Transcript A and B

WHICH OF THE FOLLOW STATISTICS IS
NOT TRUE?
1.
2.
3.
4.
More than 50% of American
25%
25%
adults are single.
For the average individual, the
number of years spent unmarried
now outweighs the number of
years married.
Compared with their married
counterparts, single people are
more likely to eat out and
exercise, go to art and music
classes, attend public events and
lectures, and volunteer.
None of the above.
1
2
25%
25%
Slide
1- 1
3
4
UPCOMING IN CLASS

Wednesday’s Quiz 2, covers HW3 of the material
learned in class. (9/11)

Sunday HW4 (9/15)

Part 2 of the data project due next Monday (9/16)
CHAPTER 13
Copyright © 2008 Pearson Education, Inc.
Publishing as Pearson Addison-Wesley
Probability General Addition and
Multiplication Rules. Non-disjoint and
dependent events.
THE GENERAL ADDITION RULE

General Addition Rule:


For any two events A and B,
P(A or B) = P(A) + P(B) – P(A and B)
The following Venn diagram shows a situation in
which we would use the general addition rule:
Slide
1- 4
HOMEWORK PROBLEM

A check of dorm rooms on a large college campus
revealed that
39% had refrigerators
 60% had TVs
 20% had both a refrigerator and TV

WHAT’S THE PROBABILITY THAT A RANDOMLY
SELECTED DORM ROOM HAS A REFRIGERATOR
BUT NO
TV?
25%
1.
2.
3.
4.
25%
25%
25%
.39+.60-.20
.39+.40-.20
.39-.20
.39-.60
Slide
1- 6
1
2
3
4
WHAT’S THE PROBABILITY OF HAVING A
REFRIGERATOR OR A TV BUT NOT BOTH?
1.
2.
3.
4.
.39+.60-.20-.20
.39+.60-.20
.39-.20
.39-.60
25%
25%
25%
25%
Slide
1- 7
1
2
3
4
Slide 1- 7
WHAT’S THE PROBABILITY OF HAVING
NEITHER A REFRIGERATOR NOR A TV?
25%
1.
2.
3.
4.
25%
25%
25%
1-.60
1-.39
1-(.39+.60)
1-(.39+.60-.20)
Slide
1- 8
1
2
3
4
Slide 1- 8
THE GENERAL ADDITION RULE

General Addition Rule:


For any two events A and B,
P(A or B) = P(A) + P(B) – P(A and B)
For disjoint events P(A and B) =0.
Slide
1- 9
MULTIPLICATION RULE FOR
INDEPENDENT EVENTS
When two events A and B are independent, we
can use the multiplication rule for independent
events from Chapter 14:
P(A and B) = P(A) x P(B)
 However, when our events are not independent,
this earlier multiplication rule does not work.
Thus, we need the General Multiplication Rule.

Slide
1- 10
THE GENERAL MULTIPLICATION RULE
We encountered the general multiplication rule
in the form of conditional probability.
 Rearranging the equation in the definition for
conditional probability, we get the General
Multiplication Rule:


For any two events A and B,
P(A and B) = P(A) x P(B|A)
or
P(A and B) = P(B) x P(A|B)
Slide
1- 11
IT DEPENDS…
Back in Chapter 3, we looked at contingency
tables and talked about conditional distributions.
 When we want the probability of an event from a
conditional distribution, we write P(B|A) and
pronounce it “the probability of B given A.”
 A probability that takes into account a given
condition is called a conditional probability.

Slide
1- 12
CONDITIONAL DISTRIBUTIONS

A conditional distribution shows the distribution
of one variable for just the individuals who
satisfy some condition on another variable.

The following is the conditional distribution of ticket
Class, conditional on having survived:
CONDITIONAL DISTRIBUTIONS (CONT.)

The following is the conditional distribution of ticket
Class, conditional on having perished:
Slide
3- 14
CONDITIONAL PROBABILITIES

You draw a card at random from a standard deck
of 52 cards.





4 suits (hearts, clubs, spades, diamond)
26 red
26 black
12 face cards
4 Aces
Slide
1- 15
WHAT’S THE PROBABILITY THAT A CARD IS
A HEART, GIVEN THAT IT IS RED?
1.
2.
3.
4.
26/52
13/52
13/26
1/26
25%
25%
25%
25%
Slide
1- 16
1
2
3
4
WHAT’S THE PROBABILITY THAT THE CARD
IS RED, GIVEN THAT IT IS A HEART?
1.
2.
3.
13/26
13/52
13/13
33%
33%
33%
Slide
1- 17
1
2
3
WHAT’S THE PROBABILITY THAT THE CARD
IS A TEN, GIVEN THAT IT IS RED?
25%
1.
2.
3.
4.
25%
25%
25%
1/13
2/26
2/13
4/26
Slide
1- 18
1
2
3
4
Slide 1- 18
WHAT’S THE PROBABILITY THAT THE CARD
IS A KING, GIVEN THAT IT IS A FACE CARD?
25%
1.
2.
3.
4.
25%
25%
25%
4/52
4/26
4/13
4/12
Slide
1- 19
1
2
3
4
THE GENERAL MULTIPLICATION RULE
We encountered the general multiplication rule
in the form of conditional probability.
 Rearranging the equation in the definition for
conditional probability, we get the General
Multiplication Rule:


For any two events A and B,
P(A and B) = P(A) x P(B|A)
or
P(A and B) = P(B) x P(A|B)
Slide
1- 20
DRAWING WITHOUT REPLACEMENT
Sampling without replacement means that once
one object is drawn it doesn’t go back into the
pool.
 We often sample without replacement, which
doesn’t matter too much when we are dealing
with a large population.
 However, when drawing from a small
population, we need to take note and adjust
probabilities accordingly.
 Drawing without replacement is just another
instance of working with conditional
probabilities.

