Statistics 221

1
Statistics 221

• Chapter 4 Part B
• Probability

2
Combining probabilities

• To determine the probability of some events,
  probabilities of individual outcomes need to be
  added or multiplied.
• Here is an example where adding probabilities is
  appropriate If you roll a die, what is the
  probability of getting an even number (2, 4 or
  6)?
• To answer that question, youd add P(2) P(4)
  P(6) 1/6 1/6 1/6 3/6 or ½.
• There is a 50 chance of getting an even value on
  the die.

3
Combining probabilities

• Here is an example where multiplying
  probabilities is appropriate If you roll a die
  three times, what is the probability of getting
  HHH?
• To answer that question, you would multiply ½ ½
  ½ 1/8.
• There is a 12.5 (1 in 8) chance of rolling a die
  and getting heads all three times.

4
When to add probabilities

• When there is more than one sample point that
  satisfies the criteria needed to say that the
  event has occurred, you add probabilities.
• The question of what is the probability of
  getting an even number on the die can be framed
  as an OR question
• What is P(2) OR P(4) OR P(6)?
• The answer is determined by adding the
  probabilities
• P(even) 1/6 1/6 1/6 3/6 or ½.

5
When to multiply probabilities

• When each sample point (outcome) for that event
  is a collection of individual outcomes treated as
  a single outcome.
• For example, rolling a die three times, the
  outcome is made up of three individual outcomes
  such as HHH or HTH or THT, etc. but each
  set is considered a single outcome for the
  event of rolling three times.
• The question of what is the probability of
  getting three heads can be framed as an AND
  question
• What is P(H) AND P(H) AND P(H)?
• The answer is determined by multiplying the
  probabilities P(HHH) ½ ½ ½ 1/8.

6
What is the probability of rolling doubles?

• That question can be framed as
• What is P(1 and 1) or (2 and 2) or (3 and 3) or
  (4 and 4) or (5 and 5) or (6 and 6)
• Since the probability of any one particular value
  on the die is 1/6, the answer to this question
  can be calculated as
• (1/6 1/6) (1/6 1/6) (1/6 1/6) (1/6
  1/6) (1/6 1/6) (1/6 1/6) or 5/36

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The Addition Principle

• The formula for calculating the probability of an
  event consisting of two or more events joined by
  OR is derived from the Addition Principle. The
  formula is this
• P(A or B) P(A) P(B)
• It says The probability of (A or B) is the
  probability of event A occurring plus the
  probability of event B occurring.

8
Asking the probability of at least one is a
union question

• When we ask, what is the probability of A OR B,
  we are actually asking
• What is P(A ? B)? where the ? symbol means union.
• A Venn diagram can be used to represent a union
  visually.

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A Venn diagram to represent the Union of A ? B

• P(2) or P(4) or P(6) P(2 ? 4 ? 6)
• P(even) is represented visually as the sum of the
  areas in the green circles as a percentage of the
  total area in the blue box.

5
4
1
6
2
3
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The Multiplication Principle

• The formula for calculating the probability of an
  event consisting of two events joined by AND is
  derived from the Multiplication Principle. The
  formula is this
• P(A and B) P(A) P(B)
• It says The probability of (A and B) is the
  probability of event A occurring times the
  probability of event B occurring.

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Intersections

• When we ask, what is the probability of A AND
  B?, we are actually asking
• What is P(A ? B)? where the ? symbol means
  intersection.
• VENN diagrams can be used to visualize the
  concepts of union and intersection.

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The expression (A and B) can be expressed as the
Intersection of A ? B

• P(H) and P(H) and P(H) P(H ? H ? H)
• P(HHH) is represented visually as the overlapping
  area in the green circles as a percentage of the
  total area in the blue box.

Overlapping area
H
H
H
13
Another example

• Lets say you draw a card from a deck. What is
  the probability that you draw an ACE or a KING or
  a QUEEN or a JACK?
• A drawing an ACE
• B drawing an KING
• C drawing an QUEEN
• D drawing an JACK
• And
• P(A) 4/52, P(B) 4/52, P(C) 4/52, P(D)
  4/52
• Therefore
• P(A or B or C or D) 4/52 4/52 4/52 4/52
  16/52 or 30.77

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Venn Diagram for P(Ace or King or Queen or Jack)
The blue box represents the sample space all
cards in the deck.
Ace
King
Jack
Queen
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Lets change the question

• Lets say you draw a card from a deck. What is
  the probability that you draw an ACE or a RED
  card?
• A drawing an ACE
• B drawing a RED (??)
• And
• P(A) 4/52, P(B) 26/52
• Therefore, is this true?
• P(A or B) 4/52 26/52 or 30/52? No!

