DO YOU KNOW WHEN EXAM 1 IS?
1.
2.
3.
Yes
No
I need a break.
0%
0%
0% Slide
1- 1
1
2
3
UPCOMING IN CLASS

Homework 5 due Sunday 9/23

Quiz 3 next Wednesday – Sept 26th
 (HW4

and HW5)
Exam 1 – Oct. 10th
CHAPTER 15
Probability Rules!
THE GENERAL ADDITION RULE

For disjoint events :
P(A or B) = P(A) + P(B)

For any two events A and B,
P(A or B) = P(A) + P(B) – P(A and B)
Slide
1- 4
THE GENERAL MULTIPLICATION RULE

For independent events
P(A and B) = P(A) x P(B)

For any two events A and B,
P(A and B) = P(A) x P(B|A)
Slide
1- 5
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- 6
HOMEWORK PROBLEM 7 – DEALING
WITHOUT REPLACEMENT

You are dealt a hand of three cards, one at a
time.





4 suits (hearts, clubs, spades, diamond)
26 red
26 black
12 face cards
4 Aces
Slide
1- 7
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- 8
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- 9
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- 10
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- 11
1
2
3
4
BATTERY PROBLEM



A junk box in your room contains 15 old
batteries, 7 of which are totally dead.
You start picking batteries one at a time and
testing them.
Find the probability of each outcome
Slide
1- 12
THE FIRST 2 YOU CHOOSE ARE GOOD.
1.
2.
3.
4.
7/15 * 7/15
7/15 * 6/14
8/15 * 8/15
8/15 * 7/14
0%
1
0%
2
0%
3
Slide
0%
1- 13
4
AT LEAST ONE OF THE FIRST 4 BATTERIES
WORK.
1.
2.
3.
4.
8/15 * 8/15 * 8/15 * 8/15
1-8/15 * 8/15 * 8/15 * 8/15
1- 8/15 * 7/14 * 6/13 * 5/12
1- 7/15 * 6/14 * 5/13 * 4/12
0%
0%
0%
Slide
0%
1- 14
1
2
3
4
YOU HAVE TO PICK 5 BATTERIES
ONE THAT WORKS.
1.
2.
3.
4.
TO FIND
7/15 * 6/14 * 5/13 * 4/12 * 8/11
7/15 * 6/14 * 5/13 * 5/12 * 4/11
1 – 8/15 * 7/14 * 6/13 * 5/12 * 4/11
8/15 * 7/14 * 6/13 * 5/12 * 4/11
0%
0%
0%
0%Slide
1- 15
1
2
3
4
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.
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.
Slide
1- 16
DISJOINT VS. INDEPENDENCE


The prerequisite for a required course is that
students must have taken either course A or
course B.
By the time they are juniors, 57% of the students
have taken course A, 21% have had course B, and
15% have done both.
Slide
1- 17
ARE THE EVENTS OF TEST A AND B
DISJOINT?
1.
2.
3.
4.
Yes, because the outcome of one influences the
probability of the other
Yes, because the outcome of one does not
influence the probability of the other
No, because there are common outcomes
between them
Yes, because there are no common outcomes
Slide
1- 18
WHAT PERCENT OF THE JUNIORS ARE
INELIGIBLE FOR THE COURSE?
1.
2.
3.
4.
57 +21
57+21-15
100-(57+21)
100-(57+21-15)
0%
0%
0%
0%Slide
1- 19
1
2
3
4
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).)
Slide
1- 20
REVERSING THE CONDITIONING
Reversing the conditioning of two events is rarely
intuitive.
 Suppose we want to know P(A|B), but we know only
P(A), P(B), and P(B|A).
 We also know P(A and B), since
P(A and B) = P(A) x P(B|A)
 From this information, we can find P(A|B):

P(A|B)  P(A and B)
P(B)
Slide
1- 21
WHAT IS THE PROBABILITY THAT A JUNIOR
WHO HAS TAKEN COURSE A HAS ALSO TAKEN
COURSE B?
1.
2.
3.
4.
(57+21-15)/(57)
(57+21-15)/(21)
(15)/(57)
(15)/(21)
0%
1
0%
2
0%
3
Slide
1- 22
0%
4
ARE THE EVENTS INDEPENDENT?
1.
2.
3.
4.
No, because the outcome of one influences the
probability of the other
Yes, because the outcome of one does not
influence the probability of the other
No, because there are common outcomes
between them
Yes, because there are no common outcomes
Slide
1- 23
TREE DIAGRAMS (CONT.)

Figure 15.4 is a
nice example of a
tree diagram and
shows how we
multiply the
probabilities of the
branches together:
Slide
1- 24
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- 25
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- 26
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- 27
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- 28
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- 29
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- 30
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- 31
1
2
3
4
UPCOMING IN CLASS



Homework 5 due Sunday 9/23
Chapter 16 – Random Variables, Expected
Values, and Variance and Standard Deviations of
Random Variables
Quiz 3 next Wednesday – Sept 26th
 (HW4

and HW5)
Exam 1 – Oct. 10th
Slide
1- 32