Transcript Chapter 14: From Randomness to Probability

Unit 4
CHAPTER 15: PROBABILITY RULES
AP Statistics
THE GENERAL ADDITION RULE
 When two events A and B are disjoint, we can use
the addition rule for disjoint events from Chapter 14:
P(A  B) = P(A) + P(B)
 However, when our events are not disjoint (not
mutually exclusive), this earlier addition rule will
double count the probability of both A and B
occurring. Thus, we need the General Addition Rule.
 Let’s look at a picture…
THE GENERAL ADDITION RULE
 General Addition Rule:
 For any two events A and B,
P(A  B) = P(A) + P(B) – P(A  B)
 The following Venn diagram shows a situation in which we
would use the general addition rule:
ADDITION RULE EXAMPLE:
 A random sample of 250 working adults found that 37%
access the Internet at work. 44% access the Internet
from home. And 21% Access the Internet at both work
and home. What is the probability that a person in this
sample will access the Internet from home or at work?
ADDITION RULE EXAMPLE 2:
B e l o w i s a Ve n n D i a g r a m o f a s a m p l e o f 2 0 0 w o m e n w h o l i ke yo g a to r e l i eve s t r e s s a n d
w h o l i ke r u n n i n g to r e l i ev e s t r e s s .
Running
Yoga
0.12
0.25
0.48
 What is the probability that a woman likes running ?
 What is the probability that a woman likes running, but doesn’t like yoga?
 What is the probability that a woman doesn’t like yoga and doesn’t like running?
 What is the probability that a woman will like running or yoga, but not both ?
CONDITIONAL PROBABILITIES
 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.
CONDITIONAL PROBABILITIES (CONT.)
 To find the probability of the event B given the event
A, we restrict our attention to the total outcomes in
A. We then find the fraction of those outcomes in B
that also occurred.
P(A

B)
P(B|A)
P(A)
 Note: P(A) cannot equal 0, since we know that A has
occurred.
CONDITIONAL PROBABILIT Y EXAMPLE:
 A random sample of 250 working adults found that 37%
access the Internet at work. 44% access the Internet
from home. And 21% Access the Internet at both work
and home. What is the probability that an adult
accesses the internet at home, given they access it at
work?
 Tell: ____% of adults who access the internet at work
also access it at home.
THE GENERAL MULTIPLICATION RULE
 When two events A and B are independent, we can
use the multiplication rule for independent events
from Chapter 14:
P(A  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.
THE GENERAL MULTIPLICATION RULE
(CONT.)
 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  B) = P(A)  P(B|A)
or
P(A  B) = P(B)  P(A|B)
INDEPENDENCE
 Independence of two events means that the
outcome of one event does not influence the
probability of the other.
 With our new notation for conditional probabilities,
we can now formalize this definition:
 Events A and B are independent whenever
P(B|A) = P(B). (Equivalently, events A and B are
independent whenever P(A|B) = P(A).)
INDEPENDENCE EXAMPLE:
 Earlier it was mentioned that 37% of working adults
access the Internet at work. 44% access the Internet
from home. And 21% Access the Internet at both
work and home.
 Are accessing the internet from work and accessing the
internet from work independent? Are they disjoint?
INDEPENDENT ≠ DISJOINT

Disjoint events cannot be independent! Well, why not?
 Since we know that disjoint events have no outcomes in common, knowing
that one occurred means the other didn’t.
 Thus, the probability of the second occurring changed based on our
knowledge that the first occurred.
 It follows, then, that the two events are not independent.
 Consider 2 disjoint events: getting an A in this course and getting a B in
this course. These are disjoint because they have nothing in common.
However, if you did earn an A in the course, does the affect your
probability of earning a B? YES! Your probability of earning a B in the
course is now 0. It cannot happen.

