Transcript Section 3.5-3.6 - Pendleton County Schools

Sta220 - Statistics
Mr. Smith
Room 310
Class #9
Section 3.5-3.6
Problem:
To develop programs for business travelers
staying at convention hotels, Hyatt Hotels Corp.
commissioned a study on executives who play
golf. The study revealed that 55% of the
executives admitted that they had cheated at
golf. Also, 20% of the executives admitted that
they had cheated at golf and had lied in
business. Given that an executive had cheated at
golf, what is the probability that they executive
also had lied in business?
The event probabilities we’ve been discussing
give the relative frequencies of the occurrences
of the events when the experiment is repeated a
very large number of times. Such probabilities
are often called unconditional probabilities.
3.5 Conditional Probability
A probability that reflects such additional
knowledge is called the conditional probability
of the event.
Example:
We’ve seen that the probability of an even
number (event A) on a toss of a fair die is ½. But
suppose we’re given the information that on a
particular throw of the die the results was a
number less than or equal to 3 (event B). Would
the probability of observing an even number on
that throw of the die still be equal to ½?
Reduced sample space for the die-toss
experiment: given that event B has occurred
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Education, Inc.. All rights
reserved.
To get the probability of event A given that
event B occurs, we divide the probability of the
part A that falls within the reduced sample
space B, namely 𝑃(𝐴 ∩ B), by the total
probability of the reduced sample space,
namely, P(B).
P(A|B) =
𝑃 𝐴∩B
𝑃 𝐵
=
𝑃 2
𝑃 1 + 2 +𝑃 3
=
1
6
3
6
=
1
3
• Note: 𝑃 𝐵 𝐴 =
𝑃 𝐴∩𝐵
𝑃 𝐴
[ We assume 𝑃(𝐴) ≠ 0]
Problem:
To develop programs for business travelers
staying at convention hotels, Hyatt Hotels Corp.
commissioned a study on executives who play
golf. The study revealed that 55% of the
executives admitted that they had cheated at
golf. Also, 20% of the executives admitted that
they had cheated at golf and had lied in
business. Given that an executive had cheated at
golf, what is the probability that they executive
also had lied in business?
Solution
Let
A: {Executive had cheated at golf}
B: {Executive had lied in business}
Then 𝑃(𝐴) = .55 and 𝑃(𝐴 ∩ 𝐵) = .20
We want P(B|A)=?
𝑃 𝐴∩𝐵
.20
𝑃 𝐵𝐴 =
=
= .364
𝑃 𝐴
.55
Thus, given that a certain executive had cheated
at golf, the probability that the executive also
had lied in business is .364.
Problem:
Many medical researchers have conducted experiments
to examine the relationship between cigarette smoking
and cancer. Consider an individual randomly selected
from the adult male population. Let A represent the
event that the individual smokes, and let 𝐴𝑐 denote the
complement of A (the even that the individual does not
smoke). Similarity, let B represent the even that the
individual develops cancer, and let 𝐵𝑐 be the
complement of that event. Then the four sample points
associated with the experiment are shown in Figure 3.15.
Use these sample point probabilities to examine the
relationship between smoking and cancer.
Solution
A method in determining whether the given
probabilities indicate that smoking and cancer
are related is compare the conditional
probability that an adult male acquires cancer
given that he smokes with the conditional
probability that an adult male acquires cancer
given that he does not smoke [ i.e. compare
𝑃(𝐵|𝐴) with 𝑃(𝐵|𝐴𝐶 )
So we are going to reduce the sample space A
corresponding to adult male smokers.
Two conditional probabilities:
𝑃 𝐵𝐴 =
𝑃 𝐴∩𝐵
𝑃 𝐴
and
𝑃 𝐵𝑐 𝐴 =
𝑃 𝐴∩𝐵𝑐
𝑃 𝐴
𝑃 𝐴 = 𝑃 𝐴 ∩ B + 𝑃 𝐴 ∩ 𝐵𝐶
= .05 + .20
= .25
So now
𝑃 𝐵𝐴 =
.80
𝑃 𝐴∩𝐵
𝑃 𝐴
=
.05
.25
= .20
and
𝑃
𝐵𝑐
𝐴 =
𝑃 𝐴∩𝐵 𝑐
𝑃 𝐴
=
.20
.25
=
These two numbers represent the probabilities that an
adult male smoker develops cancer and does not develop
cancer, respectively.
𝑃 𝐵 𝐴𝑐
𝑃 𝐵𝑐 𝐴𝑐
𝑃 𝐴𝑐 ∩ 𝐵
.03
=
=
= .04
𝑐
𝑃 𝐴
.75
𝑃 𝐴𝑐 ∩ 𝐵𝑐
.72
=
=
= .96
𝑐
𝑃 𝐴
.75
These are the conditional probabilities of an adult
male nonsmoker developing cancer and not
developing cancer.