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Disjoint events

• When two outcomes of an event can occur at the
  same time, the outcomes are NOT disjoint. When
  two outcomes of an event cannot occur at the same
  time, the outcomes are disjoint.
• In this case, A and B are not disjoint (mutually
  exclusive) events. In other words, A and B can
  occur together.
• When A and B are NOT disjoint, then to calculate
  the P(A or B), we must add the P(A) to the P(B)
  but then subtract off the probability that (A and
  B) will occur simultaneously.

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Venn Diagram for Joint Events
Sample space of all cards in the deck
Ace
Red
The red circle encompasses the points where we
drew the Ace of hearts or the Ace of diamonds.
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What is the P(A or B) now?

• If A and B are NOT disjoint, they can occur
  together, and that occurrence is represented by
  the overlap of the circles. The sample points
  that fall in that area must be subtracted out so
  that they are not counted twice.

Sample Space S
If A and B are NOT disjoint, subtract out the
sample points in the red circle (the overlap
area).
Red
Ace
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Formula refinement

• The formula for calculating the probability of a
  combination event consisting of two events joined
  by OR when events are NOT disjoint is
• P(A or B) P(A) P(B) P(A and B)
• It says The probability of (A or B) is the
  probability of event A occurring plus the
  probability of event B occurring less the
  probability that they occurred simultaneously.

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So what is P(Ace or Red)?

• A drawing an ACE
• B drawing a RED ??
• And
• P(A) 4/52, P(B) 26/52
• Therefore
• P(A or B) P(A) or P(B) P(A and B)
• P(A or B) 4/52 26/52 (4/52 26/52) .538

21
Applying the multiplication rule to obtain the
P(A ? B)

• What is the probability of rolling a dice twice
  and getting a six each time?
• A getting a 6 on the first roll
• P(A) 1/6
• B getting a 6 on the 2nd roll
• P(B) 1/6
• Therefore
• P (A and B) P(A) P(B) 1/6 1/6 1/36

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Another Example

• Lets say you draw a card from a deck 4 times.
  What is the probability that you draw an ACE all
  four times?
• A drawing an ACE on the first draw
• B drawing an ACE on the second draw
• C drawing an ACE on the third draw
• D drawing an ACE on the fourth draw
• And
• P(A) 4/52, P(B) 4/52, P(C) 4/52, P(D)
  4/52
• Therefore
• P(A and B and C and D) 4/524 or .000035

23
But what if you dont replace the card before you
draw the next one?

• Now you have a dependent events situation.
• If you replace the card each time before you draw
  the next one, then the events are independent of
  each other, and the probability of drawing an ACE
  is the same each time (4/52).
• If you do not replace the card each time before
  you draw the next one, then the events are
  dependent, and the probability of drawing an ACE
  is not the same each time.

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Dependent events

• Lets say you draw a card from a deck 4 times and
  this time you do not put the card back in the
  deck each time. What is the probability that you
  draw an ACE all four times?
• A drawing an ACE on the first draw
• B drawing an ACE on the second draw
• C drawing an ACE on the third draw
• D drawing an ACE on the fourth draw
• And now
• P(A) 4/52, P(B) 3/51, P(C) 2/50, P(D)
  1/49
• Therefore
• P(A and B and C and D) 4/52 3/51 2/50
  1/49 .0000037

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Revisiting the formula for P(A and B)

• The formula
• P(A and B) P(A) P(B)
• Is applicable only if (A) and (B) are independent
  events. That is, the occurrence of (A) does not
  effect the occurrence of (B).
• But if the outcome of As occurrence affects the
  probability of B occurring, the A and B are
  dependent events.

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When A and B are dependent

• If A and B are dependent events, the formula is
• P(A and B) P(A) P(BA)
• In English the probability of A and B both
  occurring is the probability of A occurring
  multiplied by the probability of B occurring
  assuming that A did occur.
• P(BA) is the probability of B given that A
  occurred.