A common error is to treat disjoint events as if they were independent,
and apply the Multiplication Rule for independent events —don’t make that
mistake.
EXAMPLES: INDEPENDENCE VS. DISJOINT
 The AAPOR is an association of about 1600 individuals who share
an interest in public opinion and survey research. They report that
typically as few as 10% of random phone calls result in a completed
interview. Reasons are varied, but some of the most common
include no answer, refusal to cooperate, and failure to complete the
call.
 Which of the following events are disjoint, independent, or neither?
 A = Your phone # is randomly selected.
B = You’re not at home at dinnertime when they call.
 A = As a selected subject, you complete the interview.
B = As a selected subject, you refuse to cooperate.
 A = You are not at home when they call at 11 AM.
B = You are employed full time.
DEPENDING ON INDEPENDENCE
 It’s much easier to think about independent events than
to deal with conditional probabilities.
 It seems that most people’s natural intuition for
probabilities breaks down when it comes to conditional
probabilities.
 Don’t fall into this trap: whenever you see probabilities
multiplied together, stop and ask whether you think they
are really independent.
TABLES AND CONDITIONAL
PROBABILITIES
 I t i s m u c h e a s i e r to s e e c o n d i t io n a l p r o b a b i li t i e s u s i n g c o n t i n g e n c y t a b l es :
Nursing majors
Non-Nursing Majors
Total
Males
95
1015
1110
Females
700
1727
2427
Total
795
2742
3537
•F i n d t h e p r o b a b i l i t y t h a t a s t u d e n t i s a m a l e .
•F i n d t h e p r o b a b i l i t y t h a t a s t u d e n t i s n o t a n u r s i n g m a j o r .
•F i n d t h e p r o b a b i l i t y t h a t a s t u d e n t i s a f e m a l e , g i ve n t h e s t u d e n t i s a n u r s i n g m a j o r .
•F i n d t h e p r o b a b i l i t y t h e s t u d e n t i s m a l e o r a n u r s i n g m a j o r.
•F i n d t h e p r o b a b i l i t y t h e s t u d e n t i s f e m a l e a n d n o t a n u r s i n g m a j o r.
CREATE A CONTINGENCY TABLE
 Create a contingency table for the following samples of
textbook pages:
 48% of pages had a data display
 27% of pages had an equation
 7% had both a data display and an equation
USING THE TABLE DETERMINE THE
FOLLOWING:
 What is the probability that a randomly selected sample page
with an equation also had a data display?
 Are having an equation and have a data display disjoint
events?
 Are having an equation and having a data display independent
events?
DRAWING WITHOUT REPLACEMENT
 Sampling without replacement means that once one
individual 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.
DRAWING WITHOUT REPLACEMENT
EXAMPLE
 In a box of cupcakes, there are five cupcakes with red icing, 4
with yellow icing, and 3 with green icing. If 2 of the cupcakes
are randomly selected from the box (no peeking to see which
color ), what is the probability that the first cupcake has red
icing and the second cupcake has green icing? What about
the probability that the first has red icing and the second has
red icing? (The first cupcake is not replaced before the
second cupcake is selected).
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.
 Making a tree diagram for situations with
conditional probabilities is consistent with our
“make a picture” mantra.
TREE DIAGRAMS (CONT.)
 Figure 15.5 is a nice
example of a tree diagram
and shows how we multiply
the probabilities of the
branches together.
 All the final outcomes are
disjoint and must add up to
one.
 We can add the final
probabilities to find
probabilities of compound
events.
EXAMPLE USING TREE DIAGRAM
 What is the probability that a
randomly selected student
will be a binge drinker who
has an alcohol related car
accident?
 What is the probability that a
selected student has had an
alcohol related car accident?
REVERSING THE CONDITIONING
 Reversing the conditioning of two events is rarely
intuitive.
 Suppose we want to know P(A|B), and we know only
P(A), P(B), and P(B|A).
 We also know P(A  B), since
P(A  B) = P(A) x P(B|A)
 From this information, we can find P(A|B):
P(A|B) P(A  B)
P(B)
WHAT CAN GO WRONG?
 Don’t use a simple probability rule where a general
rule is appropriate:
 Don’t assume that two events are independent or
disjoint without checking that they are.
 Don’t find probabilities for samples drawn without
replacement as if they had been drawn with
replacement.
 Don’t reverse conditioning naively.
 Don’t confuse “disjoint” with “independent.”
RECAP
 The probability rules from Chapter 14 only work in
special cases—when events are disjoint or
independent.
 We now know the General Addition Rule and General
Multiplication Rule.
 We also know about conditional probabilities and
that reversing the conditioning can give surprising
results.
RECAP (CONT.)
 Venn diagrams, tables, and tree diagrams help
organize our thinking about probabilities.
 We now know more about independence—a sound
understanding of independence will be important
throughout the rest of this course.
ASSIGNMENTS: PP. 361 – 365
 Day 1: # 1, 9, 13, 19, 31
 Day 2: # 2, 6, 8, 12, 15, 23, 27, 33, 35
 Day 3: # 4, 10, 16, 17, 21, 25, 29, 41