Suppose
Show whether or not the pair
PA = 0.4*0.7 + 0.3*0.3
PA = 0.3700
PB = 0.6*0.7 + 0.2*0.3
PB = 0.4800
PA*PB
ans = 0.1776
PAB = 0.4*0.6*0.7 + 0.3*0.2*0.3
PAB = 0.1860 % PAB not equal PA*PB; not independent
Suppose
Determine the posterior odds
Five world class sprinters are entered in a 200 meter dash. Each has a good chance to break the current track record. There is a thirty percent chance a late cold front will move in, bringing conditions that adversely affect the runners. Otherwise, conditions are expected to be favorable for an outstanding race. Their respective probabilities of breaking the record are:
The performances are (conditionally) independent, given good weather, and also, given poor weather. What is the probability that three or more will break the track record?
Hint. If B3 is the event of three or more,
PW = 0.01*[75 80 65 70 85];
PWc = 0.01*[60 65 50 55 70];
P = ckn(PW,3)*0.7 + ckn(PWc,3)*0.3
P = 0.8353
A device has five sensors connected to an alarm system. The alarm is given
if three or more of the sensors trigger a switch. If a dangerous condition is
present, each of the switches has high (but not unit) probability of activating;
if the dangerous condition does not exist, each of the switches has low (but not
zero) probability of activating (falsely). Suppose
Assume the conditional probabilities of the E1, given D, are
0.91, 0.93, 0.96, 0.87, 0.97, and given Dc, are 0.03, 0.02, 0.07, 0.04, 0.01, respectively. If
P1 = 0.01*[91 93 96 87 97];
P2 = 0.01*[3 2 7 4 1];
P = ckn(P1,3)*0.02 + (1 - ckn(P2,3))*0.98
P = 0.9997
Seven students plan to complete a term paper over the Thanksgiving recess.
They work independently; however, the likelihood of completion depends upon the weather.
If the weather is very pleasant, they are more likely to engage in outdoor activities
and put off work on the paper. Let Ei be the event the ith student completes his
or her paper, Ak be the event that k or more complete during the recess, and
W be the event the weather is highly conducive to outdoor activity. It is
reasonable to suppose
respectively, and
PW = 0.1*[4 5 3 7 5 6 2];
PWc = 0.1*[7 8 5 9 7 8 5];
PA4 = ckn(PW,4)*0.8 + ckn(PWc,4)*0.2
PA4 = 0.4993
PA5 = ckn(PW,5)*0.8 + ckn(PWc,5)*0.2
PA5 = 0.2482
A manufacturer claims to have improved the reliability of his product. Formerly, the product had probability 0.65 of operating 1000 hours without failure. The manufacturer claims this probability is now 0.80. A sample of size 20 is tested. Determine the odds favoring the new probability for various numbers of surviving units under the assumption the prior odds are 1 to 1. How many survivors would be required to make the claim creditable?
Let E1 be the event the probability is 0.80 and E2 be the
event the probability is 0.65. Assume
k = 1:20;
odds = ibinom(20,0.80,k)./ibinom(20,0.65,k);
disp([k;odds]')
- - - - - - - - - - - -
13.0000 0.2958
14.0000 0.6372
15.0000 1.3723 % Need at least 15 or 16 successes
16.0000 2.9558
17.0000 6.3663
18.0000 13.7121
19.0000 29.5337
20.0000 63.6111
A real estate agent in a neighborhood heavily populated by affluent professional persons is working with a customer. The agent is trying to assess the likelihood the customer will actually buy. His experience indicates the following: if H is the event the customer buys, S is the event the customer is a professional with good income, and E is the event the customer drives a prestigious car, then
Since buying a house and owning a prestigious car are not related for a given owner,
it seems reasonable to suppose
Assumptions amount to
In deciding whether or not to drill an oil well in a certain location, a company undertakes a geophysical survey. On the basis of past experience, the decision makers feel the odds are about four to one favoring success. Various other probabilities can be assigned on the basis of past experience. Let
The initial, or prior, odds are
Make reasonable assumptions based on the fact that the result of the geophysical survey
depends upon the geological formations and not on the presence or absence of oil. The
result of the survey is favorable. Determine the posterior odds
A software firm is planning to deliver a custom package. Past experience indicates the odds are at least four to one that it will pass customer acceptance tests. As a check, the program is subjected to two different benchmark runs. Both are successful. Given the following data, what are the odds favoring successful operation in practice? Let
Under the usual conditions, we may assume
Determine the posterior odds
A research group is contemplating purchase of a new software package to perform some specialized calculations. The systems manager decides to do two sets of diagnostic tests for significant bugs that might hamper operation in the intended application. The tests are carried out in an operationally independent manner. The following analysis of the results is made.