The conditional probability that an adult male
smoker develops cancer (.20) is five times the
probability that a nonsmoker develop cancer
(.04). This relationship does not imply that
smoking causes cancer, but it does suggest a
pronounced link between smoking and cancer.
3.6 The Multiplicative Rule and
Independent Events
The probability of an intersection of two events
can be calculated with the multiplicative rule.
Problem:
In a classic psychology study conducted in the early 1960s,
Stanley Milgram performed a series of experiment in which
a teacher is asked to shock a learner who is attempting to
memorize word pairs whenever the learner gives the wrong
answer. The shock levels increase with each successive
wrong answer. {Unknown to the teacher, the shocks are
not real.} Two events of interest are
A: {The teacher “applies” a sever shock (450 volts).}
B: {The learner protest verbally prior to receiving the shock.}
A recent application of Milgram’s shock study revealed that
P(B) = .5 and P(A|B) = .7. On the basis of the information,
what is the probability that a learner will protest verbally
and a teacher will apply a severe shock?
Solution
We want to calculate 𝑃(𝐴 ∩ B).
𝑃 𝐴 ∩ B = 𝑃 𝐵 𝑃 𝐴 𝐵 = .5 .7 = .35
Thus, about 35% of the time the learner will give
a verbal protest and the teacher will apply
severe shock.
Problem:
A county welfare agency employs 10 welfare
workers who interview prospective food stamp
recipients. Periodically, the supervisor selects, at
random, the forms completed by two workers
and subsequently audits them for illegal
deductions. Unknown to the supervisor, three of
the workers have regularly been giving illegal
deductions to applicants. What is the probability
that both of the workers chosen have been
giving illegal deductions?
Solution
Let
N: Non-Illegal
I: Illegal
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Education, Inc.. All rights
reserved.
Tree diagram
6
1
𝐼∩I=
=
90 15
Thus, there is a 1-in-15 chance that both
workers chosen by the supervisor have been
giving illegal dedication to food stamp
recipients.
We have showed that the probability of event A
may be substantially altered by the knowledge
that an event B has occurred. However, this will
not always be the case; in some instances, the
assumption that event B has occurred will not
alter the probability of event A at all.
Problem:
Consider the experiment of tossing a fair die,
and let
A = {Observe an even number.}
B = {Observe a number less than or
equal to 4.}
Are A and B independent events?
Venn diagram for die-toss experiment
Copyright © 2013 Pearson
Education, Inc.. All rights
reserved.
Solution
𝑃(𝐴) = 1/2
𝑃(𝐵) = 2/3
𝑃(𝐴 ∩ 𝐵) = 1/3
Assuming B has occurred, we see that the conditional
probability that A given B is
1
𝑃 𝐴∩𝐵
1
3
𝑃 𝐴𝐵 =
=
= = P(A)
2
𝑃 𝐵
2
3
Thus, assuming that the occurrence of event B
does not alter the probability of observing an
even number, the probability remains ½ .
Therefore, the events A and B are independent.
Problem
An experiment results in one of five sample points with the following probabilities:
𝑃(𝐸1 ) = .22, 𝑃(𝐸2 ) = .31, 𝑃(𝐸3 ) = .15, 𝑃(𝐸4 ) = .22 𝑎𝑛𝑑 𝑃(𝐸5 ) = .1. The
following events have been defined:
𝐴: {𝐸1 , 𝐸3 }
𝐵: {𝐸2 , 𝐸3 , 𝐸4 }
𝐶: {𝐸1 , 𝐸5 }
Find each of the following probabilities:
a) P(A)
b) P(B)
c) 𝑃(𝐴 ∩ 𝐵)
d) 𝑃 𝐴 𝐵
e) 𝑃(𝐵 ∪ 𝐶)
f) 𝑃(𝐶|𝐵)
g) Consider each pair of events A and B, A and C, and B and C. Are any of the pairs
of events independent? Why?
Solution
A. P(A) = .22 +.15 = .37
B. P(B) = .31+.15 +.22 = .68
C. 𝑃(𝐴 ∩ 𝐵) = .15
D. 𝑃 𝐴 𝐵 =
.15
.68
= .221
E. P(B\capC) = 0
F. P(C|B) = 0/.68 = 0
G. For any of the pairs of events to be
independent, the following must hold:
𝑃(𝐴 ∩ B) = 𝑃 𝐴 𝑃(𝐵)
𝑃(𝐴)𝑃(𝐵) = .37(.68) = .2516 ≠ 𝑃(𝐴 ∩ B) = .15
𝑃(𝐴)𝑃(𝐶) = .37(.32) = .1184 ≠ 𝑃(𝐴 ∩ C) = .22
𝑃(𝐵)𝑃(𝐶) = .68(.32) = .2176 ≠ 𝑃(𝐵 ∩ C) = 0