POD #50
12/2/2011
2007B #6b
Is this model
appropriate?
Write the
equation of
the LSRL
Stats: Modeling the World
Chapter 15
Probability Rules!
Example
On a large college campus, the RA’s randomly selected one dorm
room per floor for a total of 100 rooms. This survey revealed that
38 had refrigerators, 52 had TVs, and 21 had both a TV and a
fridge How many of the dorm rooms had a TV or a refrigerator?
(either or both)
TV
fridge
17
21
31
How many students were interviewed?
How many students liked Rock?
How many students liked Rap OR
Country?
How many students liked 2 out of 3
types?
How many students liked all three
types?
How many students liked 1 type of
music only?
Using a roster of all students in an elementary school, 35
students were randomly selected. Ten of the students had
blonde hair, 14 had brown eyes, and 4 had both blonde hair
and brown eyes.
Brown eyes
Blonde hair
6
4
10
If a child is selected at random, find the probability that the
child has blonde hair or brown eyes. (either or both)
As 400 college students walked by our interviewer, we learned that
120 are enrolled in math, 220 are enrolled in English, and 55 are
enrolled in both.
Math
65
English
55
165
Find the probability that…
a. the student is enrolled in mathematics.
b. the student is enrolled in mathematics or English.
c. the student is enrolled in either mathematics or English, but not
both.
A survey of 100 couples (taken by calling every 50th person
from the voter rolls) found the that the husband was
employed in 85 of the couples. The wife was employed in
60 of the couples, and both spouses were employed in 55 of
the couples.
Wife
Husband
30
55
Find the probability that
a. at least one of them is employed.
b. neither is employed.
5
32
12
6
6
19
21
4
Example
On a large college campus, the RA’s randomly selected one dorm
room per floor for a total of 100 rooms. This survey revealed that
38 had refrigerators, 52 had TVs, and 21 had both a TV and a
fridge How many of the dorm rooms had a TV or a refrigerator?
(either or both)
TV
fridge
17
21
31
General Addition Rule
For any two events A and B,
P(A or B) = P(A) + P(B) – P(A and B)
When the two events are
disjoint or mutually
exclusive, then
P(A and B)=0
POD #51
12/5/2011
Vocab Review
Please number your paper 1-10.
Reassessments and makeup
work should be done by
Wednesday, Dec 14!!!
Comparing Venns to Tables
A check of dorm rooms on a large college campus revealed that
38% had refrigerators, 52% had TVs, and 21% had both a TV and
a fridge What’s the probability that a randomly selected dorm
room has:
Prior to graduation a high school class was surveyed about their plans. The
table below displays the results for white and minority students.
242
42
5
17
19
268
57
325
a) What percent of the graduates are white? 268
325
b) What percent of the graduates are planning to attend a 2-year college?
42
325
c) What percent of the graduates are minority or planning to attend a 4-year
college? 255
325
Making Connections…
Prior to graduation a high school class was surveyed about their plans. The
table below displays the results for white and minority students.
242
42
5
17
19
325
268
57
Give the conditional distributions of plans for white and minority students.
White: 4-year college: 74%
2-year college: 13%
Military: 2%
Employment: 5%
Other: 6%
Minority: 4-year college: 77%
2-year college: 11%
Military: 2%
Employment: 5%
Other: 5%
Conditional Probabilities
A probability that takes into account a given
condition is called a conditional probability
This is written as P(A|B) and read as ‘the
probability of A given B.
Prior to graduation a high school class was surveyed about their plans. The
table below displays the results for white and minority students.
242
42
5
17
19
268
57
325
a) What percent of the white graduates are planning to attend a 2-year college? 36
268
b) What percent of the minority graduates are planning to attend a 4-year college?
44
57
c) What percent of the graduates pursuing the military are white?
4
5
d) What percent of the graduates pursuing a 4-year college are minority?
44
242
More Formally…
P
(A
a
n
d
B)
P (B | A ) 
P (A )
Note: P(A) cannot equal 0, since we know that
A has occurred.
General Multiplication Rule
P(A and B) = P(A) x P(B|A)
Note: There’s nothing special about which one we
write as A or B, so P(A and B) = P(B) x P(A|B)
Replacement… or not
Sampling without replacement means that once one
individual is drawn it doesn’t go back into the pool.
Drawing without replacement is just another instance of
working with conditional probabilities.
5

12
4
3
4

5
3
132
12

2

11
6
132
12
20  6  12
132
132

11
20

11

12
4

20
5
4


20
12
11
132
12
11
132
5
3
15
4
0
0
132
12
12

11


11

132

38
132
POD #52
12/6/2011
2004 #5p
1. P(Satisfied) = ?
2. P(Male OR Not Satisfied) = ?
3. P(Satisfied | Female) = ?
4. P (Male AND Satisfied) = ?
Example
A junk box in your room contains a dozen old batteries, five of which
are totally dead. You start picking batteries one at a time and testing
them. Find the probability of each outcome.
7
6
42
a. The first two you choose are both good.


12
b.
The first four you pick all work.
7
6

12
c.
11
10

4
9

132
840
11880
You have to pick 5 batteries in order to find one that works.
5
d.