Since the tests are for the presence of bugs, and are operationally
independent, it seems reasonable to assume
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Determine the posterior odds favoring H if results of both diagnostic tests are satisfactory.
with similar expressions for the other terms.
A company is considering a new product now undergoing field testing. Let
The company assumes
Since the testing of the merchandise is not affected by
market success or failure, it seems reasonable to suppose
Martha is wondering if she will get a five percent annual raise at the end of the fiscal year. She understands this is more likely if the company's net profits increase by ten percent or more. These will be influenced by company sales volume. Let
Since the prospect of a raise depends upon profits, not directly on sales, she
supposes
End of the year records show that sales increased by eighteen percent. What is the probability Martha will get her raise?
A physician thinks the odds are about 2 to 1 that a patient has
a certain disease. He seeks the “independent” advice of three specialists.
Let H be the event the disease is present, and
What is the probability of the right decision (i.e., he treats the disease if two or more think it is present, and does not if two or more think the disease is not present)?
PH = 0.01*[80 70 75];
PHc = 0.01*[85 80 70];
pH = 2/3;
P = ckn(PH,2)*pH + ckn(PHc,2)*(1 - pH)
P = 0.8577
A software company has developed a new computer game designed to appeal to teenagers and young adults. It is felt that there is good probability it will appeal to college students, and that if it appeals to college students it will appeal to a general youth market. To check the likelihood of appeal to college students, it is decided to test first by a sales campaign at Rice and University of Texas, Austin. The following analysis of the situation is made.
Since the tests are for the reception are at two separate universities and are operationally
independent, it seems reasonable to assume
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Determine the posterior odds favoring H if sales results are satisfactory at both schools.
In a region in the Gulf Coast area, oil deposits are highly likely to
be associated with underground salt domes. If H is the event that an oil deposit
is present in an area, and S is the event of a salt dome in the area, experience
indicates
Determine the posterior odds
with similar expressions for the other terms.
A sample of 150 subjects is taken from a population which has two subgroups. The subgroup membership of each subject in the sample is known. Each individual is asked a battery of ten questions designed to be independent, in the sense that the answer to any one is not affected by the answer to any other. The subjects answer independently. Data on the results are summarized in the following table:
| GROUP 1 (84 members) | GROUP 2 (66 members) | |||||
| Q | Yes | No | Unc | Yes | No | Unc |
| 1 | 51 | 26 | 7 | 27 | 34 | 5 |
| 2 | 42 | 32 | 10 | 19 | 43 | 4 |
| 3 | 19 | 54 | 11 | 39 | 22 | 5 |
| 4 | 24 | 53 | 7 | 38 | 19 | 9 |
| 5 | 27 | 52 | 5 | 28 | 33 | 5 |
| 6 | 49 | 19 | 16 | 19 | 41 | 6 |
| 7 | 16 | 59 | 9 | 37 | 21 | 8 |
| 8 | 47 | 32 | 5 | 19 | 42 | 5 |
| 9 | 55 | 17 | 12 | 27 | 33 | 6 |
| 10 | 24 | 53 | 7 | 39 | 21 | 6 |
Assume the data represent the general population consisting of these two groups, so that the data may be used to calculate probabilities and conditional probabilities.
