More on Section 6.2 - Cabarrus County Schools

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Warm-up

Define the sample space of each of the
following situations…



Choose a student in your class at random.
Ask how much time that student spent
studying during the past 24 hours.
The Physician’s Health Study asked 11,000
physicians to take an aspirin every other day
and observed how many of them had a heart
attack in a five-year period.
In a test of new package design, you drop a
carton of a dozen eggs from a height of 1 foot
and count the number of broken eggs.
Section 6.2
Independence and the
Multiplication Rule
Two Special Rules


We’ve learned the addition rule for disjoint
events: If A and B are disjoint, then P(A or B)
= P(A) + P(B).
Now we’ll learn the multiplication rule for
independent events: that if A and B are
independent, then P(A and B) = P(A)●P(B)

Remember that two events are independent if
knowing that one occurs does not change the
probability that the other occurs.
Examples of Independent Events

Toss a coin twice. Let A = first toss is a
head and B = second toss is a head.
Events A and B are independent. Thus,
the P(A∩B) = P(A)*P(B).


P(Head and Tail) = P(Head) * P(Tail) = ½ * ½ = ¼
Draw 3 cards from a deck, replacing and
shuffling in between each draw. This is
called “with replacement.”
Without Replacement

If you draw three cards from a deck
without replacing, the probabilities change
on each draw. Therefore, drawing
without replacement is NOT independent.
Cautions

The addition rule for
disjoint events and
the multiplication rule
for independent
events only work
when the criteria are
met. Resist the
temptation to use
them for events that
are not disjoint or
not independent.


You must be told or
have prior knowledge
that an event is
disjoint or
independent.
Do not confuse
disjoint with
independent.
Disjoint can be
displayed in a Venn
Diagram.
Independence can
not.
Sample Questions

Suppose that among the 6000 students at
a high school, 1500 are taking honors
courses and 1800 prefer watching
basketball to watching football. If taking
honors courses and preferring basketball
are independent, how many students are
both taking honors courses and prefer
basketball to football?
Sample Questions

Suppose that for any given year, the
probabilities that the stock market
declines is .4, and the probability that
women’s hemlines are lower is .35.
Suppose that the probability that both
events occur is .3. Are the two events
independent?
Sample Questions

In a 1974 “Dear Abby” letter, a woman
lamented that she had just given birth to her
eighth child, and all were girls! Her doctor had
assured her that the chance of the eighth child
being a girl was only 1 in 100.


A) What was the real probability that the eighth child
would be a girl?
B) Before the birth of the first child, what was the
probability that the woman would give birth to eight
girls in a row?
Section 6.3
General Probability Rules
Special Probability Rules




If A and B are disjoint, then P(AUB) =
P(A) + P(B).
This rule relies on a condition to be met.
So, what if the events are NOT disjoint?
What if there are more than 2 disjoint
events?
General Probability Rules

P(AUB) = P(A) + P(B) – P(A∩B)
If A and B are disjoint, then
what does P(A∩B) equal?

This is how the formula will appear on
your formula sheet for the exam (and thus
my tests).
Sample Question: Student
Survey







(making an A in Statistics only) = 18 students
(making an A in Calculus only) = 63 students
(making an A in Stats and in Calculus) = 27 students
S=150 students
Draw a Venn diagram.
Find the probability of making an A in Stats but not in Calc.
Find the probability of making an A in Calc but not in Stats.
Find the probability of not making an A in either subject.
Find the probability of making an A in either Calc or Stats.
Are these events independent?
Extending the Rules to Three
Events




If you have three disjoint events, then
P(A or B or C) = P(A) + P(B) + P(C)
If they are not disjoint, then P(A or B or
C) = P(A) + P(B) + P(C) – P(A∩B) –
P(A∩C) – P(B∩C)
Often it is easier to draw a Venn Diagram
than to use the formula.
Look at p365 6.51
Sample Questions

Suppose that the probability that you will receive
an A in AP Statistics is .35, the probability that
you will receive As in both AP Statistics and AP
Biology is .19, and the probability that you will
receive an A in AP Bio but not in Stats is .17.
Which is a proper conclusion?




A) P(A in AP Bio) = .36
B) P(you didn’t take Bio) = .01
C) P(not making an A in AP Stat or Bio) = .38
D) The given probabilities are impossible.
Sample Questions

If P(A) = .2 and P(B) = .1, what is P(AUB)
if A and B are independent?





A) .02
B) .28
C) .30
D) .32
E) There is insufficient information to answer
this question.
Sample Questions

Given the probabilities P(A) = .4 and P(AUB) =
.6, what is the probability P(B) if A and B are
mutually exclusive? If A and B are independent?





A)
B)
C)
D)
E)
.2, .28
.2, .33
.33, .2
.6, .33
.28, .2
Conditional Probability

When events are not independent, the
probability of one changes if we know that the
other event occurred.





I will draw two cards without replacement.
Let A = {1st card I draw is an Ace}
Let B = {2nd card I draw is an Ace}
The probability of B occurring changes
depending on whether A occurred.
The new notation P(B|A) is read “the
probability of B given A.” It asks you to find
the probability of B knowing that A has
occurred.
Let’s look at education and age
Age
25 – 34
35 – 54
55 +
Total
4,474
9,155
14,224
27,853
Completed HS
11,546
26,481
20,060
58,087
1 to 3 years
college
10,700
22,618
11,127
44,445
4+ years of
college
11,066
23,183
10,596
44,845
Education
No high
school
Total
37,786
81,435
56,008
175,230
Find these probabilities






Let A = {the person chosen is 25 – 34}
Let B = {the person chosen has 4+ years
of college}
Find P(A).
Find P(A and B).
Find P(B|A).
Given that a person has only a HS
diploma, what is the probability that the
person is 55 or older?
The general rule for
multiplication

P(A∩B) = P(A)*P(B|A)


This can be rewritten as
P(B|A) = P(A∩B)/P(A)
Example: Two cards are drawn without
replacement.
Let A = {1st card is a spade}
 Let B = {2nd card is a spade}
 Find P(B|A).
 Find P(A and B).

Next Example



Suppose the probability that the dollar
falls in value compared to the yen is 0.5.
Suppose that the probability that if the
dollar falls in value, a supplier from Japan
will renegotiate their contract is 0.7.
What is the probability that the dollar will
fall in value and the supplier demands
renegotiation?
Homework
Chapter 5# 4, 9, 40, 43, 51, 60,
83, 89, 94
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