gender is in first last columns (or left
right margins of the table) gives the
probability of each possibility for gender.
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Example contd The marginal distribution of
department is in first last rows (or top
bottom margins of the table) gives the
probability of each possibility for dept.
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Notice that when you add the numbers in the last
column or the last row, you must get one, because
youre adding all the probabilities for all the
possibilities.
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Bayesian Analysis

• Allows us to calculate some conditional
  probabilities using other conditional
  probabilities

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Example We have a population of potential
workers. We know that 40 are grade school
graduates (G), 50 are high school grads (H),
10 are college grads (C). In addition,10 of
the grade school grads are unemployed (U), 5 of
the h.s. grads are unemployed (U), 2 of the
college grads are unemployed (U).

• Convert this information into probability
  statements.
• Then determine the probability that a randomly
  selected unemployed person is a college graduate,
  that is, Pr(CU).

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40 are grade school graduates (G), 50 are high
school grads (H), 10 are college grads (C).
In addition,10 of the grade school grads are
unemployed (U), 5 of the h.s. grads are
unemployed (U), 2 of the college grads are
unemployed (U).

• Pr(G) 0.40

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40 are grade school graduates (G), 50 are high
school grads (H), 10 are college grads (C).
In addition,10 of the grade school grads are
unemployed (U), 5 of the h.s. grads are
unemployed (U), 2 of the college grads are
unemployed (U).

• Pr(G) 0.40
• Pr(H) 0.50

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40 are grade school graduates (G), 50 are high
school grads (H), 10 are college grads (C).
In addition,10 of the grade school grads are
unemployed (U), 5 of the h.s. grads are
unemployed (U), 2 of the college grads are
unemployed (U).

• Pr(G) 0.40
• Pr(H) 0.50
• Pr(C) 0.10

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40 are grade school graduates (G), 50 are high
school grads (H), 10 are college grads (C).
In addition,10 of the grade school grads are
unemployed (U), 5 of the h.s. grads are
unemployed (U), 2 of the college grads are
unemployed (U).

• Pr(G) 0.40 Pr(UG) 0.10
• Pr(H) 0.50
• Pr(C) 0.10

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40 are grade school graduates (G), 50 are high
school grads (H), 10 are college grads (C).
In addition,10 of the grade school grads are
unemployed (U), 5 of the h.s. grads are
unemployed (U), 2 of the college grads are
unemployed (U).

• Pr(G) 0.40 Pr(UG) 0.10
• Pr(H) 0.50 Pr(UH) 0.05
• Pr(C) 0.10

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40 are grade school graduates (G), 50 are high
school grads (H), 10 are college grads (C).
In addition,10 of the grade school grads are
unemployed (U), 5 of the h.s. grads are
unemployed (U), 2 of the college grads are
unemployed (U).

• Pr(G) 0.40 Pr(UG) 0.10
• Pr(H) 0.50 Pr(UH) 0.05
• Pr(C) 0.10 Pr(UC) 0.02

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In order to calculate Pr(CU), we need to
determine the probability that a randomly
selected individual is 1. a grade school grad
unemployed 2. a h.s. grad unemployed 3.
a college grad unemployed
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Recall that 40 of our population is grade school
grads, 10 of them are unemployed.

• Then 10 of 40 of our population is grade school
  grads unemployed.
• So Pr(G U) Pr(GnU)
• 0.10 x 0.40
  0.04.
• Similarly, Pr(H U) Pr(HnU)
• 0.05 x 0.50
  0.025.
• Also, Pr(C U) Pr(CnU)
• 0.02 x 0.10
  0.002.

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Given Pr(G U) Pr(G n U) 0.04, Pr(H U)
Pr(H n U) 0.025, Pr(C U) Pr(C n U)
0.002, we can calculate the probability that a
randomly selected individual is unemployed,
Pr(U).

• Pr(U) Pr(GnU) Pr(HnU) Pr(CnU)
• 0.04 0.025 0.002
• 0.067

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We can finally determine Pr(CU), using our
calculations the definition of conditional
probability.