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Applying this formula to drawing cards without
replacement

• P(drawing first ACE) 4/52 .077
• But assume that you did get an ace, so there are
  only 3 aces left in the deck which now has 51
  cards.
• P(drawing second ACE) 3/51 .059
• Again, assume that you did get an ace the second
  time, so there are only 2 aces left in the deck
  of 50 cards
• P(drawing third ACE) 2/50 .04
• P(drawing fourth ACE) 1/49 .020
• Therefore
• P(drawing 4 aces) .077 .059 .04 .020

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Practice question

• If two of the items shown below are randomly
  selected, find the probability that both items
  are colored green.
• red yellow green red blue yellow green
• Assume that the first item is replaced before the
  2nd item is selected.
• Assume that the first item is NOT replaced before
  the 2nd item is selected

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P (green and green)

• a. With replacement (independent events)
• P (A and B) P (A) P (B)
• 2/7 2/7 4/49
• 8.16
• b. Without replacement (dependent events)
• P (A and B) P (A) P(B A)
• 2/7 1/6
• 4.8

30
Practice question

• Time magazine reported that when 20,000 gas masks
  from branches of the U.S. military were tested,
  it was found that 6,850 were defective. If
  further investigation begins with the random
  selection of two gas masks from this population,
  find the probability that they are both
  defective.
• A. Assume that the first is replaced before the
  2nd is selected.
• B. Assume that the first is NOT replaced before
  the 2nd is selected.
• C. Which choice makes more sense?

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P(defective AND defective)

• a. With replacement
• P (A and B) P (A) P (B)
• 6850/20000 6850/20000
• .3425 .3425
• 11.7306
• b. Without replacement
• P (A and B) P (A) P(B A)
• 6850/20000 6849/19999
• 11.7295
• c. It makes more sense not to replace the first
  before selecting the second.

32
Conditional Probability

• Recall the formula for P(A and B) when A and B
  are dependent events
• P(A and B) P(A) P(BA)
• The expression P(BA) means the probability of
  B given that A did in fact happen.
• If A and B are related events in some way,
  knowing whether A actually did occur affects the
  probability of B occurring.

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We use conditional probability to refine our
probability estimates

• Learning about conditional probability is
  worthwhile if we want to improve our chances of
  predicting some outcome.
• If A and B are related in some way (dependent
  events), knowing that A happened can improve our
  chances of accurately predicting whether B will
  happen.

34
Examples that use conditional probability

• When a lab runs a test on a specimen or sample,
  false positives have been known to occur as
  well as true negatives.
• For example, if you get a strep test, it could
  come back negative when you really have strep and
  it could come back positive when you do not have
  strep.

35
Examples that use conditional probability

• What is the probability that a person tested
  positive, given that he has strep?
• Positive
  Negative
• Has strep 80 5
• Does not have strep 3
  11

36
What is P(positivestrep)?

• Without using formulas, common sense tells you
  that
• Take all those who truly have strep and find out
  what percent of those tested positive.
• 85 have strep (denominator)
• 80 of those tested positive (numerator)
• P (positivestrep) 80 / 85 or 94.1

The part about given that he has strep reduces
the denominator from 99 to 85.
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What is P(streppositive)?

• This is a different question. It is What is the
  probability that someone has strep given that
  they tested positive.
• Take all the ones who tested positive and find
  out what percentage of those are actually have
  strep.
• 83 tested positive
• 80 of those actually have it
• P(streppositive) 80 / 83 or 96.4

The part about given that he tested positive
reduces the denominator from 99 to 83.
38
Calculating conditional probabilities the easy
way

• 1. Create row and column totals.
• Positive Negative Total
• Has strep 80 5
  85
• Does not have strep 3 11
  14
• Total 83 16 99
• 2. To find P(AB), divide. The denominator is the
  row or column total for whatever B is, the
  numerator is the A value in that row or column
  identified as B. Examples
• P(positivestrep) 80/85
• P(streppositive) 80/83

39
Whats the conditional probability?

• If we randomly select someone who as aboard the
  Titanic, what is the probability of getting a
  man, given that the selected person died?

Results of the sinking of the Titanic
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What is P(man died)?

• Denominator 1517 died
• Numerator 1360 of those were men
• P (man died) 1360 / 1517 or 89.7

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Whats the conditional probability?

• If we randomly select someone who as aboard the
  Titanic, what is the probability of getting a boy
  or girl, given that the selected person survived?