Several persons are interviewed. The result of each interview is a “profile” of answers to the questions. The goal is to classify the person in one of the two subgroups
For the following profiles, classify each individual in one of the subgroups
% file npr05_16.m
% Data for Exercise 16
A = [51 26 7; 42 32 10; 19 54 11; 24 53 7; 27 52 5;
49 19 16; 16 59 9; 47 32 5; 55 17 12; 24 53 7];
B = [27 34 5; 19 43 4; 39 22 5; 38 19 9; 28 33 5;
19 41 6; 37 21 8; 19 42 5; 27 33 6; 39 21 6];
disp('Call for oddsdf')
npr05_16
Call for oddsdf
oddsdf
Enter matrix A of frequencies for calibration group 1 A
Enter matrix B of frequencies for calibration group 2 B
Number of questions = 10
Answers per question = 3
Enter code for answers and call for procedure "odds"
y = 1;
n = 2;
u = 3;
odds
Enter profile matrix E [y n y n y u n u y u]
Odds favoring Group 1: 3.743
Classify in Group 1
odds
Enter profile matrix E [n n u n y y u n n y]
Odds favoring Group 1: 0.2693
Classify in Group 2
odds
Enter profile matrix E [y y n y u u n n y y]
Odds favoring Group 1: 5.286
Classify in Group 1
The data of Exercise 16, above, are converted to conditional probabilities and probabilities, as follows (probabilities are rounded to two decimal places).
| GROUP 1 |
GROUP 2 |
|||||
| Q | Yes | No | Unc | Yes | No | Unc |
| 1 | 0.61 | 0.31 | 0.08 | 0.41 | 0.51 | 0.08 |
| 2 | 0.50 | 0.38 | 0.12 | 0.29 | 0.65 | 0.06 |
| 3 | 0.23 | 0.64 | 0.13 | 0.59 | 0.33 | 0.08 |
| 4 | 0.29 | 0.63 | 0.08 | 0.57 | 0.29 | 0.14 |
| 5 | 0.32 | 0.62 | 0.06 | 0.42 | 0.50 | 0.08 |
| 6 | 0.58 | 0.23 | 0.19 | 0.29 | 0.62 | 0.09 |
| 7 | 0.19 | 0.70 | 0.11 | 0.56 | 0.32 | 0.12 |
| 8 | 0.56 | 0.38 | 0.06 | 0.29 | 0.63 | 0.08 |
| 9 | 0.65 | 0.20 | 0.15 | 0.41 | 0.50 | 0.09 |
| 10 | 0.29 | 0.63 | 0.08 | 0.59 | 0.32 | 0.09 |
For the following profiles classify each individual in one of the subgroups.
npr05_17
% file npr05_17.m
% Data for Exercise 17
PG1 = 84/150;
PG2 = 66/125;
A = [0.61 0.31 0.08
0.50 0.38 0.12
0.23 0.64 0.13
0.29 0.63 0.08
0.32 0.62 0.06
0.58 0.23 0.19
0.19 0.70 0.11
0.56 0.38 0.06
0.65 0.20 0.15
0.29 0.63 0.08];
B = [0.41 0.51 0.08
0.29 0.65 0.06
0.59 0.33 0.08
0.57 0.29 0.14
0.42 0.50 0.08
0.29 0.62 0.09
0.56 0.32 0.12
0.29 0.64 0.08
0.41 0.50 0.09
0.59 0.32 0.09];
disp('Call for oddsdp')
Call for oddsdp
oddsdp
Enter matrix A of conditional probabilities for Group 1 A
Enter matrix B of conditional probabilities for Group 2 B
Probability p1 an individual is from Group 1 PG1
Number of questions = 10
Answers per question = 3
Enter code for answers and call for procedure "odds"
y = 1;
n = 2;
u = 3;
odds
Enter profile matrix E [y n y n y u n u y u]
Odds favoring Group 1: 3.486
Classify in Group 1
odds
Enter profile matrix E [n n u n y y u n n y]
Odds favoring Group 1: 0.2603
Classify in Group 2
odds
Enter profile matrix E [y y n y u u n n y y]
Odds favoring Group 1: 5.162
Classify in Group 1