• Pr(CU) Pr(CnU) / Pr(U)
• 0.002 / 0.067
• 0.030.
• So the probability that a randomly selected
  unemployed individual is a college graduate is
  0.03.

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We can also do the problem in an easily organized
table.
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We can also do the problem in an easily organized
table.
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We can also do the problem in an easily organized
table.
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We can also do the problem in an easily organized
table.
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Independence

• Two events are independent, if knowing that one
  event happened doesnt give you any information
  on whether the other happened.
• Example
• A It rained a lot in Beijing, China last year.
• B You did well in your courses last year.
• These two events are independent (unless you
  took your courses in Beijing). One of these
  events occurring tells you nothing about whether
  the other occurred.

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So in terms of probability, two events A B are
independent if and only if

• Pr(AB) Pr(A)
• Using the definition of conditional probability,
  this statement is equivalent to
• Pr(AnB) / Pr(B) Pr(A).
• Multiplying both sides by Pr(B), we have
• Pr(AnB) Pr(A) Pr(B).
• Dividing both sides by Pr(A), we have
• Pr(AnB) / Pr(A) Pr(B),
• which is equivalent to
• Pr(BA) Pr(B).
• This makes sense. If knowing about B tells us
  nothing about A, then knowing about A tells us
  nothing about B.

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We now have 3 equivalent statements for 2
independent events A B

• Pr(AB) Pr(A)
• Pr(BA) Pr(B)
• Pr(AnB) Pr(A) Pr(B).
• The last equation says that you can calculate the
  probability that both of two independent events
  occurred by multiplying the separate
  probabilities.

101
Example Toss a fair coin a fair die

• A You get a H on the coin.
• B You get a 6 on the die.
• Recall that we counted 12 possible outcomes for
  this experiment.
• Since the coin the die are fair, each outcome
  is equally likely, the probability of getting a
  H a 6 is 1/12.

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Example contd

• The probability of a H on the coin is 1/2
• The probability of a 6 on the die is 1/6.
• So Pr(H) Pr(6) (1/2)(1/6)
• 1/12
• Pr(Hn6), we can see
  that these 2 events are independent of each other.

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Mutually Exclusive

• Two events are mutually exclusive if you know
  that one occurred, then you know that the other
  could not have occurred.
• example You selected a student at random.
• A You picked a male.
• B You picked a female.
• These 2 events are mutually exclusive, because
  you know that if A occurred, B did not.

104
Mutually exclusive events are NOT independent!

• Remember that for independent events, knowing
  that one event occurred tells you nothing about
  whether the other occurred.
• For mutually exclusive events, knowing that one
  event occurred tells you that the other
  definitely did not occur!

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The Birthday Problem
106
The Birthday Problem

• What is the probability that in a group of k
  people at least two people have the same
  birthday?
• (We are going to ignore leap day, which
  complicates the analysis, but doesnt have much
  effect on the answer.)

107
For our group of k people, let p Pr(at least 2
people have the same birthday).

• At least 2 people having the same birthday is the
  complement (opposite) of no 2 people having the
  same birthday, or everyone having different
  birthdays.
• Its easier to calculate the probability of
  different birthdays.
• So we can do that then subtract the answer from
  one to get the probability we want.

108
p 1- Pr(all different birthdays)
109
p 1- Pr(all different birthdays)
110
p 1- Pr(all different birthdays)
111
p 1- Pr(all different birthdays)
112
p 1- Pr(all different birthdays)
113
p 1- Pr(all different birthdays)
114
p 1- Pr(all different birthdays)
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p 1- Pr(all different birthdays)
116
p 1- Pr(all different birthdays)

• This is very messy, but you can calculate the
  answer for any number k.
• I have the answers computed for some sample
  values.

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Birthday Problem Probabilities

• k p
• 5 0.027
• 10 0.117
• 15 0.253
• 20 0.411
• 22 0.476
• 23 0.507
• 25 0.569
• 30 0.706
• 40 0.891
• 50 0.970
• 100 0.9999997