42
What is P(boy or girl survived)?

• Denominator 706 survived
• Numerator (29 27) 56 of those were boys or
  girls
• P(boy or girl survived) 56 / 706 or 7.93

43
Generalizing the formula for conditional
probability

• In each of the previous examples, we were
  actually using the formula below.
• For example, P(positivestrep)
• Where A positive, B strep

P(A and B)
80
P (A B)

P(B)
85
Positive Negative Has strep
80
5 Does not have strep 3
11
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Bayes Theorem

• The method just presented for calculating
  conditional probabilities can be expressed using
  Bayes terminology.
• Bayes conditional probability formula is the
  same but the situation is thought of as happening
  in a time sequence where we start out with
  initial information (we call prior probabilities)
  and we then receive new information that enables
  us to refine our probability estimates to obtain
  posterior probabilities.

45
Calculating conditional probabilities using Bayes
terminology

• A manufacturing firm receives shipments of parts
  from two different suppliers, A1 and A2.
• 65 of their parts come from supplier A1 and 35
  come from supplier A2.
• Historically speaking, 98 of the parts from A1
  have been good (G) and 2 have been bad (B).
• Historically speaking, 95 of the parts from A2
  have been good (G) and 5 have been bad (B).
• Heres the question Given that a part is bad
  (B), whats the probability it came from supplier
  A?

46
Bayes terminology

• Before we know that a part is bad, there is a 65
  chance it came from A1 and a 35 chance it came
  from A2. These are called the prior
  probabilities.
• We also know that 2 of supplier A1s parts are
  bad and 5 of supplier A2s parts are bad. These
  are the conditional probabilities.
• Now if we select a part and (given) it is bad,
  what is the probability it came from supplier Ai?
  These are the posterior probabilities that we
  want to know.

47
Given that a part is bad (B), whats the
probability it came from supplier A?

• Prior probabilities We know that 65 of our
  parts come from supplier A and 35 come from
  supplier B.
• Conditional probabilities We know that 2 of
  supplier As parts are bad and 5 of supplier Bs
  parts are bad.

48
Given that a part is bad (B), whats the
probability it came from supplier A?

• If we randomly selected a part from our
  inventory, and it was bad, there is a 1.3 chance
  that it came from supplier A and its bad there
  is a 1.75 chance that it came from supplier B
  and its bad.

49
Given that a part is bad (B), whats the
probability it came from supplier A?

• 3.05 of our inventory is bad and 96.95 is good.

50
Given that a part is bad (B), whats the
probability it came from supplier A?
What is P(supplier A part is bad)? Denominator
3.05 Numerator 1.30 P(supplier A part is
bad) 1.30 / 3.05 or 42.62
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Practice Question

• Given that a part is good (G), whats the
  probability it came from supplier B?

P(Supplier B Good) .3325 / .9695 P(Supplier B
Good) 34.29
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Complements

• Sometimes its easier to calculate the probability
  of an event NOT happening than it is to calculate
  the probability of an event happening.
• If we know that
• P(A) P(not A) 1
• Then after we calculate P(not A), we can subtract
  that from 1 to find P(A).

53
The Complement of an Event

• The complement of event A is defined to be the
  event consisting of all sample points that are
  not in A.
• The complement of A is denoted by Ac.
• The Venn diagram below illustrates the concept of
  a complement.

Sample Space S
Ac
Event A
P(A) P(Ac) 1
54
Using complements to find the probability of at
least one

• The complement of at least one is none.
  Therefore
• P(at least one) 1 - P(none)

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Example

• Find the probability of a couple having at least
  one girl out of three children.
• P (none) P (boy) P (boy) P (boy)
• P (none) ½ ½ ½
• P (none) 1/8
• P (at least one girl) 1 1/8 or 7/8

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Whats the probability?

• If you run a red traffic light at an intersection
  equipped with a camera monitor, there is a .1
  probability that you will be given a ticket. If
  you run a red traffic light at this intersection
  five different times, what is the probability of
  getting at least one ticket?

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The possible outcomes

• The (6) possible outcomes are
• 0 tickets
• 1 ticket
• 2 tickets
• 3 tickets
• 4 tickets
• 5 tickets

None
At least 1
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Whats the complement?

• If A is (getting at least one ticket)
• Then not A is (getting 0 tickets)
• Rather than find the P(A), find the P(not A) and
  subtract it from 1
• P (0 tickets)
• P(.95) (.9) (.9) (.9) (.9) (.9) ) .59
• 1 - .59 .41
• Or 41

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Homework 7

• 28 on page 161
• 29 on page 161
• 32 on page 167
• 34 on page 168
• 42 on page 174