5
11
12

4
11

3
10

2
9

7
8
840

95040
At least one of the first three works.
4
3 
 5
1  P ( none )  1  


  0 . 955
 12 11 10 
Working with the Formulas
An aerospace company has submitted bids on two separate federal defense
contracts A and B. The company feels that it has a 60% chance of winning
contract A and a 30% chance of winning contract B. Given that it wins contract
B, the company believes it has an 80% chance of winning contract A.
P(A) = 0.6
P(B) = 0.3
P(A|B) = 0.8
a. What is the probability that the company will win both contracts?
P(A and B) = P(B) x P(A|B)
P(A and B) = 0.3 x 0.8 = 0.24
b. What is the probability that the company will win at least one of the two
contracts? (That means A OR B)
P(A or B) = P(A) + P(B) – P(A and B)
P(A or B) = 0.6 + 0.3 – 0.24 = 0.66
Vocab Alert…
The “rules” rely on two important ideas…
Disjoint events (Mutually Exclusive)
- No overlap in the circles… the events do not occur
together… P(A and B) = 0
Independent events
- Event A does not affect Event B
Disjoint???
A university requires its biology majors to take a course called
BioResearch. The prerequisite for this course is that students
must have taken either a Stat course or a computer course. By
the time they are juniors, 52% of Biology majors have take
Stat, 23% a computer course, and 7% both.
Are taking these two courses disjoint events?
Explain.
Not disjoint – you can take both courses
Independent???
A university requires its biology majors to take a course called
BioResearch. The prerequisite for this course is that students
must have taken either a Stat course or a computer course. By
the time they are juniors, 52% of Biology majors have take
Stat, 23% a computer course, and 7% both.
Are taking these two courses independent events?
In other words… does taking a computer course CHANGE your
probability of taking a stat course?
Independent??? How can we tell???
A university requires its biology majors to take a course called BioResearch. The
prerequisite for this course is that students must have taken either a Stat course or a
computer course. By the time they are juniors, 52% of Biology majors have take
Stat, 23% a computer course, and 7% both.
7
16
23
45
32
77
General chance of taking a stat course???
52
48
100
52%
30%
Since computers did affect your chances of stat,
these are Dependent Events!!!!
If you took computers, what’s the chance of stat???
Formal Independence
The Rule…
Events A and B are independent whenever
P(B|A) = P(B).
(Again, note that it does not matter which event is A and which is B.)
Independent??? How can we tell???
A university requires its biology majors to take a course called BioResearch. The
prerequisite for this course is that students must have taken either a Stat course or a
computer course. By the time they are juniors, 52% of Biology majors have take
Stat, 23% a computer course, and 7% both.
7
16
23
45
32
77
P(B|A) = P(B)
P(Stat|Computer) = P(Stat)
30% = 52%
52
48
100
Disjoint vs. Independence
In the real estate ads, 64% of homes have garages, 21% have
swimming pools, and 17% have both features.
17
4
21
47
32
79
64
36
100
a) Are having a garage and a pool disjoint events?
Not disjoint – you can have both
Disjoint vs. Independence
In the real estate ads, 64% of homes have garages, 21% have
swimming pools, and 17% have both features.
17
4
21
47
32
79
64
36
100
b) Are having a garage and a pool independent events?
P(B|A) = P(B)
P(Garage|Pool) = P(Garage)
81% = 64%
Dependent Events!!!!
POD #53
12/7/2011
2011 #2ab
POD #53
12/7/2011
2011 #2ab
I am NOT your mother!!!
-If you get it out, put it back where you got it
-If you make a mess, clean it up
-If you write on your desk, erase it (we now
have washcloths – use them!)
Disjoint or Not…
Ace
3
52
1
52
Hearts
Ace
King
12
52
4
52
4
52
Not Disjoint
Disjoint
Rolling Dice…
Roll a 6
P(A)=
Not a 6
P(B)=
Roll a 6
P(A|A)=
Not a 6
P(B|A)=
Roll a 6
P(A|B)=
Not a 6
P(B|B)=
Drawing Cards…
Ace
P(A)=
Not Ace
P(B)=
Ace
P(A|A)=
Not Ace
P(B|A)=
Ace
P(A|B)=
Not Ace
P(B|B)=
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.
Dan’s Diner employs three dishwashers. Al
washes 40% of dishes and breaks only 1%.
Betty and Chuck each wash 30% with Betty
breaking only 1% and Chuck breaking 3% of
his. You go to Dan’s for supper and hear a dish
break at the sink. What’s the probability that
Chuck is on the job?
A private college report contains these stats:
70% of incoming Freshmen attended public schools.
75% of public school students who enroll as freshmen eventually
graduate.
90% of other freshmen eventually graduate.
What percent of freshmen eventually graduate?
What percent of students who graduate from college attended a public high
school?
What can go wrong?




Don’t use the wrong formula! Many people use the
simple rules from Ch 14 when they should use the
general rules. They (incorrectly) assume the events are
disjoint or independent when they are not.
Watch out for replacement! When dealing with small
populations, your denominator will change!
P(B|A) is NOT the same as P(A|B)
Disjoint does NOT mean Independent!