Slide
1- 21
YOU ARE DEALT A HAND OF THREE CARDS, ONE
AT A TIME.
WHAT’S THE PROBABILITY THE FIRST RED
CARD YOU GET IS THE 3RD CARD DEALT?
1.
2.
3.
4.
½+½+½
½*½*½
26/52 * 25/51 * 26/50
26/52 * 26/52 * 26/52
0%
1
0%
2
0%
3
Slide
0%
1- 22
4
WHAT IS THE PROBABILITY THAT YOUR
CARDS ARE ALL HEARTS?
1.
2.
3.
4.
13/52 * 12/51 * 11/50
26/52 * 25/51 * 24/50
¼*¼*¼
½*½*½
0%
1
0%
2
0%
3
Slide
0%
1- 23
4
WHAT’S THE PROBABILITY THAT YOU GET
NO BLACK CARDS?
1.
2.
3.
4.
¾*¾*¾
¼*¼*¼
39/52 * 38/51 * 37/50
26/52 * 25/51 * 24/50
0%
0%
0%
0%Slide
1- 24
1
2
3
4
WHAT’S THE PROBABILITY THAT YOU HAVE
AT LEAST ONE SPADE?
1.
2.
3.
4.
1- 13/52 * 12/51 * 11/50
1- ¼ * ¼ * ¼
1- ¼
1- 39/52 * 38/51 * 37/50
0%
0%
0%
0%Slide
1- 25
1
2
3
4
TREE DIAGRAMS
A tree diagram helps us
think through
conditional probabilities
by showing sequences of
events as paths that
look like branches of a
tree.
 Figure 15.4 is a nice
example of a tree
diagram and shows how
we multiply the
probabilities of the
branches together:

Slide
1- 26
HOMEWORK PROBLEM

Mary is flying from Boston to Denver with a
connection in Chicago. The probability her first
flight leaves on time is 0.2. If the flight is on
time, the probability that her luggage will make
the connecting flight is 0.95, but if the flight is
delayed, the probability that the luggage will
make it is only 0.55.
Slide
1- 27
ARE THE FOLLOWING EVENTS INDEPENDENT FLIGHT LEAVING ON TIME AND THE LUGGAGE
MAKING IT TO
1.
2.
DENVER WITH HER?
Yes, the luggage is equally
likely to arrive with her
whether or not the flight is
on time.
No, the probability that the
luggage arrives with her
depends on whether the
flight is on time.
50%
50%
Slide
1- 28
1
2
WHAT’S THE PROBABILITY THAT HER
LUGGAGE ARRIVES IN DENVER WITH HER?
1.
2.
3.
.19
.44
.19+.44
33%
33%
33%
Slide
1- 29
1.
2.
3
IT DEPENDS… (CONT.)

To find the probability of the event B given the
event A, we restrict our attention to the outcomes
in A. We then find in what fraction of those
outcomes B also occurred.
P(B| A)  P(A and B)
P(A)

Note: P(A) cannot equal 0, since we know that A
has occurred.
Slide
1- 30
SOBRIETY CHECKPOINT PROBLEM
Police establish a sobriety checkpoint where they
detain drivers whom they suspect have been
drinking and release those who have not. The
police detain 81% of drivers who have been
drinking and release 81% of drivers who have
not.
 Assume that 10% of drivers have been drinking.

Slide
1- 31
WHAT’S THE PROBABILITY OF ANY GIVEN
DRIVER WILL BE DETAINED?
1.
2.
3.
4.
.10*.81+.9*.19
.10*.19+.9*.81
.10*.81
.9*.81
0%
0%
0%
0%Slide
1- 32
1
2
3
4
WHAT’S THE PROBABILITY THAT A DRIVER WHO
IS DETAINED HAS ACTUALLY BEEN DRINKING?
1.
2.
3.
4.
25.2 / 10
8.1 / 25.2
10 / 25.2
81 / 25.2
0%
1
0%
2
0%
3
Slide
0%
1- 33
4
WHAT’S THE PROBABILITY THAT A DRIVER WHO
WAS RELEASED HAD ACTUALLY BEEN
DRINKING?
1.
2.
3.
4.
1.9 / 25.2
72.9 / 25.2
1.9 / 74.8
72.9 / 25.2
0%
1
0%
2
0%
3
Slide
0%
1- 34
4
ONE MORE TREE PROBLEM

Dan’s Diner employs three dishwashers.
Al washes 60% of the dishes and breaks only 2% of
those he handles.
 Betty and Chuck each wash 20% of the dishes, and
Betty breaks only 2% of hers, but Chuck breaks 4% of
the dishes he washes.


You go to Dan’s for supper one night and hear a
dish break at the sink.
Slide
1- 35
WHAT’S THE PROBABILITY CHUCK BROKE
THE DISH?
1.
2.
3.
4.
.008/(.008+.004+.012)
.012/(.008+.004+.012)
.008/(.04)
.008/(.96)
0%
0%
0%
0%Slide
1- 36
1
2
3
4
WHAT’S THE PROBABILITY THAT AL BROKE
THE DISH?
1.
2.
3.
4.
.008/(.008+.004+.012)
.012/(.008+.004+.012)
.008/(.04)
.008/(.96)
0%
0%
0%
0%Slide
1- 37
1
2
3
4
UPCOMING IN CLASS

Wednesday’s Quiz 2, covers HW3 of the material
learned in class. (9/11)

Sunday HW4 (9/15)

Part 2 of the data project due next Monday (9/16)